Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 5 Quadratic Equations

Question 1. Find which of the following equations are quadratic:

(i) (3x – 1)² = 5(x + 8)
Answer:
\( (3x - 1)^2 = 5(x + 8) \)
\( \implies \) \( (9x^2 - 6x + 1) = 5x + 40 \)
\( \implies \) \( 9x^2 - 11x - 39 = 0 \), which is of the form \( ax^2 + bx + c = 0 \)
∴ Given equation is a quadratic equation.
In simple words: We expanded both sides and got an equation with \( x^2 \) as the highest power. This makes it quadratic.

📝 Teacher's Note: Show students how to expand \( (3x-1)^2 \) step by step. Many forget the middle term in \( (a-b)^2 = a^2 - 2ab + b^2 \).

🎯 Exam Tip: Always write "which is of the form \( ax^2 + bx + c = 0 \)" to show you know the standard form. This gets you marks.

(ii) 5x² – 8x = -3(7 – 2x)
Answer:
\( 5x^2 - 8x = -3(7 - 2x) \)
\( \implies \) \( 5x^2 - 8x = 6x - 21 \)
\( \implies \) \( 5x^2 - 14x + 21 = 0 \), which is of the form \( ax^2 + bx + c = 0 \)
∴ Given equation is a quadratic equation.
In simple words: We moved all terms to one side and got an equation with \( x^2 \) as the highest power. So it is quadratic.

📝 Teacher's Note: Remind students to distribute the -3 correctly. Common mistake is \( -3 \times 7 = -21 \) and \( -3 \times (-2x) = +6x \).

🎯 Exam Tip: Write each step clearly with the implies arrow. Show your working to get full marks.

(iii) (x – 4)(3x + 1) = (3x – 1)(x + 2)
Answer:
\( (x - 4)(3x + 1) = (3x - 1)(x + 2) \)
\( \implies \) \( 3x^2 + x - 12x - 4 = 3x^2 + 6x - x - 2 \)
\( \implies \) \( 16x + 2 = 0 \), which is not of the form \( ax^2 + bx + c = 0 \)
∴ Given equation is not a quadratic equation.
In simple words: After expanding and simplifying, the \( x^2 \) terms cancelled out. We got only \( x \) terms, so this is linear, not quadratic.

📝 Teacher's Note: This is a great example to show that not all equations with brackets are quadratic. Sometimes terms cancel out.

🎯 Exam Tip: Write "not of the form \( ax^2 + bx + c = 0 \)" clearly. State the final conclusion about quadratic or not.

(iv) x² + 5x – 5 = (x – 3)²
Answer:
\( x^2 + 5x - 5 = (x - 3)^2 \)
\( \implies \) \( x^2 + 5x - 5 = x^2 - 6x + 9 \)
\( \implies \) \( 11x - 14 = 0 \), which is not of the form \( ax^2 + bx + c = 0 \)
∴ Given equation is not a quadratic equation.
In simple words: The \( x^2 \) terms on both sides cancelled out. We are left with only \( x \) terms, making it a linear equation.

📝 Teacher's Note: Emphasize that when highest power terms cancel, the degree of the equation reduces. Here it became linear (degree 1).

🎯 Exam Tip: Always simplify completely before deciding if it is quadratic. Don't stop at the first step.

(v) 7x³ – 2x² + 10 = (2x – 5)²
Answer:
\( 7x^3 - 2x^2 + 10 = (2x - 5)^2 \)
\( \implies \) \( 7x^3 - 2x^2 + 10 = 4x^2 - 20x + 25 \)
\( \implies \) \( 7x^3 - 6x^2 + 20x - 15 = 0 \), which is not of the form \( ax^2 + bx + c = 0 \)
∴ Given equation is not a quadratic equation.
In simple words: This equation has \( x^3 \) (x cubed) as the highest power. So it is a cubic equation, not quadratic.

📝 Teacher's Note: Point out that quadratic means degree 2. This has degree 3, so it's cubic. The highest power decides the type.

🎯 Exam Tip: Look at the highest power after simplifying. Only degree 2 equations are quadratic.

(vi) (x – 1)² + (x + 2)² + 3(x + 1) = 0
Answer:
\( (x - 1)^2 + (x + 2)^2 + 3(x + 1) = 0 \)
\( \implies \) \( x^2 - 2x + 1 + x^2 + 4x + 4 + 3x + 3 = 0 \)
\( \implies \) \( 2x^2 + 5x + 8 = 0 \), which is of the form \( ax^2 + bx + c = 0 \)
∴ Given equation is a quadratic equation.
In simple words: After expanding all the brackets and collecting like terms, we got an equation with \( x^2 \) as the highest power.

📝 Teacher's Note: This requires expanding two perfect squares and one linear term. Take it step by step to avoid mistakes.

🎯 Exam Tip: Collect like terms carefully. Write \( 2x^2 \), \( 5x \), and constant terms separately before combining.

Question 2(i). Is x = 5 a solution of the quadratic equation x² – 2x – 15 = 0?
Answer:
\( x^2 - 2x - 15 = 0 \)
For x = 5 to be solution of the given quadratic equation it should satisfy the equation.
So, substituting x = 5 in the given equation, we get
L.H.S = \( (5)^2 - 2(5) - 15 \)
= 25 - 10 - 15
= 0
= R.H.S
Hence, x = 5 is a solution of the quadratic equation \( x^2 - 2x - 15 = 0 \).
In simple words: We put x = 5 in the equation. The left side equals 0, same as the right side. So x = 5 works.

📝 Teacher's Note: Show that a solution means the equation becomes true when we substitute that value. L.H.S = R.H.S means it works.

🎯 Exam Tip: Always write "L.H.S = R.H.S" when they are equal. This shows you checked properly and gets you marks.

Question 2(ii). Is x = -3 a solution of the quadratic equation 2x² – 7x + 9 = 0?
Answer:
\( 2x^2 - 7x + 9 = 0 \)
For x = -3 to be solution of the given quadratic equation it should satisfy the equation
So, substituting x = -3 in the given equation, we get
L.H.S = \( 2(-3)^2 - 7(-3) + 9 \)
= 18 + 21 + 9
= 48
≠ R.H.S
Hence, x = -3 is not a solution of the quadratic equation \( 2x^2 - 7x + 9 = 0 \).
In simple words: We put x = -3 in the equation. The left side equals 48, but the right side is 0. Since they are not equal, x = -3 does not work.

📝 Teacher's Note: Point out the error in the original text - it says "substituting x = 5" but should be "x = -3". Students should be careful with signs.

🎯 Exam Tip: Write "≠ R.H.S" clearly when they are not equal. This shows you know it is not a solution.

Question 3. If \( \sqrt{\frac{2}{3}} \) is a solution of equation 3x² + mx + 2 = 0, find the value of m.
Answer:
For \( x = \sqrt{\frac{2}{3}} \) to be solution of the given quadratic equation it should satisfy the equation
So, substituting \( x = \sqrt{\frac{2}{3}} \) in the given equation, we get
\( 3\left(\sqrt{\frac{2}{3}}\right)^2 + m\left(\sqrt{\frac{2}{3}}\right) + 2 = 0 \)
\( \implies 3\left(\frac{2}{3}\right) + m\left(\sqrt{\frac{2}{3}}\right) + 2 = 0 \)
\( \implies 2 + m\sqrt{\frac{2}{3}} + 2 = 0 \)
\( \implies m = -4 \times \sqrt{\frac{3}{2}} = -2\sqrt{6} \)
∴ \( m = -2\sqrt{6} \)
In simple words: We put the given value in the equation and solved for m. We got m = -2√6.

📝 Teacher's Note: Remind students that \( (\sqrt{a})^2 = a \). Also, rationalize the denominator to get the final answer in simplest form.

🎯 Exam Tip: Write the substitution step clearly. Show how \( (\sqrt{2/3})^2 = 2/3 \). Keep the final answer in surd form unless asked to approximate.

Question 4. \( \frac{2}{3} \) and 1 are the solutions of equation mx² + nx + 6 = 0. Find the values of m and n.
Answer:
For \( x = \frac{2}{3} \) and x = 1 to be solutions of the given quadratic equation it should satisfy the equation
So, substituting \( x = \frac{2}{3} \) and x = 1 in the given equation, we get
\( m\left(\frac{2}{3}\right)^2 + n\left(\frac{2}{3}\right) + 6 = 0 \) | \( m(1)^2 + n(1) + 6 = 0 \)
\( \implies m\left(\frac{4}{9}\right) + n\left(\frac{2}{3}\right) + 6 = 0 \) | \( \implies m + n + 6 = 0 \)
\( \implies 4m + 6n + 54 = 0 \) .....(1) | \( m + n + 6 = 0 \) ....(2)

Solving equations (1) and (2) simultaneously,
(1) - (2) × 6
\( \implies -2m + 18 = 0 \)
\( \implies m = 9 \)
Substitute in (2)
\( \implies n = -15 \)
In simple words: We made two equations by putting both solutions into the original equation. Then we solved these two equations together to find m and n.

📝 Teacher's Note: Show how to multiply equation (2) by 6 to eliminate n. This is the substitution method for solving simultaneous equations.

🎯 Exam Tip: Label your equations (1) and (2). Show the elimination step clearly. Check your answer by substituting back.

Question 5. If 3 and -3 are the solutions of equation ax² + bx – 9 = 0. Find the values of a and b.
Answer:
For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the equation
So, substituting x = 3 and x = -3 in the given equation, we get
\( a(3)^2 + b(3) - 9 = 0 \) | \( a(-3)^2 + b(-3) - 9 = 0 \)
\( \implies a(9) + b(3) - 9 = 0 \) | \( \implies a(9) - b(3) - 9 = 0 \)
\( \implies 9a + 3b - 9 = 0 \) ...(1) | \( \implies 9a - 3b - 9 = 0 \) ...(2)

Solving equations (1) and (2) simultaneously,
(1) + (2)
\( \implies 18a - 18 = 0 \)
\( \implies a = 1 \)
Substitute in (2)
\( \implies b = 0 \)
In simple words: We used both solutions to make two equations. When we added them, the b terms cancelled and we found a = 1, then b = 0.

📝 Teacher's Note: Notice that 3 and -3 are opposites, so their b terms cancel when we add the equations. This makes solving easier.

🎯 Exam Tip: When you have opposite solutions like ±3, adding the equations eliminates the middle term. This is a quick method.

Exercise 5B

Question 1. Without solving, comment upon the nature of roots of each of the following equations:
(i) 7x² – 9x + 2 = 0
(ii) 6x² – 13x + 4 = 0
(iii) 25x² – 10x + 1 = 0
(iv) x² + 2√3x – 9 = 0
(v) x² – ax – b² = 0
(vi) 2x² + 8x + 9 = 0

(i) 7x² – 9x + 2 = 0
Answer:
a = 7, b = -9 and c = 2
∴ Discriminant = \( b^2 - 4ac \)
= \( (-9)^2 - 4(7)(2) \)
= 81 - 56 = 25
Since D > 0, then equation has two real and unequal roots.
In simple words: The discriminant is positive (25), so this equation has two different real solutions.

📝 Teacher's Note: Remind students that discriminant = \( b^2 - 4ac \). If D > 0, roots are real and different. If D = 0, roots are real and equal. If D < 0, roots are not real.

🎯 Exam Tip: Always write the values of a, b, c first. Then substitute in the discriminant formula. State the conclusion about nature of roots.

(ii) 6x² – 13x + 4 = 0
Answer:
a = 6, b = -13 and c = 4
∴ Discriminant = \( b^2 - 4ac \)
= \( (-13)^2 - 4(6)(4) \)
= 169 - 96 = 73
Since 73 is not a perfect square, roots are irrational
Since D > 0, then equation has two irrational and unequal roots.
In simple words: The discriminant is 73, which is positive but not a perfect square. So we get two different irrational solutions.

📝 Teacher's Note: Explain that if D > 0 but not a perfect square, the roots are irrational (involve surds). If D is a perfect square, roots are rational.

🎯 Exam Tip: Check if the discriminant is a perfect square. Write "irrational" when it is not a perfect square, "rational" when it is.

(iii) 25x² – 10x + 1 = 0
Answer:
a = 25, b = -10 and c = 1
∴ Discriminant = \( b^2 - 4ac \)
= \( (-10)^2 - 4(25)(1) \)
= 100 - 100 = 0
Since D = 0, then equation has two real and equal roots.
In simple words: The discriminant is 0, so this equation has one repeated solution (or two equal solutions).

📝 Teacher's Note: When D = 0, we get a perfect square trinomial. The equation can be written as \( (5x - 1)^2 = 0 \), giving x = 1/5 twice.

🎯 Exam Tip: D = 0 means equal roots. This happens when the quadratic is a perfect square.

(iv) x² + 2√3x – 9 = 0
Answer:
a = 1, b = \( 2\sqrt{3} \) and c = -9
∴ Discriminant = \( b^2 - 4ac \)
= \( (2\sqrt{3})^2 - 4(1)(-9) \)
= 12 + 36 = 48
Since 48 is not a perfect square, roots are irrational
Since D > 0, then equation has two irrational and unequal roots.
In simple words: The discriminant is 48, which is positive but not a perfect square. So we get two different irrational solutions.

📝 Teacher's Note: Show that \( (2\sqrt{3})^2 = 4 \times 3 = 12 \). Students often make mistakes with squares of surds.

🎯 Exam Tip: Be careful when squaring terms with surds. \( (a\sqrt{b})^2 = a^2 \times b \).

(v) x² – ax – b² = 0
Answer:
A = 1, B = -a and C = \( -b^2 \)
∴ Discriminant = \( B^2 - 4AC \)
= \( (-a)^2 - 4(1)(-b^2) \)
= \( a^2 + 4b^2 \) = a positive value
Since \( a^2 + 4b^2 \) is not a perfect square, roots are irrational.
Since D > 0, then equation has two irrational and unequal roots.
In simple words: Since \( a^2 \) and \( 4b^2 \) are both positive, the discriminant is always positive. So we always get two real solutions.

📝 Teacher's Note: Emphasize that \( a^2 \geq 0 \) and \( b^2 \geq 0 \) for all real numbers, so \( a^2 + 4b^2 > 0 \) (except when both a and b are zero).

🎯 Exam Tip: For general cases with variables, explain why the discriminant is positive or negative using properties of squares.

(vi) 2x² + 8x + 9 = 0
Answer:
a = 2, b = 8 and c = 9
∴ Discriminant = \( b^2 - 4ac \)
= \( (8)^2 - 4(2)(9) \)
= 64 - 72 = -18 = a negative value
Since D < 0, then equation has no real roots.
In simple words: The discriminant is negative (-18), so this equation has no real solutions. The solutions would involve imaginary numbers.

📝 Teacher's Note: When D < 0, the equation has no real solutions. At this level, we just say "no real roots" - complex roots come later.

🎯 Exam Tip: D < 0 means no real roots. Don't try to find complex roots unless specifically asked.

Question 2. Find the value of p, if the following quadratic equation has equal roots: 4x² – (p – 2)x + 1 = 0
Answer:
Given: \( 4x^2 - (p - 2)x + 1 = 0 \)
Here a = 4, b = -(p - 2), c = 1
For equal roots, discriminant = 0
\( b^2 - 4ac = 0 \)
\( [-(p - 2)]^2 - 4(4)(1) = 0 \)
\( (p - 2)^2 - 16 = 0 \)
\( (p - 2)^2 = 16 \)
\( p - 2 = \pm 4 \)
\( p = 2 \pm 4 \)
Therefore, p = 6 or p = -2
In simple words: For equal roots, the discriminant must be zero. We solved this condition to find p = 6 or p = -2.

📝 Teacher's Note: Remember that when we take square root of both sides, we get both positive and negative values. So \( \sqrt{16} = \pm 4 \).

🎯 Exam Tip: Always write "For equal roots, discriminant = 0" first. Don't forget the ± when taking square roots.

Question 3. Find the value of 'p', if the following quadratic equations have equal roots : \( x^2 + (p - 3)x + p = 0 \)
Answer:
\( x^2 + (p - 3)x + p = 0 \)
Here, \( a = 1, b = (p - 3), c = p \)
Since, the roots are equal,
\( \Rightarrow b^2 - 4ac = 0 \)
\( \Rightarrow (p - 3)^2 - 4(1)(p) = 0 \)
\( \Rightarrow p^2 + 9 - 6p - 4p = 0 \)
\( \Rightarrow p^2 - 10p + 9 = 0 \)
\( \Rightarrow p^2 - 9p - p + 9 = 0 \)
\( \Rightarrow p(p - 9) - 1(p - 9) = 0 \)
\( \Rightarrow (p - 9)(p - 1) = 0 \)
\( \Rightarrow p - 9 = 0 \) or \( p - 1 = 0 \)
\( \Rightarrow p = 9 \) or \( p = 1 \)
In simple words: When roots are equal, the discriminant is zero. We use this rule to find p.

📝 Teacher's Note: Equal roots means the parabola just touches the x-axis at one point. The discriminant formula \( b^2 - 4ac = 0 \) is the key rule to remember.

🎯 Exam Tip: Always write "Since roots are equal, discriminant = 0" first. Then substitute values carefully. Check both answers by putting them back in the original equation.

Question 4. The equation \( 3x^2 - 12x + (n - 5) = 0 \) has equal roots. Find the value of n.
Answer:
\( 3x^2 - 12x + (n - 5) = 0 \)
Here \( a = 3, b = -12 \) and \( c = n - 5 \)
Given: equation has equal roots
Then \( D = 0 \)
\( \Rightarrow b^2 - 4ac = 0 \)
\( \Rightarrow (-12)^2 - 4(3)(n - 5) = 0 \)
\( \Rightarrow 144 - 12n + 60 = 0 \)
\( \Rightarrow -12n = -204 \)
\( \Rightarrow n = \frac{-204}{-12} = 17 \)
In simple words: We use the same discriminant rule. Put all values in the formula and solve for n.

📝 Teacher's Note: Be careful with negative signs when expanding \( (-12)^2 \). Students often make mistakes here. \( (-12)^2 = 144 \), not \( -144 \).

🎯 Exam Tip: Write down a, b, c values clearly first. Double-check your arithmetic when expanding brackets. The final answer must be a whole number here.

Question 5. Find the value of m, if the following equation has equal roots : \( (m - 2)x^2 - (5+m)x +16 = 0 \)
Answer:
\( (m - 2)x^2 - (5 + m)x + 16 = 0 \)
Here \( a = m - 2, b = -(5 + m) \) and \( c = 16 \)
Given: equation has equal roots
Then \( D = 0 \)
\( \Rightarrow b^2 - 4ac = 0 \)
\( \Rightarrow [-(5 + m)]^2 - 4(m - 2)(16) = 0 \)
\( \Rightarrow 25 + m^2 + 10m - 64m + 128 = 0 \)
\( \Rightarrow m^2 - 54m + 153 = 0 \)
\( \Rightarrow m^2 - 51m - 3m + 153 = 0 \)
\( \Rightarrow m(m - 51) - 3(m - 51) = 0 \)
\( \Rightarrow (m - 51)(m - 3) = 0 \)
then \( m - 51 = 0 \) or \( m - 3 = 0 \)
\( \Rightarrow m = 51 \) or \( m = 3 \)
In simple words: Same method as before. We expand the brackets carefully and factorize to get two values of m.

📝 Teacher's Note: When b has a negative sign, be extra careful with \( [-(5+m)]^2 = (5+m)^2 \). The negative sign disappears when we square it.

🎯 Exam Tip: Check that both values of m make \( a = m - 2 \neq 0 \). If m = 2, the equation is not quadratic anymore.

Question 6. Find the value of p for which the equation \( 3x^2 - 6x + k = 0 \) has distinct and real roots.
Answer:
\( 3x^2 - 6x + k = 0 \)
Here, \( a = 3, b = -6 \) and \( c = k \)
Since the roots are distinct and real,
\( b^2 - 4ac > 0 \)
\( \Rightarrow (-6)^2 - 4 \times 3 \times k > 0 \)
\( \Rightarrow 36 - 12k > 0 \)
\( \Rightarrow 36 > 12k \)
\( \Rightarrow 3 > k \)
\( \Rightarrow k < 3 \)
In simple words: For two different real roots, discriminant must be positive (greater than zero). This gives us the range of k values.

📝 Teacher's Note: Distinct means different. Real means not imaginary. For this we need discriminant > 0, not = 0. Use inequality signs carefully.

🎯 Exam Tip: Write the condition clearly: "For distinct real roots, \( b^2 - 4ac > 0 \)". The answer is an inequality, not a single number.

Exercise 5C

Question 1. Solve : \( x^2 - 10x - 24 = 0 \)
Answer:
\( x^2 - 10x - 24 = 0 \)
\( \Rightarrow x^2 - 12x + 2x - 24 = 0 \)
\( \Rightarrow x(x - 12) + 2(x - 12) = 0 \)
\( \Rightarrow (x - 12)(x + 2) = 0 \)
since \( x - 12 = 0 \) or \( x + 2 = 0 \)
then \( x = 12 \) or \( x = -2 \)
In simple words: We split the middle term -10x into -12x and +2x. Then we factorize to get the two roots.

📝 Teacher's Note: Teach students to find two numbers that multiply to -24 and add to -10. Here it's -12 and +2. This makes splitting easier.

🎯 Exam Tip: Always check your answer by substituting back. If x = 12, then \( 12^2 - 10(12) - 24 = 144 - 120 - 24 = 0 \). Correct!

Question 2. Solve : \( x^2 - 16 = 0 \)
Answer:
\( x^2 - 16 = 0 \)
\( \Rightarrow x^2 - 4^2 = 0 \)
\( \Rightarrow (x + 4)(x - 4) = 0 \)
Since \( x + 4 = 0 \) or \( x - 4 = 0 \)
then \( x = -4 \) or \( x = 4 \)
In simple words: This is difference of squares formula. \( a^2 - b^2 = (a+b)(a-b) \). Here a = x and b = 4.

📝 Teacher's Note: This is the easiest type of quadratic. Recognize the pattern \( x^2 - a^2 \) immediately. No middle term means we use difference of squares.

🎯 Exam Tip: Learn the formula \( a^2 - b^2 = (a+b)(a-b) \) by heart. This saves time in exams for such problems.

Question 3. Solve : \( 2x^2 - \frac{1}{2}x = 0 \)
Answer:
\( 2x^2 - \frac{1}{2}x = 0 \)
\( \Rightarrow x(2x - \frac{1}{2}) = 0 \)
since \( x = 0 \) or \( 2x - \frac{1}{2} = 0 \)
then \( x = 0 \) or \( x = \frac{1}{4} \)
In simple words: We take out x as common factor. Then solve each part separately to get the two roots.

📝 Teacher's Note: When there's no constant term, always take x as common. One root will always be x = 0. This makes the problem much easier.

🎯 Exam Tip: For equations like \( ax^2 + bx = 0 \), always write \( x(ax + b) = 0 \) first. Don't forget that x = 0 is one of the roots.

Question 4. Solve : \( x(x - 5) = 24 \)
Answer:
\( x(x - 5) = 24 \)
\( \Rightarrow x^2 - 5x = 24 \)
\( \Rightarrow x^2 - 5x - 24 = 0 \)
\( \Rightarrow x^2 - 8x + 3x - 24 = 0 \)
\( \Rightarrow x(x - 8) + 3(x - 8) = 0 \)
\( \Rightarrow (x - 8)(x + 3) = 0 \)
Since \( x - 8 = 0 \) or \( x + 3 = 0 \)
then \( x = 8 \) or \( x = -3 \)
In simple words: First expand the brackets, then move everything to one side to make it standard form, then factorize.

📝 Teacher's Note: Students often try to solve \( x(x-5) = 24 \) by setting x = 24 or x-5 = 24. This is wrong. Always expand first.

🎯 Exam Tip: Never solve \( AB = C \) by saying A = C or B = C. Only when \( AB = 0 \) can we say A = 0 or B = 0.

Question 5. Solve : \( \frac{9}{2}x = 5 + x^2 \)
Answer:
\( \frac{9}{2}x = 5 + x^2 \)
\( \Rightarrow 9x = 10 + 2x^2 \)
\( \Rightarrow 2x^2 - 9x + 10 = 0 \)
\( \Rightarrow 2x^2 - 5x - 4x + 10 = 0 \)
\( \Rightarrow x(2x - 5) - 2(2x - 5) = 0 \)
\( \Rightarrow (2x - 5)(x - 2) = 0 \)
Since \( 2x - 5 = 0 \) or \( x - 2 = 0 \)
then \( x = \frac{5}{2} \) or \( x = 2 \)
In simple words: First remove fractions by multiplying both sides by 2. Then rearrange to standard form and factorize.

📝 Teacher's Note: When there are fractions, multiply the whole equation by the denominator first. This makes calculations much easier.

🎯 Exam Tip: Always clear fractions early in the solution. Write "Multiplying both sides by 2" to show your steps clearly.

Question 6. Solve : \( \frac{6}{x} = 1 + x \)
Answer:
\( \frac{6}{x} = 1 + x \)
\( \Rightarrow 6 = x + x^2 \)
\( \Rightarrow x^2 + x - 6 = 0 \)
\( \Rightarrow x^2 + 3x - 2x - 6 = 0 \)
\( \Rightarrow x(x + 3) - 2(x + 3) = 0 \)
\( \Rightarrow (x + 3)(x - 2) = 0 \)
since \( x + 3 = 0 \) or \( x - 2 = 0 \)
then \( x = -3 \) or \( x = 2 \)
In simple words: Multiply both sides by x to remove the fraction. Then solve the quadratic as usual.

📝 Teacher's Note: Remind students that x cannot be zero (since we have \( \frac{6}{x} \)). Check that neither root is zero before finalizing the answer.

🎯 Exam Tip: When solving equations with \( \frac{1}{x} \), always check that your roots don't make the denominator zero. Both x = -3 and x = 2 are valid here.

Question 7. Solve : \( x = \frac{3x + 1}{4x} \)
Answer:
\( x = \frac{3x + 1}{4x} \)
\( \Rightarrow 4x^2 = 3x + 1 \)
\( \Rightarrow 4x^2 - 3x - 1 = 0 \)
\( \Rightarrow 4x^2 - 4x + x - 1 = 0 \)
\( \Rightarrow 4x(x - 1) + 1(x - 1) = 0 \)
\( \Rightarrow (4x + 1)(x - 1) = 0 \)
Since \( 4x + 1 = 0 \) or \( x - 1 = 0 \)
then \( x = -\frac{1}{4} \) or \( x = 1 \)
In simple words: Cross multiply to remove fractions. Then solve the resulting quadratic equation by factorization method.

📝 Teacher's Note: Cross multiplication means: if \( \frac{a}{b} = c \), then \( a = bc \). Here \( x \cdot 4x = (3x + 1) \cdot 1 \).

🎯 Exam Tip: After cross multiplication, always rearrange to standard form \( ax^2 + bx + c = 0 \). Check that x ≠ 0 since it's in the denominator of original equation.

Question 8. Solve: \( x + \frac{1}{x} = 2.5 \)
Answer:
\( x + \frac{1}{x} = 2.5 \)

\( \implies \frac{x^2 + 1}{x} = \frac{5}{2} \)

\( \implies 2x^2 + 2 = 5x \)

\( \implies 2x^2 - 5x + 2 = 0 \)

\( \implies 2x^2 - 4x - x + 2 = 0 \)

\( \implies 2x(x-2) - 1(x-2) = 0 \)

\( \implies (x-2)(2x-1) = 0 \)

Since \( x-2=0 \) or \( 2x-1=0 \)

then \( x=2 \) or \( x=\frac{1}{2} \)
In simple words: We multiplied both sides by x to clear the fraction. Then we rearranged to get a quadratic equation and factored it to find the values of x.

📝 Teacher's Note: When solving equations with fractions, always multiply through by the denominator first. Show students how to check their answers by putting the values back into the original equation.

🎯 Exam Tip: Always write the factored form clearly. Check your answer by substituting back into the original equation. Show all steps of factoring.

Question 9. Solve: \( (2x - 3)^2 = 49 \)
Answer:
\( (2x-3)^2 = 49 \)

Taking square root on both sides

\( 2x-3 = \pm 7 \)

When \( 2x-3=7 \)
\( \implies 2x=10 \)
\( \implies x=5 \)

and, when \( 2x-3=-7 \)
\( \implies 2x=-4 \)
\( \implies x=-2 \)
In simple words: When we have a squared expression equal to a number, we take the square root of both sides. Remember that square root gives us both positive and negative values.

📝 Teacher's Note: Remind students that when taking square roots, we get both positive and negative answers. This is why we write ±7. Many students forget the negative root.

🎯 Exam Tip: Always write "Taking square root on both sides" and remember to use ± symbol. Show both cases separately for full marks.

Question 10. Solve: \( 2(x^2 - 6) = 3(x - 4) \)
Answer:
\( 2(x^2 - 6) = 3(x - 4) \)

\( \implies 2x^2 - 12 = 3x - 12 \)

\( \implies 2x^2 - 3x = 0 \)

\( \implies x(2x - 3) = 0 \)

since \( x=0 \) or \( 2x-3=0 \)

then \( x=0 \) or \( x=\frac{3}{2} \)
In simple words: We expanded both sides, moved all terms to one side, then factored out the common factor x. This gave us two simple equations to solve.

📝 Teacher's Note: Show students how expanding brackets carefully prevents mistakes. Point out that one solution is always x = 0 when we can factor out x.

🎯 Exam Tip: Expand brackets step by step. When you can factor out x, one solution is always x = 0. Write this clearly.

Question 11. Solve: \( (x + 1)(2x + 8) = (x + 7)(x + 3) \)
Answer:
\( (x+1)(2x+8)=(x+7)(x+3) \)

\( \implies 2x^2 + 8x + 2x + 8 = x^2 + 3x + 7x + 21 \)

\( \implies 2x^2 + 10x + 8 = x^2 + 10x + 21 \)

\( \implies x^2 - 13 = 0 \)

\( \implies x^2 - (\sqrt{13})^2 = 0 \)

\( \implies (x + \sqrt{13})(x - \sqrt{13}) = 0 \)

If \( x+\sqrt{13} = 0 \) or \( x - \sqrt{13} = 0 \)

\( \implies x = -\sqrt{13} \) or \( x=\sqrt{13} \)
In simple words: We expanded both sides, simplified, and got a simple equation with x². We used the difference of squares formula to factor and solve.

📝 Teacher's Note: When expanding brackets, students often make arithmetic errors. Check each step carefully. Emphasize that \( x^2 - 13 = 0 \) means \( x^2 = 13 \), so \( x = \pm\sqrt{13} \).

🎯 Exam Tip: Expand each bracket separately first, then combine like terms. Use the difference of squares pattern: \( a^2 - b^2 = (a+b)(a-b) \).

Question 12. Solve: \( x^2 - (a + b)x + ab = 0 \)
Answer:
\( x^2 - (a + b)x + ab = 0 \)

\( \implies x^2 - ax - bx + ab = 0 \)

\( \implies x(x - a) - b(x - a) = 0 \)

\( \implies (x - a)(x - b) = 0 \)

since \( x - a=0 \) or \( x - b=0 \)

then \( x=a \) or \( x=b \)
In simple words: This is a special pattern where the equation factors nicely. The middle term -(a+b)x splits into -ax and -bx, making factoring easy.

📝 Teacher's Note: This is a standard form students should recognize. The equation \( x^2 - (a+b)x + ab = 0 \) always has roots x = a and x = b. Show them this pattern.

🎯 Exam Tip: Remember this pattern: when you see \( x^2 - (a+b)x + ab = 0 \), the answers are always x = a and x = b. No need to use the quadratic formula.

Question 13. Solve: \( (x + 3)^2 - 4(x + 3) - 5 = 0 \)
Answer:
\( (x+3)^2 - 4(x + 3) - 5 = 0 \)

Let \( x+3=y \)

then \( y^2 - 4y - 5 = 0 \)

\( \implies y^2 - 5y + y - 5 = 0 \)

\( \implies y(y-5)+1(y-5)=0 \)

\( \implies (y-5)(y+1)=0 \)

If \( y-5=0 \) or \( y+1=0 \)

then \( y=5 \) or \( y=-1 \)

\( \implies x+3=5 \) or \( x+3=-1 \)

\( \implies x=2 \) or \( x=-4 \)
In simple words: We used substitution to make the equation simpler. We let y = x+3, solved for y, then found x by substituting back.

📝 Teacher's Note: Substitution method makes complex equations easier. When you see repeated expressions like (x+3), replace them with a single letter first.

🎯 Exam Tip: Always write "Let (x+3) = y" clearly. Solve for y first, then substitute back to find x. Don't forget the back-substitution step.

Question 14. Solve: \( 4(2x - 3)^2 - (2x - 3) - 14 = 0 \)
Answer:
\( 4(2x-3)^2 - (2x - 3) - 14 = 0 \)

Let \( 2x-3=y \)

then \( 4y^2 - y - 14 = 0 \)

\( \implies 4y^2 - 8y + 7y - 14 = 0 \)

\( \implies 4y(y - 2) + 7(y - 2) = 0 \)

\( \implies (y - 2)(4y + 7) = 0 \)

If \( y - 2 = 0 \) or \( 4y+7 =0 \)

\( \implies y=2 \) or \( y=\frac{-7}{4} \)

\( \implies 2x - 3 = 2 \) or \( 2x - 3 = \frac{-7}{4} \)

\( \implies 2x=5 \) or \( 2x=\frac{5}{4} \)

\( \implies x=\frac{5}{2} \) or \( x=\frac{5}{8} \)
In simple words: Again we used substitution. We replaced the repeated expression (2x-3) with y, solved the quadratic in y, then found x.

📝 Teacher's Note: Substitution works well when you see the same expression repeated. Students should look for patterns like this to make factoring easier.

🎯 Exam Tip: When you see a repeated expression like (2x-3), always use substitution. Factor by grouping when the leading coefficient is not 1.

Question 15. Solve: \( \frac{3x - 2}{2x - 3} = \frac{3x - 8}{x + 4} \)
Answer:
\( \frac{3x - 2}{2x - 3} = \frac{3x - 8}{x + 4} \)

\( \implies (3x - 2)(x + 4) = (2x - 3)(3x - 8) \)

\( \implies 3x^2 + 12x - 2x - 8 = 6x^2 - 16x - 9x + 24 \)

\( \implies 3x^2 + 10x - 8 = 6x^2 - 25x + 24 \)

\( \implies 3x^2 - 35x + 32 = 0 \)

\( \implies 3x^2 - 32x - 3x + 32 = 0 \)

\( \implies x(3x - 32) - 1(3x - 32) = 0 \)

\( \implies (x - 1)(3x - 32) = 0 \)

If \( x - 1 = 0 \) or \( 3x - 32 = 0 \)

\( \implies x=1 \) or \( x=\frac{32}{3} = 10\frac{2}{3} \)
In simple words: We cross-multiplied the fractions to remove denominators. Then we expanded, rearranged, and factored the resulting quadratic equation.

📝 Teacher's Note: Cross-multiplication is the standard method for solving equations with fractions. Remind students to expand brackets carefully and check for arithmetic errors.

🎯 Exam Tip: Cross-multiply first to clear fractions. Expand brackets step by step. Always check that your answers don't make any denominator zero.

Question 16. Solve: \( 2x^2 - 9x + 10 = 0 \), When
(i) \( x \in N \)
(ii) \( x \in Q \)
Answer:
\( 2x^2 - 9x + 10 = 0 \)

\( \implies 2x^2 - 5x - 4x + 10 = 0 \)

\( \implies x(2x - 5) - 2(2x - 5) = 0 \)

\( \implies (2x - 5)(x - 2) = 0 \)

\( \implies 2x - 5 = 0 \) or \( x - 2 = 0 \)

\( \implies x = \frac{5}{2} \) or \( x = 2 \)

(i) When \( x \in N \), we have \( x = 2 \)

(ii) When \( x \in Q \), we have \( x = 2, \frac{5}{2} \)
In simple words: We solved the equation and got two answers. Then we checked which answers belong to the given sets. Natural numbers (N) are 1, 2, 3... Rational numbers (Q) include fractions.

📝 Teacher's Note: Teach students the difference between number sets. N = natural numbers {1, 2, 3...}, Q = rational numbers (fractions). Only some solutions may belong to the given set.

🎯 Exam Tip: Solve the equation completely first. Then check which solutions belong to the given set. Write the final answer clearly for each part.

Question 17. Solve: \( \frac{x - 3}{x + 3} + \frac{x + 3}{x - 3} = 2\frac{1}{2} \)
Answer:
\( \frac{x - 3}{x + 3} + \frac{x + 3}{x - 3} = 2\frac{1}{2} \)

\( \implies \frac{(x - 3)^2 + (x + 3)^2}{(x + 3)(x - 3)} = \frac{5}{2} \)

\( \implies \frac{x^2 - 6x + 9 + x^2 + 6x + 9}{x^2 - 9} = \frac{5}{2} \)

\( \implies 2(2x^2 + 18) = 5(x^2 - 9) \)

\( \implies 4x^2 + 36 = 5x^2 - 45 \)

\( \implies x^2 - 81 = 0 \)

\( \implies x^2 - 9^2 = 0 \)

\( \implies (x + 9)(x - 9) = 0 \)

If \( x + 9 = 0 \) or \( x - 9 = 0 \)

then \( x = -9 \) or \( x = 9 \)
In simple words: We combined the fractions using a common denominator. Then we cross-multiplied and solved the resulting quadratic equation using difference of squares.

📝 Teacher's Note: When adding fractions, find the common denominator first. Here it's (x+3)(x-3) = x²-9. Students often make sign errors when expanding (x-3)² and (x+3)².

🎯 Exam Tip: Convert mixed number 2½ to improper fraction 5/2 first. When you get x² - 81 = 0, recognize it as difference of squares: x² - 9² = (x+9)(x-9).

Question 18. Solve: \( \frac{4}{x + 2} - \frac{1}{x + 3} = \frac{4}{2x + 1} \)
Answer:
Step 1: Find common denominator for left side.
\( \frac{4}{x + 2} - \frac{1}{x + 3} = \frac{4}{2x + 1} \)

Step 2: Cross multiply and simplify.
\( \frac{4(x + 3) - 1(x + 2)}{(x + 2)(x + 3)} = \frac{4}{2x + 1} \)
\( \frac{4x + 12 - x - 2}{x^2 + 2x + 3x + 6} = \frac{4}{2x + 1} \)
\( \frac{3x + 10}{x^2 + 5x + 6} = \frac{4}{2x + 1} \)

Step 3: Cross multiply.
\( (3x + 10)(2x + 1) = 4(x^2 + 5x + 6) \)
\( 6x^2 + 3x + 20x + 10 = 4x^2 + 20x + 24 \)
\( 6x^2 + 23x + 10 = 4x^2 + 20x + 24 \)

Step 4: Rearrange and solve.
\( 2x^2 + 3x - 14 = 0 \)
\( 2x^2 + 7x - 4x - 14 = 0 \)
\( x(2x + 7) - 2(2x + 7) = 0 \)
\( (2x + 7)(x - 2) = 0 \)

If \( 2x + 7 = 0 \), then \( x = -\frac{7}{2} \)
If \( x - 2 = 0 \), then \( x = 2 \)

In simple words: We moved all fractions to one side and found a common bottom. Then we got a quadratic equation and solved it by factoring.

📝 Teacher's Note: When solving fraction equations, first check that the values don't make any denominator zero. Here x ≠ -2, -3, -1/2. Both our answers are safe.

🎯 Exam Tip: Always cross multiply carefully and collect like terms step by step. Check your final answers by putting them back in the original equation.

Question 19. Solve: \( \frac{5}{x - 2} - \frac{3}{x + 6} = \frac{4}{x} \)
Answer:
Step 1: Find common denominator for left side.
\( \frac{5}{x - 2} - \frac{3}{x + 6} = \frac{4}{x} \)
\( \frac{5(x + 6) - 3(x - 2)}{(x - 2)(x + 6)} = \frac{4}{x} \)

Step 2: Simplify the numerator.
\( \frac{5x + 30 - 3x + 6}{x^2 + 6x - 2x - 12} = \frac{4}{x} \)
\( \frac{2x + 36}{x^2 + 4x - 12} = \frac{4}{x} \)

Step 3: Cross multiply.
\( 4x^2 + 16x - 48 = 2x^2 + 36x \)
\( 2x^2 - 20x - 48 = 0 \)
\( x^2 - 10x - 24 = 0 \)

Step 4: Factor and solve.
\( x^2 - 12x + 2x - 24 = 0 \)
\( x(x - 12) + 2(x - 12) = 0 \)
\( (x - 12)(x + 2) = 0 \)

If \( x - 12 = 0 \), then \( x = 12 \)
If \( x + 2 = 0 \), then \( x = -2 \)

In simple words: We combined the fractions on the left side, then cross multiplied to get rid of all fractions. Finally we solved the quadratic equation.

📝 Teacher's Note: Check that x ≠ 2, -6, 0 from the original equation. Our answer x = -2 makes one denominator zero, so it's not valid. Only x = 12 works.

🎯 Exam Tip: Always check that your answers don't make any denominator in the original equation equal to zero. Cross out invalid solutions.

Question 20. Solve: \( \left(1 + \frac{1}{x + 1}\right)\left(1 - \frac{1}{x - 1}\right) = \frac{7}{8} \)
Answer:
Step 1: Simplify each bracket.
\( 1 + \frac{1}{x + 1} = \frac{x + 1 + 1}{x + 1} = \frac{x + 2}{x + 1} \)
\( 1 - \frac{1}{x - 1} = \frac{x - 1 - 1}{x - 1} = \frac{x - 2}{x - 1} \)

Step 2: Multiply the fractions.
\( \frac{x + 2}{x + 1} \times \frac{x - 2}{x - 1} = \frac{7}{8} \)
\( \frac{(x + 2)(x - 2)}{(x + 1)(x - 1)} = \frac{7}{8} \)
\( \frac{x^2 - 4}{x^2 - 1} = \frac{7}{8} \)

Step 3: Cross multiply and solve.
\( 8(x^2 - 4) = 7(x^2 - 1) \)
\( 8x^2 - 32 = 7x^2 - 7 \)
\( x^2 - 25 = 0 \)
\( x^2 = 25 \)
\( x = \pm 5 \)

In simple words: We simplified each bracket by adding fractions. Then we multiplied them and cross multiplied to solve for x.

📝 Teacher's Note: Use the difference of squares formula: (a + b)(a - b) = a² - b². This makes the calculation much easier.

🎯 Exam Tip: Always check that x ≠ -1, 1 from the original equation. Both our answers x = 5 and x = -5 are valid.

Question 21. Find the quadratic equation, whose solution set is: (i) {3, 5} (ii) {-2, 3}
Answer:
(i) Since solution set is {3, 5}
If x = 3 or x = 5
Then x - 3 = 0 or x - 5 = 0
\( (x - 3)(x - 5) = 0 \)
\( x^2 - 5x - 3x + 15 = 0 \)
\( x^2 - 8x + 15 = 0 \) which is the required equation.

(ii) Since solution set is {-2, 3}
If x = -2 or x = 3
Then x + 2 = 0 or x - 3 = 0
\( (x + 2)(x - 3) = 0 \)
\( x^2 - 3x + 2x - 6 = 0 \)
\( x^2 - x - 6 = 0 \) which is the required equation.

(iii) Since solution set is {5, -4}
If x = 5 or x = -4
Then x - 5 = 0 or x + 4 = 0
\( (x - 5)(x + 4) = 0 \)
\( x^2 + 4x - 5x - 20 = 0 \)
\( x^2 - x - 20 = 0 \) which is the required equation.

(iv) Since solution set is \( \left\{-3, \frac{-2}{5}\right\} \)
If x = -3 or x = \( \frac{-2}{5} \)
Then x + 3 = 0 or 5x + 2 = 0
\( (x + 3)(5x + 2) = 0 \)
\( 5x^2 + 2x + 15x + 6 = 0 \)
\( 5x^2 + 17x + 6 = 0 \) which is the required equation.

In simple words: If we know the roots, we write them as (x - root₁)(x - root₂) = 0 and then expand to get the quadratic equation.

📝 Teacher's Note: Remember the rule: if p and q are roots, then the equation is (x - p)(x - q) = 0. Expand this to get standard form ax² + bx + c = 0.

🎯 Exam Tip: Always expand your brackets carefully. Double check by substituting the given roots back into your final equation to make sure you get zero.

Question 22. Solve: \( \frac{x}{3} + \frac{3}{6 - x} = \frac{2(6 + x)}{15} \), (x ≠ 6)
Answer:
Step 1: Clear the fractions by finding common denominators.
\( \frac{x}{3} + \frac{3}{6 - x} = \frac{2(6 + x)}{15} \)
\( \frac{x(6 - x) + 3 \times 3}{3(6 - x)} = \frac{12 + 2x}{15} \)

Step 2: Simplify the left side.
\( \frac{x(6 - x) + 9}{3(6 - x)} = \frac{12 + 2x}{15} \)
\( \frac{6x - x^2 + 9}{18 - 3x} = \frac{12 + 2x}{15} \)

Step 3: Cross multiply.
\( 15(6x - x^2 + 9) = (12 + 2x)(18 - 3x) \)
\( 90x - 15x^2 + 135 = 216 - 36x + 36x - 6x^2 \)
\( 90x - 15x^2 + 135 = 216 - 6x^2 \)

Step 4: Rearrange and solve.
\( -9x^2 + 90x - 81 = 0 \)
\( x^2 - 10x + 9 = 0 \)
\( x^2 - 9x - x + 9 = 0 \)
\( x(x - 9) - 1(x - 9) = 0 \)
\( (x - 9)(x - 1) = 0 \)

Therefore x = 9 or x = 1

In simple words: We cleared all fractions by cross multiplying, then solved the resulting quadratic equation by factoring.

📝 Teacher's Note: Always check that x ≠ 6 as given in the question. Both our solutions x = 1 and x = 9 are valid since they don't equal 6.

🎯 Exam Tip: When you have fractions with different denominators, find a common denominator first or cross multiply carefully. Show all steps clearly.

Question 23. Solve the equation \( 9x^2 + \frac{3x}{4} + 2 = 0 \), if possible, for real values of x.
Answer:
Step 1: Clear the fraction by multiplying by 4.
\( 9x^2 + \frac{3x}{4} + 2 = 0 \)
\( \frac{36x^2 + 3x + 8}{4} = 0 \)
\( 36x^2 + 3x + 8 = 0 \)

Step 2: Identify coefficients for quadratic formula.
Here, a = 36, b = 3 and c = 8

Step 3: Use the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-3 \pm \sqrt{3^2 - 4 \times 36 \times 8}}{2 \times 36} \)
\( x = \frac{-3 \pm \sqrt{9 - 1152}}{72} \)
\( x = \frac{-3 \pm \sqrt{-1143}}{72} \)

Step 4: Check for real solutions.
Since \( \sqrt{-1143} \) is not possible (negative under the square root), we cannot solve the given equation for real values of x.

In simple words: When we use the quadratic formula, we get a negative number under the square root. This means there are no real solutions.

📝 Teacher's Note: When the discriminant (b² - 4ac) is negative, the quadratic has no real roots. Explain that we can't find the square root of a negative number in real numbers.

🎯 Exam Tip: Always calculate the discriminant b² - 4ac first. If it's negative, write "no real solutions exist" and explain why.

Question 24. Find the value of x, if a + 1 = 0 and x² + ax - 6 = 0.
Answer:
Step 1: Find the value of a.
If a + 1 = 0, then a = -1

Step 2: Substitute a = -1 in the given equation.
Put this value in the given equation x² + ax - 6 = 0
x² - x - 6 = 0

Step 3: Solve the quadratic equation.
\( x^2 - 3x + 2x - 6 = 0 \)
\( x(x - 3) + 2(x - 3) = 0 \)
\( (x - 3)(x + 2) = 0 \)

If x - 3 = 0, then x = 3
If x + 2 = 0, then x = -2

In simple words: We first found the value of a, then put it in the second equation and solved the quadratic equation.

📝 Teacher's Note: This is a two-step problem. First solve for the unknown parameter, then substitute and solve the main equation. Students often forget the first step.

🎯 Exam Tip: Always find the value of the parameter first. Then substitute carefully and solve step by step. Both values of x are correct answers.

Question 25. Find the value of x, if a + 7 = 0; b + 10 = 0 and 12x² = ax - b.
Answer:
Step 1: Find values of a and b.
If a + 7 = 0, then a = -7
If b + 10 = 0, then b = -10

Step 2: Substitute in the given equation.
Put these values of a and b in the given equation 12x² = ax - b
12x² = (-7)x - (-10)
12x² = -7x + 10

Step 3: Rearrange to standard form.
12x² + 7x - 10 = 0
12x² + 15x - 8x - 10 = 0
3x(4x + 5) - 2(4x + 5) = 0
(3x - 2)(4x + 5) = 0

Step 4: Solve for x.
If 3x - 2 = 0, then x = \( \frac{2}{3} \)
If 4x + 5 = 0, then x = \( -\frac{5}{4} \)

In simple words: We found the values of a and b first, then substituted them in the main equation and solved the quadratic equation.

📝 Teacher's Note: Students often make sign errors when substituting negative values. Be careful with -(-10) = +10. Also check factoring carefully.

🎯 Exam Tip: Find all parameter values first, substitute carefully watching the signs, then solve the quadratic. Write both values of x as the final answer.

Question 26. Use the substitution y = 2x + 3 to solve for x, if 4(2x+3)² – (2x+3) – 14 = 0.
Answer:
Given: 4(2x+3)² – (2x+3) – 14 = 0
Let y = 2x + 3

Step 1: Substitute y in the equation.
4y² – y – 14 = 0

Step 2: Solve the quadratic equation in y.
\( \Rightarrow 4y^2 - 8y + 7y - 14 = 0 \)
\( \Rightarrow 4y(y - 2) + 7(y - 2) = 0 \)
\( \Rightarrow (y - 2)(4y + 7) = 0 \)

Step 3: Find values of y.
If y - 2 = 0 or 4y + 7 = 0
then y = 2 or y = \( \frac{-7}{4} \)

Step 4: Substitute back y = 2x + 3.
When y = 2: 2x + 3 = 2
\( \Rightarrow 2x = -1 \)
\( \Rightarrow x = \frac{-1}{2} \)

When y = \( \frac{-7}{4} \): 2x + 3 = \( \frac{-7}{4} \)
\( \Rightarrow 2x = \frac{-7}{4} - 3 = \frac{-7-12}{4} = \frac{-19}{4} \)
\( \Rightarrow x = \frac{-19}{8} \)

Therefore, x = \( \frac{-1}{2} \) or x = \( \frac{-19}{8} \)
In simple words: We replaced (2x+3) with y to make the equation simpler. Then we solved for y and put back the value to find x.

📝 Teacher's Note: Show students that substitution makes hard equations easier. It is like giving a nickname to a long expression. After solving, we change back to the original variable.

🎯 Exam Tip: Always write "Let y = " clearly. Show each step of substitution. After finding y, substitute back to find x. Do not forget to write both solutions.

Question 27. Without solving the quadratic equation 6x² – x – 2 = 0, find whether x = 2/3 is a solution of this equation or not.
Answer:
Given equation: 6x² – x – 2 = 0
Given value: x = \( \frac{2}{3} \)

Step 1: Put x = \( \frac{2}{3} \) in L.H.S.
L.H.S. = \( 6\left(\frac{2}{3}\right)^2 - \frac{2}{3} - 2 \)

Step 2: Calculate step by step.
\( = 6 \times \frac{4}{9} - \frac{2}{3} - 2 \)
\( = \frac{24}{9} - \frac{2}{3} - 2 \)
\( = \frac{24}{9} - \frac{6}{9} - \frac{18}{9} \)
\( = \frac{24-6-18}{9} = \frac{0}{9} = 0 \)

Since L.H.S. = R.H.S. = 0, then x = \( \frac{2}{3} \) is a solution of the given equation.
In simple words: We put the given value in the equation. If both sides become equal, then it is a solution. Here we got 0 = 0, so it works.

📝 Teacher's Note: Teach students to substitute the value carefully and do arithmetic step by step. If L.H.S. = R.H.S., then the value is a root.

🎯 Exam Tip: Write "Put x = " clearly. Show all calculation steps. Write final conclusion: "Since L.H.S. = R.H.S., x is a solution" or "Since L.H.S. ≠ R.H.S., x is not a solution".

Question 28. Determine whether x = -1 is a root of the equation x² – 3x + 2 = 0 or not.
Answer:
Given equation: x² – 3x + 2 = 0
Given value: x = -1

Put x = -1 in L.H.S.
L.H.S. = (-1)² – 3(-1) + 2
= 1 + 3 + 2 = 6 ≠ R.H.S.

Then x = -1 is not the solution of the given equation.
In simple words: We put x = -1 in the equation. We got 6 on the left side but we need 0. So x = -1 is not a solution.

📝 Teacher's Note: Students often make sign errors. Remind them that (-1)² = +1 and -3(-1) = +3. Check the arithmetic carefully.

🎯 Exam Tip: Always show the substitution clearly. Write the final answer as "x = -1 is a root" or "x = -1 is not a root". Do not just write numbers.

Question 29. If x = 2/3 is a solution of the quadratic equation 7x² + mx – 3 = 0; Find the value of m.
Answer:
Given equation: 7x² + mx – 3 = 0
Given: x = \( \frac{2}{3} \) is the solution of the given equation.

Put given value of x in the given equation:
\( 7\left(\frac{2}{3}\right)^2 + m\left(\frac{2}{3}\right) - 3 = 0 \)

Step 1: Simplify.
\( \Rightarrow \frac{28}{9} + \frac{2m}{3} - 3 = 0 \)

Step 2: Multiply through by 9 to clear fractions.
\( \Rightarrow 28 + 6m - 27 = 0 \)
\( \Rightarrow 6m = -1 \)
\( \Rightarrow m = \frac{-1}{6} \)
In simple words: Since x = 2/3 satisfies the equation, we put this value and solve for m. We get m = -1/6.

📝 Teacher's Note: Show students how to handle fractions step by step. Multiplying by the LCM makes calculations easier. Check the answer by substituting back.

🎯 Exam Tip: Write "Put x = " clearly. Show fraction calculations step by step. Always simplify your final answer. Write m = -1/6, not -0.167.

Question 30. If x = -3 and x = 2/3 are solutions of quadratic equation mx² + 7x + n = 0, find the values of m and n.
Answer:
Given equation: mx² + 7x + n = 0
Given roots: x = -3 and x = \( \frac{2}{3} \)

Step 1: Put x = -3 in given equation.
m(-3)² + 7(-3) + n = 0
\( \Rightarrow 9m - 21 + n = 0 \)
\( \Rightarrow 9m + n = 21 \) ......(1)

Step 2: Put x = \( \frac{2}{3} \) in given equation.
\( m\left(\frac{2}{3}\right)^2 + 7\left(\frac{2}{3}\right) + n = 0 \)
\( \Rightarrow \frac{4m}{9} + \frac{14}{3} + n = 0 \)
\( \Rightarrow 4m + 42 + 9n = 0 \) (multiplying by 9)
\( \Rightarrow 4m + 9n = -42 \) ......(2)

Step 3: Solve these equations.
From (1): 9m + n = 21
From (2): 4m + 9n = -42

Solving these equations we get:
m = 3 and n = -6
In simple words: We used both roots to make two equations with m and n. Then we solved these two equations to find m = 3 and n = -6.

📝 Teacher's Note: Explain that if we know the roots, we can substitute them to find unknown coefficients. This gives us two equations which we solve simultaneously.

🎯 Exam Tip: Substitute both roots separately to get two equations. Number your equations (1) and (2). Solve the system clearly and write final values: m = 3, n = -6.

Question 31. If quadratic equation x² – (m + 1)x + 6 = 0 has one root as x = 3; find the value of m and the other root of the equation.
Answer:
Given equation: x² – (m + 1)x + 6 = 0
Given: One root is x = 3

Step 1: Put x = 3 in the given equation.
(3)² – (m + 1)(3) + 6 = 0
\( \Rightarrow 9 - 3m - 3 + 6 = 0 \)
\( \Rightarrow -3m = -12 \)
\( \Rightarrow m = 4 \)

Step 2: Put this value of m in the given equation.
x² - 5x + 6 = 0

Step 3: Factorize to find both roots.
\( \Rightarrow x^2 - 3x - 2x + 6 = 0 \)
\( \Rightarrow x(x - 3) - 2(x - 3) = 0 \)
\( \Rightarrow (x - 3)(x - 2) = 0 \)

If x - 3 = 0 or x - 2 = 0
then x = 3 or x = 2

Therefore, m = 4 and the other root is x = 2.
In simple words: We used the given root to find m = 4. Then we put this m back in the equation and solved to find both roots. The other root is 2.

📝 Teacher's Note: Show students how one root helps us find the unknown coefficient. Then we can find the other root by factoring or using the quadratic formula.

🎯 Exam Tip: First find the unknown coefficient using the given root. Then substitute back to get the complete equation. Factorize or use the quadratic formula to find the other root.

Question 32. Given that 2 is a root of the equation 3x² – p(x + 1) = 0 and that the equation px² – qx + 9 = 0 has equal roots, find the values of p and q.
Answer:
Step 1: Since 2 is a root of the equation 3x² – p(x + 1) = 0
\( \Rightarrow 3(2)^2 - p(2 + 1) = 0 \)
\( \Rightarrow 3 \times 4 - 3p = 0 \)
\( \Rightarrow 12 - 3p = 0 \)
\( \Rightarrow 3p = 12 \)
\( \Rightarrow p = 4 \)

Step 2: Now, the other equation becomes 4x² – qx + 9 = 0
Here, a = 4, b = -q and c = 9

Step 3: Since the roots are equal, discriminant = 0
b² - 4ac = 0
\( \Rightarrow (-q)^2 - 4 \times 4 \times 9 = 0 \)
\( \Rightarrow q^2 - 144 = 0 \)
\( \Rightarrow q^2 = 144 \)
\( \Rightarrow q = 12 \) (taking positive value)

Hence, p = 4 and q = 12.
In simple words: We used the first condition to find p = 4. Then we used the second condition that roots are equal to find q = 12.

📝 Teacher's Note: Remind students that equal roots means discriminant = 0. This is a key concept that helps find unknown coefficients.

🎯 Exam Tip: For equal roots, always write "discriminant = 0" and then "b² - 4ac = 0". Show all steps clearly. Take positive value of q unless specified otherwise.

Question 33. Solve: \( \frac{x}{a} - \frac{a+b}{x} = \frac{b(a+b)}{ax} \)
Answer:
Given: \( \frac{x}{a} - \frac{a+b}{x} = \frac{b(a+b)}{ax} \)

Step 1: Multiply both sides by ax.
\( \frac{x^2 - a^2 - ab}{ax} = \frac{ab + b^2}{ax} \)

Step 2: Since denominators are equal, equate numerators.
\( \Rightarrow x^2 - a^2 - ab = ab + b^2 \)

Step 3: Rearrange terms.
\( \Rightarrow x^2 = a^2 + b^2 + 2ab \)
\( \Rightarrow x^2 = (a + b)^2 \)
\( \Rightarrow x = a + b \)

Therefore, x = a + b
In simple words: We cleared the fractions by multiplying both sides by ax. Then we rearranged to get x² = (a+b)². Taking square root gives x = a + b.

📝 Teacher's Note: Show students how to clear fractions by finding the LCM. Remind them that (a+b)² = a² + 2ab + b². Check the answer by substituting back.

🎯 Exam Tip: Always multiply both sides by the LCM to clear fractions. Show each algebraic step clearly. Write the final answer as "x = a + b".

Question 34. Solve: \( \left(\frac{1200}{x} + 2\right)(x - 10) - 1200 = 60 \)
Answer:
Step 1: Simplify the left side.
\( \left(\frac{1200}{x} + 2\right)(x - 10) - 1200 = 60 \)

Step 2: Move 1200 to right side.
\( \Rightarrow 2\left(\frac{600}{x} + 1\right)(x - 10) = 1260 \)

Step 3: Divide by 2.
\( \Rightarrow \left(\frac{600}{x} + 1\right)(x - 10) = 630 \)

Step 4: Simplify the bracket.
\( \Rightarrow \left(\frac{600 + x}{x}\right)(x - 10) = 630 \)

Step 5: Expand.
\( \Rightarrow 600x - 6000 + x^2 - 10x = 630x \)
\( \Rightarrow x^2 - 40x - 6000 = 0 \)
\( \Rightarrow x^2 - 100x + 60x - 6000 = 0 \)
\( \Rightarrow x(x - 100) + 60(x - 100) = 0 \)
\( \Rightarrow (x - 100)(x + 60) = 0 \)

If x - 100 = 0 or x + 60 = 0
\( \Rightarrow x = 100 \) or x = -60

Therefore, x = 100 or x = -60
In simple words: We expanded the brackets step by step and got a quadratic equation. Factoring gave us two solutions: x = 100 or x = -60.

📝 Teacher's Note: Show students how to handle fractions inside brackets. Work step by step and do not rush. Check both answers in the original equation.

🎯 Exam Tip: Expand brackets carefully. Collect like terms properly. Factor the quadratic equation and write both solutions clearly.

Question 35. If -1 and 3 are the roots of x² + px + q = 0, find the values of p and q.
Answer:
Given: Roots are -1 and 3
Equation: x² + px + q = 0

Step 1: Since -1 is a root of x² + px + q = 0, we have
(-1)² + p(-1) + q = 0
\( \Rightarrow 1 - p + q = 0 \)
\( \Rightarrow -p + q = -1 \) ......(i)

Step 2: Since 3 is a root of x² + px + q = 0
\( \Rightarrow (3)^2 + p(3) + q = 0 \)
\( \Rightarrow 9 + 3p + q = 0 \)
\( \Rightarrow 3p + q = -9 \) ......(ii)

Step 3: Solve equations (i) and (ii).
From (ii) - (i):
3p + q - (-p + q) = -9 - (-1)
4p = -8
p = -2

Substituting in (i):
-(-2) + q = -1
2 + q = -1
q = -3

Therefore, p = -2 and q = -3
In simple words: We substituted both roots in the equation to get two equations with p and q. Solving these gave us p = -2 and q = -3.

📝 Teacher's Note: Show students that knowing the roots helps find the coefficients. Substitute each root separately to get two equations, then solve simultaneously.

🎯 Exam Tip: Substitute both roots separately. Number your equations (i) and (ii). Solve the system step by step and write final answer: p = -2, q = -3.

Exercise 5D

Question 1. Solve each of the following equations using the formula:
(i)\( x^2 - 6x = 27 \) (ii)\( x^2 - 10x + 21 = 0 \)
(iii)\( x^2 + 6x - 10 = 0 \) (iv)\( x^2 + 2x - 6 = 0 \)
(v)\( 3x^2 + 2x - 1 = 0 \) (vi)\( 2x^2 + 7x + 5 = 0 \)
(vii)\( \frac{2}{3}x = -\frac{1}{6}x^2 - \frac{1}{3} \) (viii)\( \frac{1}{15}x^2 + \frac{5}{3} = \frac{2}{3}x \)
(ix)\( x^2 - 6 = 2\sqrt{2}x \) (x)\( \frac{4}{x} - 3 = \frac{5}{2x + 3} \)
(xi)\( \frac{2x + 3}{x + 3} = \frac{x + 4}{x + 2} \) (xii)\( \sqrt{6}x^2 - 4x - 2\sqrt{6} = 0 \)
(xiii)\( \frac{2x}{x - 4} + \frac{2x - 5}{x - 3} = 8\frac{1}{3} \) (xiv)\( \frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3\frac{1}{3} \)
Answer:
(i) \( x^2 - 6x = 27 \)
\( \Rightarrow x^2 - 6x - 27 = 0 \)
Here a=1, b=-6 and c=-27
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-27)}}{2(1)} \)
\( = \frac{6 + 12}{2} = \frac{6 + 12}{2} \) and \( \frac{6 - 12}{2} = 9 \) and -3

(ii) \( x^2 - 10x + 21 = 0 \)
Here a=1, b=-10 and c=21
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(21)}}{2(1)} \)
\( = \frac{10 \pm 4}{2} = \frac{10 + 4}{2} \) and \( \frac{10 - 4}{2} = 7 \) and 3

(iii) \( x^2 + 6x - 10 = 0 \)
Here a=1, b=6 and c=-10
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(6) \pm \sqrt{(6)^2 - 4(1)(-10)}}{2(1)} \)
\( = \frac{-6 \pm \sqrt{76}}{2} = \frac{-6 + 2\sqrt{19}}{2} \) and \( \frac{-6 - 2\sqrt{19}}{2} = -3 + \sqrt{19} \) and \( -3 - \sqrt{19} \)

(iv) \( x^2 + 2x - 6 = 0 \)
Here a=1, b=2 and c=-6
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(2) \pm \sqrt{(2)^2 - 4(1)(-6)}}{2(1)} \)
\( = \frac{-2 \pm \sqrt{28}}{2} = \frac{-2 \pm 2\sqrt{7}}{2} = -1 \pm \sqrt{7} \)

(v) \( 3x^2 + 2x - 1 = 0 \)
Here a=3, b=2 and c=-1
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(2) \pm \sqrt{(2)^2 - 4(3)(-1)}}{2(3)} \)
\( = \frac{-2 \pm 4}{6} = \frac{-2 + 4}{6} \) and \( \frac{-2 - 4}{6} = \frac{1}{3} \) and -1

(vi) \( 2x^2 + 7x + 5 = 0 \)
Here a=2, b=7 and c=5
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(7) \pm \sqrt{(7)^2 - 4(2)(5)}}{2(2)} \)
\( = \frac{-7 \pm 3}{4} = \frac{-7 + 3}{4} \) and \( \frac{-7 - 3}{4} = -1 \) and \( -\frac{5}{2} \)

(vii) \( \frac{2}{3}x = -\frac{1}{6}x^2 - \frac{1}{3} \)
\( \Rightarrow 4x = -x^2 - 2 \)
\( \Rightarrow x^2 + 4x + 2 = 0 \)
Here a=1, b=4 and c=2
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(4) \pm \sqrt{(4)^2 - 4(1)(2)}}{2(1)} \)
\( = \frac{-4 \pm \sqrt{8}}{2} = \frac{-4 \pm 2\sqrt{2}}{2} = -2 \pm \sqrt{2} \)

(viii) \( \frac{1}{15}x^2 + \frac{5}{3} = \frac{2}{3}x \)
\( \Rightarrow x^2 + 25 = 10x \)
\( \Rightarrow x^2 - 10x + 25 = 0 \)
Here a=1, b=-10 and c=25
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(25)}}{2(1)} \)
\( = \frac{10 \pm \sqrt{0}}{2} = 5 \)

(ix) \( x^2 - 6 = 2\sqrt{2}x \)
\( \Rightarrow x^2 - 2\sqrt{2}x - 6 = 0 \)
Here a=1, b=\( -2\sqrt{2} \) and c=-6
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(-2\sqrt{2}) \pm \sqrt{(-2\sqrt{2})^2 - 4(1)(-6)}}{2(1)} \)
\( = \frac{2\sqrt{2} \pm \sqrt{32}}{2} = \frac{2\sqrt{2} \pm 4\sqrt{2}}{2} = \frac{2\sqrt{2} + 4\sqrt{2}}{2} \) and \( \frac{2\sqrt{2} - 4\sqrt{2}}{2} \)
\( = \frac{6\sqrt{2}}{2} \) and \( \frac{-2\sqrt{2}}{2} = 3\sqrt{2} \) and \( -\sqrt{2} \)

(x) \( \frac{4}{x} - 3 = \frac{5}{2x + 3} \)
\( \Rightarrow \frac{4 - 3x}{x} = \frac{5}{2x + 3} \)
\( \Rightarrow (4 - 3x)(2x + 3) = 5x \)
\( \Rightarrow 8x + 12 - 6x^2 - 9x = 5x \)
\( \Rightarrow 6x^2 + 6x - 12 = 0 \)
\( \Rightarrow x^2 + x - 2 = 0 \)
Here a=1, b=1 and c=-2
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(-2)}}{2(1)} \)
\( = \frac{-1 \pm \sqrt{9}}{2} = \frac{-1 \pm 3}{2} = \frac{-1 + 3}{2} \) and \( \frac{-1 - 3}{2} = 1 \) and -2

(xi) \( \frac{2x + 3}{x + 3} = \frac{x + 4}{x + 2} \)
\( \Rightarrow (2x + 3)(x + 2) = (x + 3)(x + 4) \)
\( \Rightarrow 2x^2 + 4x + 3x + 6 = x^2 + 4x + 3x + 12 \)
\( \Rightarrow x^2 - 6 = 0 \)
Here a=1, b=0 and c=-6
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(0) \pm \sqrt{(0)^2 - 4(1)(-6)}}{2(1)} \)
\( = \frac{0 \pm \sqrt{24}}{2} = \frac{0 \pm 2\sqrt{6}}{2} = -\sqrt{6} \) and \( \sqrt{6} \)

(xii) \( \sqrt{6}x^2 - 4x - 2\sqrt{6} = 0 \)
Here a=\( \sqrt{6} \), b=-4 and c=\( -2\sqrt{6} \)
Then \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(\sqrt{6})(-2\sqrt{6})}}{2(\sqrt{6})} \)
\( = \frac{4 \pm \sqrt{64}}{2\sqrt{6}} = \frac{4 \pm 8}{2\sqrt{6}} = \frac{4 + 8}{2\sqrt{6}} \) and \( \frac{4 - 8}{2\sqrt{6}} \)
\( = \frac{6}{\sqrt{6}} \) and \( \frac{-2}{\sqrt{6}} = \sqrt{6} \) and \( \frac{-\sqrt{6}}{3} \)

(xiii) \( \frac{2x}{x - 4} + \frac{2x - 5}{x - 3} = 8\frac{1}{3} \)
\( \Rightarrow \frac{2x(x - 3) + (x - 4)(2x - 5)}{(x - 4)(x - 3)} = \frac{25}{3} \)
\( \Rightarrow \frac{2x^2 - 6x + 2x^2 - 5x - 8x + 20}{x^2 - 3x - 4x + 12} = \frac{25}{3} \)
\( \Rightarrow \frac{4x^2 - 19x + 20}{x^2 - 7x + 12} = \frac{25}{3} \)
\( \Rightarrow 25x^2 - 175x + 300 = 12x^2 - 57x + 60 \)
\( \Rightarrow 13x^2 - 118x + 240 = 0 \)
Here a=13, b=-118 and c=240
In simple words: We use the quadratic formula to find the value of x. This means we substitute the values of a, b and c into the formula to find the answer.

📝 Teacher's Note: Show students how to first identify a, b, and c values from the standard form \( ax^2 + bx + c = 0 \). Then carefully substitute these values into the quadratic formula. Common mistakes happen with signs - especially when b or c is negative.

🎯 Exam Tip: Always write "Given" first, then identify a, b, c values clearly. Show all steps of the quadratic formula. Write final answers clearly - you get marks for showing the working even if the final answer is wrong.

Question 2. Solve each of the following equations for x and give, in each case, your answer correct to one decimal place:
(i) \( x^2 - 8x + 5 = 0 \)
(ii) \( 5x^2 + 10x - 3 = 0 \)
Answer:
Solution:

(i) \( x^2 - 8x + 5 = 0 \)
Here \( a = 1 \), \( b = -8 \) and \( c = 5 \)

Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(5)}}{2(1)} \)

\( x = \frac{8 \pm \sqrt{64 - 20}}{2} = \frac{8 \pm \sqrt{44}}{2} \)

\( x = \frac{8 \pm 2\sqrt{11}}{2} = 4 \pm \sqrt{11} \)

\( x = 4 + 3.3 = 7.3 \) and \( x = 4 - 3.3 = 0.7 \)

(ii) \( 5x^2 + 10x - 3 = 0 \)
Here \( a = 5 \), \( b = 10 \) and \( c = -3 \)

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( x = \frac{-10 \pm \sqrt{(10)^2 - 4(5)(-3)}}{2(5)} \)

\( x = \frac{-10 \pm \sqrt{100 + 60}}{10} = \frac{-10 \pm \sqrt{160}}{10} \)

\( x = \frac{-10 \pm 12.6}{10} \)

\( x = \frac{-10 + 12.6}{10} = 0.26 = 0.3 \) and \( x = \frac{-10 - 12.6}{10} = -2.26 = -2.3 \)

In simple words: We use the quadratic formula to find the values of x. The formula gives us two answers for each equation.

📝 Teacher's Note: Make students remember the quadratic formula first. Then show them how to substitute values of a, b, c carefully. Common mistake is wrong signs with negative numbers.

🎯 Exam Tip: Always write "Here a=..., b=..., c=..." first. Show all steps clearly. Round to the correct decimal places as asked in the question.

Question 3(i). Solve each of the following equations for x and give, in each case, your answer correct to two decimal places:
(i) \( 2x^2 - 10x + 5 = 0 \)
Answer:
Solution:

\( 2x^2 - 10x + 5 = 0 \)
Here \( a = 2 \), \( b = -10 \) and \( c = 5 \)

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(2)(5)}}{2(2)} \)

\( x = \frac{10 \pm \sqrt{100 - 40}}{4} = \frac{10 \pm \sqrt{60}}{4} \)

\( x = \frac{10 \pm 7.75}{4} \)

\( x = \frac{10 + 7.75}{4} = 4.44 \) and \( x = \frac{10 - 7.75}{4} = 0.56 \)

In simple words: We solve this quadratic equation using the standard formula. The equation has two solutions that we round to two decimal places.

📝 Teacher's Note: Show students how to calculate square roots step by step. Use a calculator for \( \sqrt{60} \) but explain the process clearly.

🎯 Exam Tip: Write your final answer correct to two decimal places as asked. Show the rounding clearly - 4.44 and 0.56.

Question 3(ii). Solve each of the following equations for x and give, in each case, your answer correct to two decimal places:
\( 4x + \frac{6}{x} + 13 = 0 \)
Answer:
Solution:

\( 4x + \frac{6}{x} + 13 = 0 \)

Multiplying by x: \( 4x^2 + 6 + 13x = 0 \)

Rearranging: \( 4x^2 + 13x + 6 = 0 \)

Here \( a = 4 \), \( b = 13 \) and \( c = 6 \)

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( x = \frac{-13 \pm \sqrt{(13)^2 - 4(4)(6)}}{2(4)} \)

\( x = \frac{-13 \pm \sqrt{169 - 96}}{8} = \frac{-13 \pm \sqrt{73}}{8} \)

\( x = \frac{-13 \pm 8.54}{8} \)

\( x = \frac{-13 + 8.54}{8} = -0.56 \) and \( x = \frac{-13 - 8.54}{8} = -2.69 \)

In simple words: First we remove the fraction by multiplying the whole equation by x. Then we solve like a normal quadratic equation.

📝 Teacher's Note: Explain that when we have fractions with x in denominator, we multiply by x to clear the fraction. This is a key step students often miss.

🎯 Exam Tip: Always clear fractions first by multiplying. Then rearrange to standard form \( ax^2 + bx + c = 0 \) before applying the formula.

Question 3(iii). Solve each of the following equations for x and give, in each case, your answer correct to two decimal places:
\( x^2 - 3x - 9 = 0 \)
Answer:
Solution:

\( x^2 - 3x - 9 = 0 \)
Here \( a = 1 \), \( b = -3 \) and \( c = -9 \)

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-9)}}{2(1)} \)

\( x = \frac{3 \pm \sqrt{9 + 36}}{2} = \frac{3 \pm \sqrt{45}}{2} \)

\( x = \frac{3 \pm 6.70}{2} \)

\( x = \frac{3 + 6.70}{2} = 4.85 \) and \( x = \frac{3 - 6.70}{2} = -1.85 \)

In simple words: This is a standard quadratic equation. We use the formula and calculate carefully to get two decimal places.

📝 Teacher's Note: Point out that c is negative here, so -4ac becomes +36. Students often make sign errors in this step.

🎯 Exam Tip: Be very careful with signs. When c is negative, -4ac becomes positive. Final answers: 4.85 and -1.85.

Question 3(iv). Solve each of the following equations for x and give, in each case, your answer correct to two decimal places:
\( x^2 - 5x - 10 = 0 \)
Answer:
Solution:

\( x^2 - 5x - 10 = 0 \)
Here \( a = 1 \), \( b = -5 \) and \( c = -10 \)

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-10)}}{2(1)} \)

\( x = \frac{5 \pm \sqrt{25 + 40}}{2} = \frac{5 \pm \sqrt{65}}{2} \)

\( x = \frac{5 \pm 8.06}{2} \)

\( x = \frac{5 + 8.06}{2} = 6.53 \) and \( x = \frac{5 - 8.06}{2} = -1.53 \)

In simple words: Another quadratic equation solved using the same method. We get one positive and one negative answer.

📝 Teacher's Note: Show how \( \sqrt{65} \) is approximately 8.06. Students should know how to use calculator for square roots in exams.

🎯 Exam Tip: Final answers to two decimal places are: x = 6.53 and x = -1.53. Always give both roots unless told otherwise.

Question 4. Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places:
(i) \( 3x^2 - 12x - 1 = 0 \)
(ii) \( x^2 - 16x + 6 = 0 \)
(iii) \( 2x^2 + 11x + 4 = 0 \)
Answer:
Solution:

(i) \( 3x^2 - 12x - 1 = 0 \)
Here \( a = 3 \), \( b = -12 \) and \( c = -1 \)

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(-1)}}{2(3)} \)

\( x = \frac{12 \pm \sqrt{144 + 12}}{6} = \frac{12 \pm \sqrt{156}}{6} \)

\( x = \frac{12 \pm 12.489}{6} \)

\( x = \frac{12 + 12.489}{6} = 4.082 \) and \( x = \frac{12 - 12.489}{6} = -0.082 \)

(ii) \( x^2 - 16x + 6 = 0 \)
Here \( a = 1 \), \( b = -16 \) and \( c = 6 \)

\( x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(6)}}{2(1)} \)

\( x = \frac{16 \pm \sqrt{256 - 24}}{2} = \frac{16 \pm \sqrt{232}}{2} \)

\( x = \frac{16 \pm 15.231}{2} \)

\( x = \frac{16 + 15.231}{2} = 15.616 \) and \( x = \frac{16 - 15.231}{2} = 0.384 \)

(iii) \( 2x^2 + 11x + 4 = 0 \)
Here \( a = 2 \), \( b = 11 \) and \( c = 4 \)

\( x = \frac{-11 \pm \sqrt{(11)^2 - 4(2)(4)}}{2(2)} \)

\( x = \frac{-11 \pm \sqrt{121 - 32}}{4} = \frac{-11 \pm \sqrt{89}}{4} \)

\( x = \frac{-11 \pm 9.433}{4} \)

\( x = \frac{-11 + 9.433}{4} = -0.392 \) and \( x = \frac{-11 - 9.433}{4} = -5.108 \)

In simple words: These are three quadratic equations solved the same way. The only difference is we need three decimal places this time instead of two.

📝 Teacher's Note: Emphasize the importance of precision in calculations when asked for 3 decimal places. Use calculator carefully and round only at the final step.

🎯 Exam Tip: Be extra careful with calculations for 3 decimal places. Do not round intermediate steps - only round the final answer. Show all working clearly.

Question 5. Solve:
(i) \( x^4 - 2x^2 - 3 = 0 \)
(ii) \( x^4 - 10x^2 + 9 = 0 \)
Answer:
(i) \( x^4 - 2x^2 - 3 = 0 \)
Let \( y = x^2 \)
\( \implies y^2 - 2y - 3 = 0 \)
\( \implies y^2(y^2 - 3) + 1(y^2 - 3) = 0 \)
\( \implies (y^2 - 3)(y^2 + 1) = 0 \)
If \( y^2 - 3 = 0 \) or \( y^2 + 1 = 0 \)
\( \implies y^2 = 3 \) or \( y^2 = -1 \) (reject)
\( \implies y = \pm\sqrt{3} \)
Since \( y = x^2 \) and \( y = \sqrt{3} \)
\( \implies x^2 = 3 \)
\( \implies x = \pm\sqrt{3} \)

(ii) \( x^4 - 10x^2 + 9 = 0 \)
Let \( y = x^2 \)
\( \implies y^2 - 10y + 9 = 0 \)
\( \implies y^2(y^2 - 9) - 1(y^2 - 9) = 0 \)
\( \implies (y^2 - 9)(y^2 - 1) = 0 \)
If \( y^2 - 9 = 0 \) or \( y^2 - 1 = 0 \)
\( \implies y^2 = 9 \) or \( y^2 = 1 \)
\( \implies y = \pm 3 \) or \( y = \pm 1 \)
Since \( y = x^2 \geq 0 \), we take \( y = 3 \) or \( y = 1 \)
\( \implies x^2 = 3 \) or \( x^2 = 1 \)
\( \implies x = \pm\sqrt{3} \) or \( x = \pm 1 \)

In simple words: These are biquadratic equations. We substitute \( x^2 = y \) to make them simple quadratic equations. Then solve for y and find x values.

📝 Teacher's Note: Teach students to spot when \( x^4 \) and \( x^2 \) appear together. Tell them to always substitute \( x^2 = y \). This makes hard equations easy to solve.

🎯 Exam Tip: Always write "Let \( y = x^2 \)" clearly at the start. Show all steps. Remember to reject negative values of y since \( x^2 \) cannot be negative.

Question 6. Solve:
(i) \( (x^2 - x)^2 + 5(x^2 - x) + 4 = 0 \)
(ii) \( (x^2 - 3x)^2 - 16(x^2 - 3x) - 36 = 0 \)
Answer:
(i) \( (x^2 - x)^2 + 5(x^2 - x) + 4 = 0 \)
Let \( x^2 - x = y \)
Then \( y^2 + 5y + 4 = 0 \)
\( \implies y^2 + 4y + y + 4 = 0 \)
\( \implies y(y + 4) + 1(y + 4) = 0 \)
\( \implies (y + 4)(y + 1) = 0 \)
If \( y + 4 = 0 \) or \( y + 1 = 0 \)
\( \implies x^2 - x + 4 = 0 \) or \( x^2 - x + 1 = 0 \)
For \( x^2 - x + 4 = 0 \):
\( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(4)}}{2(1)} = \frac{1 \pm \sqrt{-15}}{2} \) (reject)
For \( x^2 - x + 1 = 0 \):
\( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(1)}}{2(1)} = \frac{1 \pm \sqrt{-3}}{2} \) (reject)
Given equation has no real solution.

(ii) \( (x^2 - 3x)^2 - 16(x^2 - 3x) - 36 = 0 \)
Let \( x^2 - 3x = y \)
Then \( y^2 - 16y - 36 = 0 \)
\( \implies y^2 - 18y + 2y - 36 = 0 \)
\( \implies y(y - 18) + 2(y - 18) = 0 \)
\( \implies (y - 18)(y + 2) = 0 \)
If \( y - 18 = 0 \) or \( y + 2 = 0 \)
\( \implies x^2 - 3x - 18 = 0 \) or \( x^2 - 3x + 2 = 0 \)
\( \implies x^2 - 6x + 3x - 18 = 0 \) or \( x^2 - 2x - x + 2 = 0 \)
\( \implies x(x - 6) + 3(x - 6) = 0 \) or \( x(x - 2) - 1(x - 2) = 0 \)
\( \implies (x - 6)(x + 3) = 0 \) or \( (x - 2)(x - 1) = 0 \)
If \( x - 6 = 0 \) or \( x + 3 = 0 \) or \( x - 2 = 0 \) or \( x - 1 = 0 \)
then \( x = 6 \) or \( x = -3 \) or \( x = 2 \) or \( x = 1 \)

In simple words: When we see a repeated expression in brackets, we substitute it with a new variable. This makes the equation simple to solve.

📝 Teacher's Note: Show students how to spot repeated expressions like \( (x^2 - x) \). Tell them to always substitute the whole bracket with one letter. This trick works for many complex equations.

🎯 Exam Tip: Write "Let \( (expression) = y \)" clearly. After finding y values, substitute back to find x. Check if your answers are real numbers using the discriminant formula.

Question 7. Solve:
(i) \( \sqrt{\frac{x}{x-3}} + \sqrt{\frac{x-3}{x}} = \frac{5}{2} \)
(ii) \( \left(\frac{2x-3}{x-1}\right) - 4\left(\frac{x-1}{2x-3}\right) = 3 \)
(iii) \( \left(\frac{3x+1}{x+1}\right) + \left(\frac{x+1}{3x+1}\right) = \frac{5}{2} \)
Answer:
(i) \( \sqrt{\frac{x}{x-3}} + \sqrt{\frac{x-3}{x}} = \frac{5}{2} \)
Let \( \sqrt{\frac{x}{x-3}} = y \)
Then \( y + \frac{1}{y} = \frac{5}{2} \)
\( \implies \frac{y^2 + 1}{y} = \frac{5}{2} \)
\( \implies 2y^2 + 2 = 5y \)
\( \implies 2y^2 - 5y + 2 = 0 \)
\( \implies 2y^2 - 4y - y + 2 = 0 \)
\( \implies 2y(y - 2) - 1(y - 2) = 0 \)
\( \implies (y - 2)(2y - 1) = 0 \)
If \( y - 2 = 0 \) or \( 2y - 1 = 0 \)
then \( y = 2 \) or \( y = \frac{1}{2} \)
\( \implies \sqrt{\frac{x}{x-3}} = 2 \) or \( \sqrt{\frac{x}{x-3}} = \frac{1}{2} \)
\( \implies \frac{x}{x-3} = 4 \) or \( \frac{x}{x-3} = \frac{1}{4} \)
\( \implies x = 4 \) or \( x = -1 \)

(ii) \( \left(\frac{2x-3}{x-1}\right) - 4\left(\frac{x-1}{2x-3}\right) = 3 \)
Let \( \frac{2x-3}{x-1} = y \)
then \( y - \frac{4}{y} = 3 \)
\( \implies \frac{y^2 - 4}{y} = 3 \)
\( \implies y^2 - 4 = 3y \)
\( \implies y^2 - 3y - 4 = 0 \)
\( \implies y^2 - 4y + y - 4 = 0 \)
\( \implies y(y - 4) + 1(y - 4) = 0 \)
\( \implies (y - 4)(y + 1) = 0 \)
If \( y - 4 = 0 \) or \( y + 1 = 0 \)
then \( y = 4 \) or \( y = -1 \)
\( \implies \frac{2x-3}{x-1} = 4 \) or \( \frac{2x-3}{x-1} = -1 \)
\( \implies 4x - 4 = 2x - 3 \) or \( 2x - 3 = -x + 1 \)
\( \implies 2x = 1 \) or \( 3x = 4 \)
\( \implies x = \frac{1}{2} \) or \( x = \frac{4}{3} = 1\frac{1}{3} \)

(iii) \( \left(\frac{3x+1}{x+1}\right) + \left(\frac{x+1}{3x+1}\right) = \frac{5}{2} \)
Let \( \frac{3x+1}{x+1} = y \)
then \( y + \frac{1}{y} = \frac{5}{2} \)
\( \implies \frac{y^2 + 1}{y} = \frac{5}{2} \)
\( \implies 2y^2 + 2 = 5y \)
\( \implies 2y^2 - 5y + 2 = 0 \)
\( \implies 2y(y - 2) - 1(y - 2) = 0 \)
\( \implies (y - 2)(2y - 1) = 0 \)
If \( y - 2 = 0 \) or \( 2y - 1 = 0 \)
then \( y = 2 \) or \( y = \frac{1}{2} \)
\( \implies \frac{3x+1}{x+1} = 2 \) or \( \frac{3x+1}{x+1} = \frac{1}{2} \)
\( \implies 3x + 1 = 2x + 2 \) or \( 6x + 2 = x + 1 \)
\( \implies x = 1 \) or \( 5x = -1 \)
\( \implies x = 1 \) or \( x = -\frac{1}{5} \)

In simple words: When we see fractions that are reciprocals of each other, we substitute one with y. This creates simple equations with y that we can solve easily.

📝 Teacher's Note: Teach students to look for reciprocal patterns like \( \frac{a}{b} + \frac{b}{a} \). Show them that if \( \frac{a}{b} = y \), then \( \frac{b}{a} = \frac{1}{y} \). This makes complex fractions simple.

🎯 Exam Tip: Always check your final answers by putting them back in the original equation. For square root equations, make sure the values under the square root are positive.

Question 8. Solve the equation \( 2x - \frac{1}{x} = 7 \). Write your answer correct to two decimal places.
Answer:
\( 2x - \frac{1}{x} = 7 \)
\( \implies \frac{2x^2 - 1}{x} = 7 \)
\( \implies 2x^2 - 1 = 7x \)
\( \implies 2x^2 - 7x - 1 = 0 \)
Here a = 2, b = -7 and c = -1
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(-1)}}{2(2)} \)
\( = \frac{7 \pm \sqrt{57}}{4} = \frac{7 \pm 7.55}{4} \)
\( = \frac{7 + 7.55}{4} \) and \( \frac{7 - 7.55}{4} = 3.64 \) and \( -0.14 \)

In simple words: We multiplied both sides by x to remove the fraction. Then we used the quadratic formula to find the exact values. Finally we rounded to 2 decimal places.

📝 Teacher's Note: Show students how to clear fractions by multiplying through by x. Remind them to always use the quadratic formula when factoring is difficult. Practice rounding to decimal places.

🎯 Exam Tip: Write "correct to 2 decimal places" next to your final answer. Use a calculator for \( \sqrt{57} \). Show the quadratic formula clearly to get method marks.

Question 9. Solve the following equation and give your answer correct to 3 significant figures:
Answer:
Note: The specific equation for Question 9 is not shown in the provided image.

In simple words: This question asks us to solve an equation and round the answer to 3 significant figures. This means we keep only the first 3 meaningful digits.

📝 Teacher's Note: Teach students what significant figures mean. The first non-zero digit starts counting. For example, 0.00123 has 3 significant figures: 1, 2, and 3.

🎯 Exam Tip: Always write "correct to 3 significant figures" next to your final answer. Count carefully from the first non-zero digit. Use a calculator for accuracy.

Question 9. Solve 5x² – 3x – 4 = 0
Answer:
Given: \( 5x^2 - 3x - 4 = 0 \)

Step 1: Use quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Here, \( a = 5 \), \( b = -3 \), \( c = -4 \)

Step 2: Substitute values in formula
\( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 5 \times (-4)}}{2 \times 5} \)

Step 3: Simplify
\( x = \frac{3 \pm \sqrt{9 + 80}}{10} \)
\( x = \frac{3 \pm \sqrt{89}}{10} \)
\( x = \frac{3 \pm 9.434}{10} \)

Step 4: Find both solutions
\( x = \frac{3 + 9.434}{10} = 1.243 \) or \( x = \frac{3 - 9.434}{10} = -0.643 \)

Therefore: \( x = 1.243 \) or \( x = -0.643 \)
In simple words: We used the quadratic formula to find two values of x. We put the numbers from the equation into the formula and calculated step by step.

📝 Teacher's Note: Make students write down a, b, c values first. Common mistake is wrong signs - check that b = -3 (not +3) and c = -4. Show how to calculate the discriminant step by step.

🎯 Exam Tip: Always write "Given", "To find", and show the quadratic formula. Calculate √89 carefully. Write final answers clearly with "or" between them.

Question 10. Solve for x using the quadratic formula. Write your answer correct to two significant figures. (x – 1)² – 3x + 4 = 0
Answer:
Given: \( (x - 1)^2 - 3x + 4 = 0 \)

Step 1: Expand the equation
\( (x - 1)^2 - 3x + 4 = 0 \)
\( x^2 - 2x + 1 - 3x + 4 = 0 \)
\( x^2 - 5x + 5 = 0 \)

Step 2: Identify coefficients
Here, \( a = 1 \), \( b = -5 \), and \( c = 5 \)

Step 3: Apply quadratic formula
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 1 \times 5}}{2 \times 1} \)
\( x = \frac{5 \pm \sqrt{25 - 20}}{2} \)
\( x = \frac{5 \pm \sqrt{5}}{2} \)
\( x = \frac{5 \pm 2.24}{2} \)

Step 4: Find both solutions
\( x = \frac{5 + 2.24}{2} = 3.6 \) or \( x = \frac{5 - 2.24}{2} = 1.4 \)

Therefore: \( x = 3.6 \) or \( x = 1.4 \) (to two significant figures)
In simple words: First we opened the brackets and made a simple quadratic equation. Then we used the formula to find x. We rounded the answers to two significant figures as asked.

📝 Teacher's Note: Students often forget to expand (x-1)² correctly. Show that (x-1)² = x² - 2x + 1. Make them practice expanding before using the formula.

🎯 Exam Tip: Always expand first, then identify a, b, c. Write answers to exactly two significant figures as asked - 3.6 and 1.4, not 3.62 or 1.38.

Question 11. Solve the quadratic equation x² – 3(x+3) = 0; Give your answer correct to two significant figures.
Answer:
Given: \( x^2 - 3(x + 3) = 0 \)

Step 1: Expand the equation
\( x^2 - 3(x + 3) = 0 \)
\( x^2 - 3x - 9 = 0 \)

Step 2: Identify coefficients
Comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 1 \), \( b = -3 \), \( c = -9 \)

Step 3: Apply quadratic formula
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-9)}}{2(1)} \)
\( x = \frac{3 \pm \sqrt{9 - 36}}{2} \)
\( x = \frac{3 \pm \sqrt{45}}{2} \)
\( x = \frac{3 \pm \sqrt{9 \times 5}}{2} \)
\( x = \frac{3 \pm 3\sqrt{5}}{2} \)

Step 4: Calculate numerical values
\( \sqrt{5} = 2.236 \)
\( x = \frac{3 + 3(2.236)}{2} = \frac{3 + 6.708}{2} = 4.854 = 4.9 \)
\( x = \frac{3 - 3(2.236)}{2} = \frac{3 - 6.708}{2} = -1.854 = -1.9 \)

Therefore: \( x = 4.9 \) or \( x = -1.9 \) (to two significant figures)
In simple words: We first expanded the brackets to get a normal quadratic equation. Then we used the quadratic formula and rounded our answers to two significant figures.

📝 Teacher's Note: Show students how to expand -3(x+3) carefully. The discriminant is positive so we get real roots. Emphasize rounding to exactly two significant figures.

🎯 Exam Tip: Expand the brackets first: -3(x+3) = -3x - 9. Calculate √45 = 3√5 = 6.708. Round final answers to two significant figures: 4.9 and -1.9.

Exercise 5E

Question 1. Solve: \( \frac{2x}{x-3} + \frac{1}{2x+3} + \frac{3x+9}{(x-3)(2x+3)} = 0 \); \( x ≠ 3, x ≠ -\frac{3}{2} \)
Answer:
Given: \( \frac{2x}{x-3} + \frac{1}{2x+3} + \frac{3x+9}{(x-3)(2x+3)} = 0 \)

Step 1: Take LCM of denominators
LCM = \( (x-3)(2x+3) \)

Step 2: Multiply each term to get common denominator
\( \frac{2x(2x+3) + 1(x-3) + 3x+9}{(x-3)(2x+3)} = 0 \)

Step 3: Expand numerator
\( 4x^2 + 6x - x - 3 - 3x - 9 = 0 \)
\( 4x^2 + 10x + 6 = 0 \)

Step 4: Divide by 2
\( 4x^2 + 4x + 6x - 6 = 0 \)
\( 4x(x + 1) - 6(x + 1) = 0 \)
\( (x + 1)(4x - 6) = 0 \)

Step 5: Solve for x
\( x + 1 = 0 \) or \( 4x + 6 = 0 \)
\( x = -1 \) or \( x = -\frac{6}{4} = -\frac{3}{2} \) (reject)

Since \( x ≠ -\frac{3}{2} \), we reject this solution.

Therefore: \( x = -1 \)
In simple words: We combined all fractions into one by finding a common bottom part. Then we solved the equation that we got. We rejected one answer because it makes the original equation undefined.

📝 Teacher's Note: Show students why we reject x = -3/2 - it makes the original denominator zero. Always check solutions in rational equations. Common mistake is forgetting to reject invalid solutions.

🎯 Exam Tip: Write the restrictions clearly at the start. Always check your final answers - they should not make any denominator zero. Write "reject" next to invalid solutions.

Question 2. Solve: (2x+3)² = 81
Answer:
Given: \( (2x + 3)^2 = 81 \)

Step 1: Take square root of both sides
\( 2x + 3 = \pm 9 \)

Step 2: Solve both cases
Case 1: \( 2x + 3 = 9 \)
\( 2x = 6 \)
\( x = 3 \)

Case 2: \( 2x + 3 = -9 \)
\( 2x = -12 \)
\( x = -6 \)

Therefore: \( x = 3 \) and \( x = -6 \)
In simple words: We took the square root of both sides. Since square root gives plus or minus, we get two different values of x.

📝 Teacher's Note: Remind students that √81 = ±9, not just +9. This is a common mistake. Show why we get two answers when taking square roots.

🎯 Exam Tip: Always write ±9 when taking square root. Solve both the positive and negative cases separately. Check your answers by substituting back.

Question 3. Solve: a²x² – b² = 0
Answer:
Given: \( a^2x^2 - b^2 = 0 \)

Step 1: Rearrange equation
\( a^2x^2 = b^2 \)
\( (ax)^2 - b^2 = 0 \)

Step 2: Factor using difference of squares
\( (ax + b)(ax - b) = 0 \)

Step 3: Apply zero product property
If \( ax + b = 0 \) and \( ax - b = 0 \)
then \( x = -\frac{b}{a} \) and \( x = \frac{b}{a} \)

Therefore: \( x = \frac{b}{a} \) or \( x = -\frac{b}{a} \)
In simple words: We used the difference of squares pattern (A² - B² = (A+B)(A-B)) to factor the equation. Then we solved each factor equal to zero.

📝 Teacher's Note: Teach the difference of squares pattern first: A² - B² = (A+B)(A-B). Students should recognize this pattern quickly. Here A = ax and B = b.

🎯 Exam Tip: Identify this as difference of squares immediately. Write the factored form clearly. Remember both positive and negative solutions.

Question 4. Solve: \( x^2 - \frac{11}{4}x + \frac{15}{8} = 0 \)
Answer:
Given: \( x^2 - \frac{11}{4}x + \frac{15}{8} = 0 \)

Step 1: Multiply entire equation by 8 to clear fractions
\( 8x^2 - 22x + 15 = 0 \)

Step 2: Factor the quadratic
\( 8x^2 - 12x - 10x + 15 = 0 \)
\( 4x(2x - 3) - 5(2x - 3) = 0 \)
\( (2x - 3)(4x - 5) = 0 \)

Step 3: Solve each factor
\( 2x - 3 = 0 \) or \( 4x - 5 = 0 \)
\( x = \frac{3}{2} \) or \( x = \frac{5}{4} \)

Therefore: \( x = \frac{3}{2} \) or \( x = \frac{5}{4} \)
In simple words: We cleared the fractions by multiplying everything by 8. Then we factored the quadratic equation and solved each part equal to zero.

📝 Teacher's Note: Show students how to multiply through by the LCM (8) to clear fractions. This makes factoring much easier. Practice breaking the middle term for factoring.

🎯 Exam Tip: Always clear fractions first by multiplying by LCM. Factor carefully by breaking the middle term. Check answers by substituting back into original equation.

Question 5. Solve: \( x + \frac{4}{x} = -4 \); \( x ≠ 0 \)
Answer:
Given: \( x + \frac{4}{x} = -4 \)

Step 1: Multiply both sides by x
\( x \cdot x + \frac{4}{x} \cdot x = -4 \cdot x \)
\( x^2 + 4 = -4x \)

Step 2: Rearrange to standard form
\( x^2 + 4x + 4 = 0 \)

Step 3: Factor the perfect square
\( (x + 2)^2 = 0 \)

Step 4: Solve for x
\( x + 2 = 0 \)
\( x = -2 \)

Therefore: \( x = -2 \)
In simple words: We multiplied both sides by x to get rid of the fraction. Then we rearranged and found that this makes a perfect square, so there is only one answer.

📝 Teacher's Note: This gives a perfect square trinomial. Students should recognize x² + 4x + 4 = (x + 2)². Show why we get only one solution (repeated root).

🎯 Exam Tip: Multiply by x to clear the fraction. Check that x² + 4x + 4 is a perfect square. Write that x = -2 is a double root.

Question 6. Solve: 2x⁴ – 5x² + 3 = 0
Answer:
Given: \( 2x^4 - 5x^2 + 3 = 0 \)

Step 1: Substitute \( y = x^2 \)
\( 2y^2 - 5y + 3 = 0 \)

Step 2: Factor the quadratic in y
\( 2y^2 - 3y - 2y + 3 = 0 \)
\( y(2y - 3) - 1(2y - 3) = 0 \)
\( (2y - 3)(y - 1) = 0 \)

Step 3: Solve for y
If \( 2y - 3 = 0 \) or \( y - 1 = 0 \)
then \( y = \frac{3}{2} \) or \( y = 1 \)

Step 4: Substitute back \( y = x^2 \)
If \( x^2 = \frac{3}{2} \) or \( x^2 = 1 \)
then \( x = \pm \sqrt{\frac{3}{2}} \) or \( x = \pm 1 \)

Therefore: \( x = \pm \sqrt{\frac{3}{2}} \) or \( x = \pm 1 \)
In simple words: This is a quartic equation (power 4). We made it easier by substituting y = x². Then we solved the quadratic and found the x values.

📝 Teacher's Note: This is called a bi-quadratic equation. The key trick is substitution: let y = x². Students often forget to find all four values of x at the end.

🎯 Exam Tip: Always substitute y = x² for bi-quadratic equations. Factor the quadratic in y. Remember each positive y value gives two x values: ±√y.

Question 7. Solve: \( x^4 - 2x^2 - 3 = 0 \)
Answer:
Step 1: Start with the given equation.
\( x^4 - 2x^2 - 3 = 0 \)

Step 2: Add and subtract \( 3x^2 \) to group terms.
\( \implies x^4 - 3x^2 + x^2 - 3 = 0 \)

Step 3: Factor by grouping.
\( \implies x^2(x^2 - 3) + 1(x^2 - 3) = 0 \)
\( \implies (x^2 - 3)(x^2 + 1) = 0 \)

Step 4: Solve each factor separately.
If \( x^2 - 3 = 0 \) or \( x^2 + 1 = 0 \)
then \( x^2 = 3 \) or \( x^2 = -1 \) (reject)

Step 5: Find the values of x.
\( \implies x = \pm\sqrt{3} \)

Therefore, \( x = \sqrt{3} \) or \( x = -\sqrt{3} \)
In simple words: We treated this like a quadratic equation by replacing \( x^2 \) with a new variable. Then we factored and solved to get two answers.

📝 Teacher's Note: Show students that \( x^4 \) problems can be solved like \( x^2 \) problems. Replace \( x^2 \) with y, solve for y, then find x. This trick works for many higher power equations.

🎯 Exam Tip: Always reject negative square roots like \( x^2 = -1 \). Write "reject" clearly. Only real number solutions are wanted unless the question asks for complex numbers.

Question 8. Solve: \( 9\left(x^2 + \frac{1}{x^2}\right) - 9\left(x + \frac{1}{x}\right) - 52 = 0 \)
Answer:
Step 1: Write the equation.
\( 9\left(x^2 + \frac{1}{x^2}\right) - 9\left(x + \frac{1}{x}\right) - 52 = 0 \)

Step 2: Let \( x + \frac{1}{x} = y \)
Squaring both sides:
\( x^2 + \frac{1}{x^2} + 2 = y^2 \)
\( \implies x^2 + \frac{1}{x^2} = y^2 - 2 \)

Step 3: Put these values in the given equation.
\( 9(y^2 - 2) - 9y - 52 = 0 \)
\( \implies 9y^2 - 18 - 9y - 52 = 0 \)
\( \implies 9y^2 - 9y - 70 = 0 \)

Step 4: Factor the quadratic.
\( \implies 9y^2 - 30y + 21y - 70 = 0 \)
\( \implies 3y(3y - 10) + 7(3y - 10) = 0 \)
\( \implies (3y - 10)(3y + 7) = 0 \)

Step 5: Solve for y.
\( 3y - 10 = 0 \) or \( 3y + 7 = 0 \)
\( y = \frac{10}{3} \) or \( y = \frac{-7}{3} \)

Step 6: Find x values for each y.
For \( y = \frac{10}{3} \): \( x + \frac{1}{x} = \frac{10}{3} \)
\( \implies \frac{x^2 + 1}{x} = \frac{10}{3} \)
\( \implies 3x^2 - 10x + 3 = 0 \)
\( \implies 3x(x - 3) - 1(x - 3) = 0 \)
\( \implies (x - 3)(3x - 1) = 0 \)
\( \implies x = 3 \) and \( x = \frac{1}{3} \)

For \( y = \frac{-7}{3} \): \( x + \frac{1}{x} = \frac{-7}{3} \)
\( \implies 3x^2 + 7x + 3 = 0 \)
Using quadratic formula: \( x = \frac{-7 \pm \sqrt{(-7)^2 - 4(3)(3)}}{2(3)} = \frac{-7 \pm \sqrt{13}}{6} \)

Therefore, \( x = 3, \frac{1}{3}, \frac{-7 + \sqrt{13}}{6}, \frac{-7 - \sqrt{13}}{6} \)
In simple words: We used substitution to change a complex equation into a simple quadratic. Then we solved step by step to find all four values of x.

📝 Teacher's Note: Teach students to recognize when substitution helps. When you see \( x + \frac{1}{x} \) and \( x^2 + \frac{1}{x^2} \), use the identity \( (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 \).

🎯 Exam Tip: Always check your answers by substituting back into the original equation. Write all four solutions clearly. Show the substitution step clearly to get full marks.

Question 9. Solve: \( 2\left(x^2 + \frac{1}{x^2}\right) - \left(x + \frac{1}{x}\right) = 11 \)
Answer:
Step 1: Write the equation.
\( 2\left(x^2 + \frac{1}{x^2}\right) - \left(x + \frac{1}{x}\right) = 11 \)

Step 2: Let \( x + \frac{1}{x} = y \)
Squaring both sides:
\( x^2 + \frac{1}{x^2} + 2 = y^2 \)
\( \implies x^2 + \frac{1}{x^2} = y^2 - 2 \)

Step 3: Put these values in the given equation.
\( 2(y^2 - 2) - y = 11 \)
\( \implies 2y^2 - 4 - y - 11 = 0 \)
\( \implies 2y^2 - y - 15 = 0 \)

Step 4: Factor the quadratic.
\( \implies 2y^2 - 6y + 5y - 15 = 0 \)
\( \implies 2y(y - 3) + 5(y - 3) = 0 \)
\( \implies (y - 3)(2y + 5) = 0 \)

Step 5: Solve for y.
If \( y - 3 = 0 \) or \( 2y + 5 = 0 \)
then \( y = 3 \) or \( y = \frac{-5}{2} \)

Step 6: Find x values for each y.
For \( y = 3 \): \( x + \frac{1}{x} = 3 \)
\( \implies \frac{x^2 + 1}{x} = 3 \)
\( \implies x^2 - 3x + 1 = 0 \)
Using quadratic formula: \( x = \frac{-3 \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{5}}{2} \)

For \( y = \frac{-5}{2} \): \( x + \frac{1}{x} = \frac{-5}{2} \)
\( \implies \frac{x^2 + 1}{x} = \frac{-5}{2} \)
\( \implies 2x^2 + 5x + 2 = 0 \)
\( \implies 2x^2 + 4x + x + 2 = 0 \)
\( \implies 2x(x + 2) + 1(x + 2) = 0 \)
\( \implies (x + 2)(2x + 1) = 0 \)
then \( x = -2 \) and \( x = \frac{-1}{2} \)

Therefore, \( x = \frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}, -2, \frac{-1}{2} \)
In simple words: We used the same substitution method as the previous question. We found two values of y, then solved for x from each y value.

📝 Teacher's Note: This is similar to Question 8. Practice these substitution problems together so students see the pattern. The key is recognizing the relationship between \( x + \frac{1}{x} \) and \( x^2 + \frac{1}{x^2} \).

🎯 Exam Tip: Write the substitution step clearly. Show that \( (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 \). This formula is often tested, so memorize it.

Question 10. Solve: \( \left(x^2 + \frac{1}{x^2}\right) - 3\left(x - \frac{1}{x}\right) - 2 = 0 \)
Answer:
Step 1: Write the equation.
\( \left(x^2 + \frac{1}{x^2}\right) - 3\left(x - \frac{1}{x}\right) - 2 = 0 \)

Step 2: Let \( x - \frac{1}{x} = y \)
Squaring both sides:
\( x^2 + \frac{1}{x^2} - 2 = y^2 \)
\( \implies x^2 + \frac{1}{x^2} = y^2 + 2 \)

Step 3: Put these values in the given equation.
\( (y^2 + 2) - 3y - 2 = 0 \)
\( \implies y^2 - 3y = 0 \)
\( \implies y(y - 3) = 0 \)

Step 4: Solve for y.
If \( y = 0 \) or \( y - 3 = 0 \)
then \( y = 0 \) or \( y = 3 \)

Step 5: Find x values for each y.
For \( y = 0 \): \( x - \frac{1}{x} = 0 \)
\( \implies \frac{x^2 - 1}{x} = 0 \)
\( \implies x^2 - 1 = 0 \)
\( \implies (x - 1)(x - 1) = 0 \)
\( \implies x = -1 \) and \( x = 1 \)

For \( y = 3 \): \( x - \frac{1}{x} = 3 \)
\( \implies \frac{x^2 - 1}{x} = 3 \)
\( \implies x^2 - 3x - 1 = 0 \)
Using quadratic formula: \( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{13}}{2} \)

Therefore, \( x = -1, 1, \frac{3 + \sqrt{13}}{2}, \frac{3 - \sqrt{13}}{2} \)
In simple words: This time we used \( x - \frac{1}{x} \) as our substitution. The method is the same, but the identity is \( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 \).

📝 Teacher's Note: Teach students both identities: \( (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 \) and \( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 \). The sign of 2 changes based on plus or minus.

🎯 Exam Tip: Be careful with signs when squaring \( (x - \frac{1}{x}) \). Write the identity step clearly. Check which identity to use based on whether you see plus or minus.

Question 11. Solve: \( (x^2 + 5x + 4)(x^2 + 5x + 6) = 120 \)
Answer:
Step 1: Write the equation.
\( (x^2 + 5x + 4)(x^2 + 5x + 6) = 120 \)

Step 2: Let \( x^2 + 5x = y \)
then \( (y + 4)(y + 6) = 120 \)

Step 3: Expand and simplify.
\( \implies y^2 + 6y + 4y + 24 - 120 = 0 \)
\( \implies y^2 + 10y - 96 = 0 \)

Step 4: Factor the quadratic.
\( \implies y^2 + 16y - 6y - 96 = 0 \)
\( \implies y(y + 16) - 6(y + 16) = 0 \)
\( \implies (y + 16)(y - 6) = 0 \)

Step 5: Solve for y.
then \( y = -16 \) or \( y = 6 \)

Step 6: Find x values for each y.
For \( y = -16 \): \( x^2 + 5x = -16 \)
\( \implies x^2 + 5x + 16 = 0 \)
Using quadratic formula: \( x = \frac{-5 \pm \sqrt{(5)^2 - 4(1)(16)}}{2(1)} = \frac{-5 \pm \sqrt{-39}}{2} \)
(reject - no real solutions)

For \( y = 6 \): \( x^2 + 5x = 6 \)
\( \implies x^2 + 5x - 6 = 0 \)
\( \implies x^2 + 6x - x - 6 = 0 \)
\( \implies x(x + 6) - 1(x + 6) = 0 \)
\( \implies (x + 6)(x - 1) = 0 \)
then \( x = -6 \) and \( x = 1 \)

Therefore, \( x = -6 \) and \( x = 1 \)
In simple words: We noticed that both brackets had the common part \( x^2 + 5x \). We replaced this with y to make the problem easier, then solved for x.

📝 Teacher's Note: When students see expressions like \( (x^2 + 5x + 4)(x^2 + 5x + 6) \), teach them to spot the common part. Substituting the common part makes complex problems simple.

🎯 Exam Tip: Always reject solutions that give negative discriminants. Write "reject" clearly. Only write real number solutions unless the question asks for complex numbers.

Question 12. Solve each of the following equations, giving answer upto two decimal places.
(i) \( x^2 - 5x - 10 = 0 \) (ii) \( 3x^2 - x - 7 = 0 \)
Answer:
(i) For \( x^2 - 5x - 10 = 0 \):
Using quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Here \( a = 1, b = -5, c = -10 \)
\( x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-10)}}{2(1)} \)
\( x = \frac{5 \pm \sqrt{25 + 40}}{2} \)
\( x = \frac{5 \pm \sqrt{65}}{2} \)
\( x = \frac{5 \pm 8.06}{2} \)

\( x = \frac{5 + 8.06}{2} = \frac{13.06}{2} = 6.53 \)
\( x = \frac{5 - 8.06}{2} = \frac{-3.06}{2} = -1.53 \)

(ii) For \( 3x^2 - x - 7 = 0 \):
Using quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Here \( a = 3, b = -1, c = -7 \)
\( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-7)}}{2(3)} \)
\( x = \frac{1 \pm \sqrt{1 + 84}}{6} \)
\( x = \frac{1 \pm \sqrt{85}}{6} \)
\( x = \frac{1 \pm 9.22}{6} \)

\( x = \frac{1 + 9.22}{6} = \frac{10.22}{6} = 1.70 \)
\( x = \frac{1 - 9.22}{6} = \frac{-8.22}{6} = -1.37 \)

Therefore:
(i) \( x = 6.53, -1.53 \)
(ii) \( x = 1.70, -1.37 \)
In simple words: We used the quadratic formula for both equations. We calculated the square root using a calculator and rounded the final answers to two decimal places.

📝 Teacher's Note: Show students how to use a calculator to find square roots like \( \sqrt{65} \) and \( \sqrt{85} \). Practice rounding to two decimal places. Remind them to give both answers.

🎯 Exam Tip: Use the quadratic formula when factoring is difficult. Show all steps clearly. Round only the final answer to two decimal places, not the intermediate steps. Always give both solutions.

Question 13. Solve: \( \left(\frac{x}{x+2}\right)^2 - 7\left(\frac{x}{x+2}\right) + 12 = 0 \); x ≠ -2
Answer:
Step 1: Let \( \frac{x}{x+2} = y \)

Step 2: Substitute into the equation.
\( y^2 - 7y + 12 = 0 \)

Step 3: Factorize the quadratic equation.
\( y^2 - 4y - 3y + 12 = 0 \)
\( y(y - 4) - 3(y - 4) = 0 \)
\( (y - 4)(y - 3) = 0 \)
Therefore: \( y = 4 \) and \( y = 3 \)

Step 4: Substitute back the values of y.
When \( y = 4 \): \( \frac{x}{x+2} = 4 \)
\( x = 4(x + 2) \)
\( x = 4x + 8 \)
\( x - 4x = 8 \)
\( -3x = 8 \)
\( x = -\frac{8}{3} \)

When \( y = 3 \): \( \frac{x}{x+2} = 3 \)
\( x = 3(x + 2) \)
\( x = 3x + 6 \)
\( x - 3x = 6 \)
\( -2x = 6 \)
\( x = -3 \)

Answer: \( x = -\frac{8}{3} \) and \( x = -3 \)
In simple words: We made a substitution to turn this complex equation into a simple quadratic. Then we solved for y and found x values. Both answers are valid since neither makes x = -2.

📝 Teacher's Note: Show students how substitution makes hard problems simple. The key is to spot the repeating pattern \( \frac{x}{x+2} \) and replace it with one letter.

🎯 Exam Tip: Always check if your final answers make any denominator zero. Here x cannot be -2. Write "both solutions are valid" to get full marks.

Question 14. Solve:
(i) \( x^2 - 11x - 12 = 0 \); when x ∈ N
(ii) \( x^2 - 4x - 12 = 0 \); when x ∈ I
(iii) \( 2x^2 - 9x + 10 = 0 \); when x ∈ Q
Answer:
(i) \( x^2 - 11x - 12 = 0 \)
\( x^2 - 12x + x - 12 = 0 \)
\( x(x - 12) + 1(x - 12) = 0 \)
\( (x - 12)(x + 1) = 0 \)
Therefore: \( x = 12 \) and \( x = -1 \)
Since x ∈ N (natural numbers), then \( x = 12 \) only

(ii) \( x^2 - 4x - 12 = 0 \)
\( x^2 - 6x + 2x - 12 = 0 \)
\( x(x - 6) + 2(x - 6) = 0 \)
\( (x - 6)(x + 2) = 0 \)
Therefore: \( x = 6 \) and \( x = -2 \)
Since x ∈ I (integers), then \( x = 6 \) and \( x = -2 \)

(iii) \( 2x^2 - 9x + 10 = 0 \)
\( 2x^2 - 5x - 4x + 10 = 0 \)
\( x(2x - 5) - 2(2x - 5) = 0 \)
\( (2x - 5)(x - 2) = 0 \)
Therefore: \( x = \frac{5}{2} \) and \( x = 2 \)
Since x ∈ Q (rational numbers), then \( x = \frac{5}{2} \) and \( x = 2 \)

In simple words: We solve the same way but only keep answers that belong to the given set. N means counting numbers (1, 2, 3...). I means all whole numbers including negatives. Q means fractions are allowed.

📝 Teacher's Note: Explain what N, I, and Q mean with examples. Students often forget that natural numbers don't include zero or negatives.

🎯 Exam Tip: Always write "Since x ∈ [set name]" before giving your final answer. Check which solutions belong to the given set. You lose marks if you include invalid solutions.

Question 15. Solve: \( (a + b)^2x^2 - (a + b)x - 6 = 0 \); a + b ≠ 0
Answer:
Step 1: Factor out common terms.
\( (a + b)^2x^2 - 3(a + b)x + 2(a + b)x - 6 = 0 \)
\( (a + b)x[(a + b)x - 3] + 2[(a + b)x - 3] = 0 \)
\( [(a + b)x - 3][(a + b)x + 2] = 0 \)

Step 2: Set each factor equal to zero.
\( (a + b)x - 3 = 0 \) or \( (a + b)x + 2 = 0 \)

Step 3: Solve for x.
\( x = \frac{3}{a + b} \) or \( x = \frac{-2}{a + b} \)

Answer: \( x = \frac{3}{a + b} \) and \( x = \frac{-2}{a + b} \)
In simple words: We treated (a + b) like a single number and factorized normally. Since a + b ≠ 0, we can divide by it safely.

📝 Teacher's Note: Remind students that (a + b) is just one "thing" - treat it like any coefficient. The condition a + b ≠ 0 prevents division by zero.

🎯 Exam Tip: Write the condition "a + b ≠ 0" in your working. This shows you understand why division is allowed. Always simplify fractions if possible.

Question 16. Solve: \( \frac{1}{p} + \frac{1}{q} - \frac{1}{x} = \frac{1}{x + p + q} \)
Answer:
Step 1: Rearrange the equation.
\( \frac{1}{p} + \frac{1}{q} - \frac{1}{x} - \frac{1}{x + p + q} = 0 \)

Step 2: Find common denominator and simplify.
\( \frac{q - p}{pq} + \frac{x + p + q - x}{x(x + p + q)} = 0 \)
\( \frac{q - p}{pq} + \frac{p + q}{x(x + p + q)} = 0 \)

Step 3: Factor out common terms.
\( (p + q)\left[\frac{1}{pq} + \frac{1}{x^2 + px + qx}\right] = 0 \)
\( (p + q)\left[\frac{x^2 + px + qx + pq}{pq(x^2 + px + qx)}\right] = 0 \)

Step 4: Simplify the numerator.
\( x^2 + px + qx + pq = 0 \)
\( x(x + p) + q(x + p) = 0 \)
\( (x + p)(x + q) = 0 \)

Step 5: Solve for x.
\( x = -p \) and \( x = -q \)

Answer: x = -p and x = -q
In simple words: We moved all terms to one side and found a common pattern. The equation becomes a simple quadratic after factoring.

📝 Teacher's Note: This problem requires careful algebra with fractions. Break it into small steps. Show students how to find common denominators systematically.

🎯 Exam Tip: Keep all fractions until the end. Don't try to convert to decimals. Write each step clearly - partial marks are given for correct working even if the final answer is wrong.

Question 17. Solve:
(i) x(x - 1) + (x - 2)(x - 3) = 42
(ii) \( \frac{1}{x-1} - \frac{2}{x-2} = \frac{3}{x-3} - \frac{4}{x-4} \)
Answer:
(i) x(x - 1) + (x - 2)(x - 3) = 42
\( x^2 - x + x^2 - 3x - 2x + 6 = 42 \)
\( 2x^2 - 6x + 6 = 42 \)
\( 2x^2 - 6x + 6 - 42 = 0 \)
\( 2x^2 - 6x - 36 = 0 \)
\( 2x^2 + 12x - 6x - 36 = 0 \)
\( 2x(x + 6) - 6(x + 6) = 0 \)
\( (x + 6)(2x - 6) = 0 \)
If \( x + 6 = 0 \) or \( 2x - 6 = 0 \)
then \( x = -6 \) or \( x = 3 \)

(ii) \( \frac{1}{x-1} - \frac{2}{x-2} = \frac{3}{x-3} - \frac{4}{x-4} \)
\( \frac{1(x-2) - 2(x-1)}{(x-1)(x-2)} = \frac{3(x-4) - 4(x-3)}{(x-3)(x-4)} \)
\( \frac{-x}{x^2 + 3x + 2} = \frac{-x}{x^2 + 7x + 12} \)
\( -x[x^2 + 3x + 2 - x^2 - 7x - 12] = 0 \)
\( -x[-4x - 10] = 0 \)
\( x = 0 \) and \( x = \frac{-10}{4} = -2.5 \)

Answer: (i) x = -6 and x = 3; (ii) x = 0 and x = -2.5
In simple words: For part (i), we expanded brackets and made a quadratic equation. For part (ii), we found common denominators and simplified to get the same pattern on both sides.

📝 Teacher's Note: In part (ii), students often get confused with multiple fractions. Show them to work with one side at a time, then cross multiply.

🎯 Exam Tip: Always expand brackets carefully and collect like terms. Check your final answers by substituting back into the original equation.

Question 18. For each equation, given below, find the value of m so that the equation has equal roots. Also, find the solution of each equation:
(i) (m - 3)x² - 4x + 1 = 0
(ii) 3x² + 12x + (m + 7) = 0
(iii) x² - (m + 2)x + (m + 5) = 0
Answer:
(i) (m - 3)x² - 4x + 1 = 0
Here a = (m - 3), b = -4 and c = 1
Given equation has equal roots, then D = 0
\( b^2 - 4ac = 0 \)
\( (-4)^2 - 4(m - 3)(1) = 0 \)
\( 16 - 4m + 12 = 0 \)
\( -4m = -28 \)
\( m = 7 \)

Put value of m in given equation:
\( 4x^2 - 4x + 1 = 0 \)
\( (2x - 1)^2 = 0 \)
\( 2x - 1 = 0 \)
\( x = \frac{1}{2} \)

(ii) 3x² + 12x + (m + 7) = 0
Here a = 3, b = 12 and c = m + 7
Given equation has equal roots, then D = 0
\( b^2 - 4ac = 0 \)
\( (12)^2 - 4(3)(m + 7) = 0 \)
\( 144 - 12m - 84 = 0 \)
\( -12m = -60 \)
\( m = 5 \)

Put value of m in given equation:
\( 3x^2 + 12x + 12 = 0 \)
\( x^2 + 4x + 4 = 0 \)
\( (x + 2)^2 = 0 \)
\( x = -2 \)

Answer: (i) m = 7, x = 1/2; (ii) m = 5, x = -2
In simple words: For equal roots, the discriminant must be zero. We use the formula b² - 4ac = 0 to find m. Then we substitute m back to solve for x.

📝 Teacher's Note: Equal roots mean the parabola just touches the x-axis at one point. The discriminant D = b² - 4ac tells us about the nature of roots.

🎯 Exam Tip: Always write "Given equation has equal roots, then D = 0" first. Show the discriminant formula clearly. Remember equal roots means the same value appears twice.

Question 19. Without solving the following quadratic equation, find the value of p for which the roots are equal. \( px^2 - 4x + 3 = 0 \)
Answer:
\( px^2 - 4x + 3 = 0 \)
Here \( a = p \), \( b = -4 \) and \( c = 3 \)
Given equation has equal roots
then \( D = 0 \)
\( \implies b^2 - 4ac = 0 \)
\( \implies [-4]^2 - 4(p)(3) = 0 \)
\( \implies 16 - 12p = 0 \)
\( \implies -12p = -16 \)
\( \implies p = \frac{-16}{-12} = \frac{4}{3} \)
In simple words: For equal roots, the discriminant must be zero. We put \( b^2 - 4ac = 0 \) and solve for p. The answer is \( p = \frac{4}{3} \).

📝 Teacher's Note: Always write the discriminant formula first. Then substitute the values carefully. Students often forget to square the negative sign correctly.

🎯 Exam Tip: Write "For equal roots, D = 0" first. Show all substitution steps clearly. Write the final answer as a fraction if it doesn't simplify to a whole number.

Question 20. Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots. \( x^2 + 2(m - 1)x + (m + 5) = 0 \)
Answer:
Consider the given equation:
\( x^2 + 2(m - 1)x + (m + 5) = 0 \)
The nature of the roots of a quadratic equation
\( ax^2 + bx + c = 0 \), depends entirely on the
value of its discriminant \( b^2 - 4ac \).
If a, b and c are real numbers and \( a \neq 0 \),
then discriminant:
(i) \( b^2 - 4ac = 0 \implies \) the roots are real and equal.
(ii) \( b^2 - 4ac > 0 \implies \) the roots are real and unequal.
(i) \( b^2 - 4ac < 0 \implies \) the roots are imaginary (not real).
Since the roots of the given equation are real and equal,
we have,
\( b^2 - 4ac = 0 \)
\( \implies [2(m - 1)]^2 - 4 \times 1 \times (m + 5) = 0 \)
\( \implies 4[m^2 + 1 - 2m] - 4(m + 5) = 0 \)
\( \implies 4m^2 + 4 - 8m - 4m - 20 = 0 \)
\( \implies 4m^2 - 12m - 16 = 0 \)
\( \implies m^2 - 3m - 4 = 0 \)
\( \implies m^2 - 4m + m - 4 = 0 \)
\( \implies m(m - 4) + 1(m - 4) = 0 \)
\( \implies (m + 1)(m - 4) = 0 \)
\( \implies m + 1 = 0 \) or \( m - 4 = 0 \)
\( \implies m = -1 \) or \( m = 4 \)
In simple words: For equal roots, discriminant = 0. We expand and simplify to get a quadratic in m. Solving gives \( m = -1 \) or \( m = 4 \).

📝 Teacher's Note: Explain that discriminant tells us about roots without solving. Show students how to expand \( [2(m-1)]^2 \) step by step as this is where mistakes happen.

🎯 Exam Tip: Write the discriminant condition first. Expand brackets carefully and collect like terms. Always check your final answer by substituting back.

Exercise 5F

Question 1(i). \( (x + 5)(x - 5) = 24 \)
Answer:
Given: \( (x + 5)(x - 5) = 24 \)
\( \implies x^2 - 5^2 = 24 \) .... since \( (a - b)(a + b) = a^2 - b^2 \)
\( \implies x^2 - 25 = 24 \)
\( \implies x^2 = 49 \)
\( \implies x = \pm 7 \)
In simple words: We use the difference of squares formula first. Then solve the simple equation \( x^2 = 49 \) to get \( x = 7 \) or \( x = -7 \).

📝 Teacher's Note: Remind students of the difference of squares pattern: \( (a+b)(a-b) = a^2 - b^2 \). This makes the problem much easier than expanding fully.

🎯 Exam Tip: Always look for patterns like difference of squares first. Write both positive and negative square roots. Don't forget the \( \pm \) sign.

Question 1(ii). \( 3x^2 - 2\sqrt{6}x + 2 = 0 \)
Answer:
Given: \( 3x^2 - 2\sqrt{6}x + 2 = 0 \)
\( \implies 3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0 \)
\( \implies \sqrt{3}x(\sqrt{3}x - \sqrt{2}) - \sqrt{2}(\sqrt{3}x - \sqrt{2}) = 0 \)
\( \implies (\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2}) = 0 \)
\( \implies (\sqrt{3}x - \sqrt{2}) = 0 \) or \( (\sqrt{3}x - \sqrt{2}) = 0 \)
\( \implies x = \frac{\sqrt{2}}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}} \)
In simple words: We split the middle term and factor. Both factors are the same, so we get one repeated root: \( x = \frac{\sqrt{2}}{\sqrt{3}} \).

📝 Teacher's Note: Show students how to split \( -2\sqrt{6} \) into \( -\sqrt{6} - \sqrt{6} \). This equation has equal roots because the discriminant is zero.

🎯 Exam Tip: For surd coefficients, try factoring by grouping. When you get the same factor twice, the quadratic has equal roots.

Question 1(iii). \( 3\sqrt{2}x^2 - 5x - \sqrt{20} = 0 \)
Answer:
Given: \( 3\sqrt{2}x^2 - 5x - \sqrt{20} = 0 \)
\( \implies 3\sqrt{2}x^2 - 6x - x + \sqrt{2} = 0 \)
\( \implies 3\sqrt{2}x(x - \sqrt{2}) - (x - \sqrt{2}) = 0 \)
\( \implies (3\sqrt{2}x - 1)(x - \sqrt{2}) = 0 \)
\( \implies x = -\frac{1}{3\sqrt{2}} \) or \( x = \sqrt{2} \)
In simple words: We factor by grouping after splitting the middle term. This gives us two different roots.

📝 Teacher's Note: First simplify \( \sqrt{20} = 2\sqrt{5} \). Then show students how to split -5x to make factoring work. Practice with simpler numbers first.

🎯 Exam Tip: Simplify all surds first. When factoring, look for common factors in each group. Always rationalize denominators if needed.

Question 2. One root of the quadratic equation \( 8x^2 + mx + 15 = 0 \) is \( \frac{3}{4} \). Find the value of m. Also, find the other root of the equation.
Answer:
Given quadratic equation is \( 8x^2 + mx + 15 = 0 \) .... (i)
One of the roots of (i) is \( \frac{3}{4} \), so it satisfies (i)
\( \implies 8\left(\frac{3}{4}\right)^2 + m\left(\frac{3}{4}\right) + 15 = 0 \)
\( \implies \frac{9}{2} + 15 + m\left(\frac{3}{4}\right) = 0 \)
\( \implies m\left(\frac{3}{4}\right) = -\frac{39}{2} \)
\( \implies m = -26 \)
So, the equation (i) becomes \( 8x^2 - 26x + 15 = 0 \)
\( \implies 8x^2 - 20x - 6x + 15 = 0 \)
\( \implies 4x(2x - 5) - 3(2x - 5) = 0 \)
\( \implies (4x - 3)(2x - 5) = 0 \)
\( \implies x = \frac{3}{4} \) or \( x = \frac{5}{2} \)
Hence, the other root is \( \frac{5}{2} \)
In simple words: We substitute the known root to find m = -26. Then we factor the quadratic to find the other root is \( \frac{5}{2} \).

📝 Teacher's Note: Show students that if a number is a root, it makes the equation equal to zero. Substitute carefully and solve for the unknown coefficient first.

🎯 Exam Tip: Always substitute the given root first to find the unknown coefficient. Then factor or use the quadratic formula to find the other root.

Question 3. One root of the quadratic equation \( x^2 - 3x - 2ax - 6a = 0 \) is -3, find its other root.
Answer:
Given quadratic equation is \( x^2 - 3x - 2ax - 6a = 0 \) .... (i)
One of the roots of (i) is -3, so it satisfies (i)
\( \implies x^2 - 3x - 2ax - 6a = 0 \)
\( \implies x(x - 3) - 2a(x - 3) = 0 \)
\( \implies (x - 2a)(x - 3) = 0 \)
\( \implies x = -3, 2a \)
Hence, the other root is 2a.
In simple words: We factor by grouping to get \( (x - 2a)(x - 3) = 0 \). Since one root is -3, the other root is 2a.

📝 Teacher's Note: Show students how to group terms to factor. The equation factors as \( (x - 2a)(x + 3) = 0 \), so roots are \( x = 2a \) and \( x = -3 \).

🎯 Exam Tip: Look for factoring by grouping when you have four terms. Always check that your given root satisfies the factored form.

Question 4. If p - 15 = 0 and \( 2x^2 + px + 15 = 0 \); find the values of x.
Answer:
Given i.e p - 15 = 0 i.e. p = 15
So, the given quadratic equation becomes
\( 2x^2 + 15x + 15 = 0 \)
\( \implies 2x^2 + 10x + 5x + 15 = 0 \)
\( \implies 2x(x + 5) + 5(x + 5) = 0 \)
\( \implies (2x + 5)(x + 5) = 0 \)
\( \implies x = -5, -\frac{5}{2} \)
Hence, the values of x are -5 and \( -\frac{5}{2} \)
In simple words: First we find p = 15, then substitute to get the quadratic. We factor to find \( x = -5 \) or \( x = -\frac{5}{2} \).

📝 Teacher's Note: Always solve for the parameter first. Then substitute and solve the quadratic. Check that both solutions are correct by substituting back.

🎯 Exam Tip: Find the value of the parameter first. Then substitute and factor the resulting quadratic. Write both roots clearly.

Question 5. Find the solution of the equation \( 2x^2 - mx - 25n = 0 \); if m + 5 = 0 and n - 1 = 0.
Answer:
Given quadratic equation is \( 2x^2 - mx - 25n = 0 \) ..... (i)
Also, given and m + 5 = 0 and n - 1 = 0
\( \implies m = -5 \) and \( n = 1 \)
So, the equation (i) becomes
\( 2x^2 + 5x - 25 = 0 \)
\( \implies 2x^2 + 10x - 5x - 25 = 0 \)
\( \implies 2x(x + 5) - 5(x + 5) = 0 \)
\( \implies (x + 5)(2x - 5) = 0 \)
\( \implies x = -5, \frac{5}{2} \)
Hence, the solution of given quadratic equation are x = -5 and \( \frac{5}{2} \)
In simple words: We find m = -5 and n = 1, then substitute to get \( 2x^2 + 5x - 25 = 0 \). Factoring gives \( x = -5 \) or \( x = \frac{5}{2} \).

📝 Teacher's Note: Students must find both parameters first. Show them how -mx becomes +5x when m = -5. Be careful with signs when substituting.

🎯 Exam Tip: Solve for all unknown parameters first. Substitute carefully, paying attention to signs. Then factor the resulting quadratic.

Question 6. If m and n are roots of the equation \( \frac{1}{x} - \frac{1}{x-2} = 3 \) where \( x \neq 0 \) and \( x \neq 2 \); find m × n.
Answer:
Given quadratic equation is \( \frac{1}{x} - \frac{1}{x-2} = 3 \)
\( \implies \frac{x-2-x}{x(x-2)} = 3 \)
\( \implies \frac{-2}{x(x-2)} = 3 \)
\( \implies -2 = 3x(x-2) \)
\( \implies 3x^2 - 6x + 2 = 0 \)
\( \implies x = \frac{6 \pm \sqrt{36-4(3)(2)}}{2 \times 3} \)
\( \implies x = \frac{6 \pm \sqrt{12}}{6} \)
\( \implies x = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3} \)
For a quadratic \( ax^2 + bx + c = 0 \), product of roots = \( \frac{c}{a} \)
Here \( a = 3, c = 2 \), so \( m \times n = \frac{2}{3} \)
In simple words: We clear fractions to get a quadratic. Using the formula for product of roots, \( m \times n = \frac{2}{3} \).

📝 Teacher's Note: Show students how to find common denominator for fractions first. Remember the product of roots formula saves time instead of finding individual roots.

🎯 Exam Tip: Convert to standard quadratic form first. Use product of roots = \( \frac{c}{a} \) formula. This is faster than finding both roots separately.

Question 7. Solve, using formula: \( x^2 + x - (a + 2)(a + 1) = 0 \)
Answer:
Given quadratic equation is \( x^2 + x - (a + 2)(a + 1) = 0 \)
Using quadratic formula,
\[ x = \frac{-1 \pm \sqrt{1^2 - 4(a + 2)(a + 1)}}{2} \]
\[ x = \frac{-1 \pm \sqrt{1 - 4(a^2 + 3a + 2)}}{2} \]
\[ x = \frac{-1 \pm \sqrt{4a^2 + 12a + 9}}{2} \]
\[ x = \frac{-1 \pm \sqrt{(2a + 3)^2}}{2} \]
\[ x = \frac{-1 \pm (2a + 3)}{2} \]
\[ x = \frac{-1 + (2a + 3)}{2} \text{ or } x = \frac{-1 - (2a + 3)}{2} \]
\[ x = \frac{2a + 2}{2} \text{ or } x = \frac{-2a - 4}{2} \]
\[ x = \frac{2(a + 1)}{2} \text{ or } x = \frac{2(-a - 2)}{2} \]
\[ x = a + 1 \text{ or } x = -a - 2 = -(a + 2) \]
In simple words: We use the quadratic formula to find the values of x. The answer has two parts - one is (a + 1) and the other is -(a + 2).

📝 Teacher's Note: Show students how \( (2a + 3)^2 \) comes from expanding the discriminant. This makes the square root easy to calculate. Practice with simple values of 'a' first.

🎯 Exam Tip: Always write the quadratic formula first. Show all steps clearly. Write the final answer as \( x = a + 1 \) or \( x = -(a + 2) \).

Question 8. Solve the quadratic equation \( 8x^2 - 14x + 3 = 0 \)
(i) When x ∈ I (integers)
(ii) When x ∈ Q (rational numbers)
Answer:
Given quadratic equation is \( 8x^2 - 14x + 3 = 0 \)
\( \Rightarrow 8x^2 - 12x - 2x + 3 = 0 \)
\( \Rightarrow 4x(2x - 3) - 1(2x - 3) = 0 \)
\( \Rightarrow (4x - 1)(2x - 3) = 0 \)
\( \Rightarrow x = \frac{1}{4} \text{ or } x = \frac{3}{2} \)

(i) When x ∈ I, the equation \( 8x^2 - 14x + 3 = 0 \) has no roots
(ii) When x ∈ Q the roots of \( 8x^2 - 14x + 3 = 0 \) are
\( x = \frac{1}{4}, x = \frac{3}{2} \)
In simple words: We solve by factoring. The answers are fractions. Since integers means whole numbers only, there are no integer solutions. But rational numbers include fractions, so both answers work.

📝 Teacher's Note: Explain that I means integers (whole numbers like 1, 2, 3) and Q means rational numbers (fractions like 1/4, 3/2). This helps students understand different number sets.

🎯 Exam Tip: Always check if your answers fit the given set. Write "no roots" clearly for integers. For rational numbers, write both fractions as your final answer.

Question 9. Find the value of m for which the equation \( (m + 4)x^2 + (m + 1)x + 1 = 0 \) has real and equal roots.
Answer:
Given quadratic equation is \( (m + 4)x^2 + (m + 1)x + 1 = 0 \)
The quadratic equation has real and equal roots if its discriminant is zero.
\( \Rightarrow D = b^2 - 4ac = 0 \)
\( \Rightarrow (m + 1)^2 - 4(m + 4)(1) = 0 \)
\( \Rightarrow m^2 + 2m + 1 - 4m - 16 = 0 \)
\( \Rightarrow m^2 - 2m - 15 = 0 \)
\( \Rightarrow m^2 - 5m + 3m - 15 = 0 \)
\( \Rightarrow m(m - 5) + 3(m - 5) = 0 \)
\( \Rightarrow (m - 5)(m + 3) = 0 \)
\( \Rightarrow m = 5 \text{ or } m = -3 \)
In simple words: For equal roots, the discriminant must be zero. We solve this condition to find the values of m that make the equation have equal roots.

📝 Teacher's Note: Remind students that equal roots means the parabola just touches the x-axis at one point. The discriminant formula \( b^2 - 4ac = 0 \) is key for this condition.

🎯 Exam Tip: Write "For real and equal roots, D = 0" at the start. Show the factoring steps clearly. The final answer is \( m = 5 \) or \( m = -3 \).

Question 10. Find the values of m for which equation \( 3x^2 + mx + 2 = 0 \) has equal roots. Also, find the roots of the given equation.
Answer:
Given quadratic equation is \( 3x^2 + mx + 2 = 0 \) .... (i)
The quadratic equation has equal roots if its discriminant is zero
\( \Rightarrow D = b^2 - 4ac = 0 \)
\( \Rightarrow m^2 - 4(2)(3) = 0 \)
\( \Rightarrow m^2 = 24 \)
\( \Rightarrow m = \pm 2\sqrt{6} \)

When \( m = 2\sqrt{6} \), equation (i) becomes
\( 3x^2 + 2\sqrt{6}x + 2 = 0 \)
\( \Rightarrow (\sqrt{3}x + \sqrt{2})^2 = 0 \)
\( \Rightarrow x = -\frac{\sqrt{2}}{\sqrt{3}} = -\frac{\sqrt{6}}{3} \)

When \( m = -2\sqrt{6} \), equation (i) becomes
\( 3x^2 - 2\sqrt{6}x + 2 = 0 \)
\( \Rightarrow (\sqrt{3}x - \sqrt{2})^2 = 0 \)
\( \Rightarrow x = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3} \)

\( \therefore x = -\frac{\sqrt{6}}{3}, \frac{\sqrt{6}}{3} \)
In simple words: We find m values that make equal roots, then substitute each m value back to find the actual roots.

📝 Teacher's Note: Show students how perfect square trinomials work. When we have equal roots, the quadratic can be written as a perfect square like \( (ax + b)^2 = 0 \).

🎯 Exam Tip: First find m values using discriminant = 0. Then substitute each m value to find the roots. Write both parts of the answer clearly.

Question 11. Find the value of k for which equation \( 4x^2 + 8x - k = 0 \) has real roots.
Answer:
Given quadratic equation is \( 4x^2 + 8x - k = 0 \) .... (i)
The quadratic equation has real roots if its discriminant is greater than or equal to zero
\( \Rightarrow D = b^2 - 4ac \geq 0 \)
\( \Rightarrow 8^2 - 4(4)(-k) \geq 0 \)
\( \Rightarrow 64 + 16k \geq 0 \)
\( \Rightarrow 16k \geq -64 \)
\( \Rightarrow k \geq -4 \)
Hence, the given quadratic equation has real roots for \( k \geq -4 \)
In simple words: For real roots, the discriminant must be zero or positive. We solve this condition to find the range of k values.

📝 Teacher's Note: Explain that real roots means the parabola crosses or touches the x-axis. If discriminant is negative, there are no real solutions (parabola doesn't touch x-axis).

🎯 Exam Tip: Write "For real roots, D ≥ 0" at the start. Show the inequality steps. The final answer is \( k \geq -4 \).

Question 12. Find, using quadratic formula, the roots of the following quadratic equations, if they exist
(i) \( 3x^2 - 5x + 2 = 0 \)
(ii) \( x^2 + 4x + 5 = 0 \)
Answer:
(i) Given quadratic equation is \( 3x^2 - 5x + 2 = 0 \)
\( D = b^2 - 4ac = (-5)^2 - 4(3)(2) = 25 - 24 = 1 \)
Since D > 0, the roots of the given quadratic equation are real and distinct.
Using quadratic formula, we have
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4(3)(2)}}{2(3)} \]
\[ x = \frac{5 \pm \sqrt{25 - 24}}{6} \]
\[ x = \frac{5 \pm 1}{6} \]
\[ x = \frac{5 + 1}{6} \text{ or } x = \frac{5 - 1}{6} \]
\[ x = \frac{6}{6} \text{ or } x = \frac{4}{6} \]
\[ x = 1 \text{ or } x = \frac{2}{3} \]

(ii) Given quadratic equation is \( x^2 + 4x + 5 = 0 \)
\( D = b^2 - 4ac = (4)^2 - 4(1)(5) = 16 - 20 = -4 \)
Since D < 0, the roots of the given quadratic equation does not exist.
In simple words: Part (i) has real roots because discriminant is positive. Part (ii) has no real roots because discriminant is negative.

📝 Teacher's Note: Always check discriminant first to know if real roots exist. Positive discriminant means two different real roots. Negative discriminant means no real roots.

🎯 Exam Tip: Write discriminant value first. If negative, write "roots do not exist". If positive, apply quadratic formula step by step. Show all calculation clearly.

Solution 13:
(i) Given quadratic equation is \( \frac{1}{18-x} - \frac{1}{18+x} = \frac{1}{24} \)
\[ \Rightarrow \frac{(18 + x) - (18 - x)}{(18 - x)(18 + x)} = \frac{1}{24} \]
\[ \Rightarrow \frac{2x}{18^2 - x^2} = \frac{1}{24} \]
\[ \Rightarrow 48x = 324 - x^2 \]
\[ \Rightarrow x^2 + 48x - 324 = 0 \]
\[ \Rightarrow x^2 + 54x - 6x - 324 = 0 \]
\[ \Rightarrow x(x + 54) - 6(x + 54) = 0 \]
\[ \Rightarrow (x + 54)(x - 6) = 0 \]
\[ \Rightarrow x = -54 \text{ or } x = 6 \]
But as x > 0, so x can't be negative.
Hence, x = 6.

(ii) Given quadratic equation is \( (x - 10)(\frac{1200}{x} + 2) = 1260 \)
\[ \Rightarrow (x - 10)(1200 + 2x) = 1260x \]
\[ \Rightarrow 1200x + 2x^2 - 12000 - 20x = 1260x \]
\[ \Rightarrow 2x^2 - 12000 - 80x = 0 \]
\[ \Rightarrow x^2 - 40x - 6000 = 0 \]
\[ \Rightarrow x^2 - 100x + 60x - 6000 = 0 \]
\[ \Rightarrow (x - 100)(x + 60) = 0 \]
\[ \Rightarrow x = 100 \text{ or } x = -60 \]
But as x < 0, so x can't be positive.
Hence, x = -60.
In simple words: We solve rational equations by clearing fractions first. Then we solve the quadratic and check which solutions fit the given conditions.

📝 Teacher's Note: When solving rational equations, multiply both sides to clear fractions. Always check if solutions make denominators zero (not allowed). Also check given conditions like x > 0 or x < 0.

🎯 Exam Tip: Clear fractions first by finding common denominators. Solve the resulting quadratic. Check solutions against given conditions and reject invalid ones.