Maharashtra Board Class 10 Maths Chapter 2 Quadratic Equations Set 2 Solutions

Question 1. Choose the correct answers for the following questions.

(i) Which one is the quadratic equation?
(A) \(\frac{5}{x} - 3 = x^2\)
(B) \(x (x + 5) = 2\)
(C) \(n - 1 = 2n\)
(D) \(\frac{1}{x^2} (x + 2) = x\)
Answer: (B) \(x (x + 5) = 2\)
In simple words: A quadratic equation is of the form \(ax^2 + bx + c = 0\), where \(a \neq 0\). Option (B) simplifies to \(x^2 + 5x - 2 = 0\), which fits this form.

🎯 Exam Tip: Remember to simplify equations first to determine if they are quadratic. The highest power of the variable must be 2, and the coefficient of the \(x^2\) term must not be zero.

(ii) Out of the following equations which one is not a quadratic equation?
(A) \(x^2 + 4x = 11 + x^2\)
(B) \(x = 4x\)
(C) \(5x^2 = 90\)
(D) \(2x - x^2 = x^2 + 5\)
Answer: (A) \(x^2 + 4x = 11 + x^2\)
In simple words: When you simplify option (A), the \(x^2\) terms cancel out, leaving \(4x = 11\), which is a linear equation, not quadratic.

🎯 Exam Tip: Always bring all terms to one side and simplify before identifying the type of equation. If the \(x^2\) term cancels out, it's not a quadratic equation.

(iii) The roots of \(x^2 + kx + k = 0\) are real and equal, find k.
(A) 0
(B) 4
(C) 0 or 4
(D) 2
Answer: (C) 0 or 4
In simple words: For real and equal roots, the discriminant (\(b^2 - 4ac\)) must be zero. For \(x^2 + kx + k = 0\), \(a=1, b=k, c=k\). So, \(k^2 - 4(1)(k) = 0 \implies k^2 - 4k = 0 \implies k(k - 4) = 0\). This gives \(k=0\) or \(k=4\).

🎯 Exam Tip: The nature of roots (real and equal, real and unequal, or not real) is determined by the discriminant (\(\Delta = b^2 - 4ac\)). For real and equal roots, \(\Delta = 0\).

(iv) For \(\sqrt{2} x^2 - 5x + \sqrt{2} = 0\), find the value of the discriminant.
(A) -5
(B) 17
(C) \(\sqrt{2}\)
(D) \(2\sqrt{2}-5\)
Answer: (B) 17
In simple words: The discriminant is calculated as \(b^2 - 4ac\). For the given equation, \(a=\sqrt{2}\), \(b=-5\), \(c=\sqrt{2}\). So, \((-5)^2 - 4(\sqrt{2})(\sqrt{2}) = 25 - 4(2) = 25 - 8 = 17\).

🎯 Exam Tip: Carefully identify the coefficients \(a\), \(b\), and \(c\) from the standard form \(ax^2 + bx + c = 0\) before calculating the discriminant. Pay attention to signs and square roots.

(v) Which of the following quadratic equations has roots 3,5?
(A) \(x^2 - 15x + 8 = 0\)
(B) \(x^2 - 8x + 15 = 0\)
(C) \(x^2 + 3x + 5 = 0\)
(D) \(x^2 + 8x - 15 = 0\)
Answer: (B) \(x^2 - 8x + 15 = 0\)
In simple words: If roots are \(\alpha\) and \(\beta\), the equation is \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\). Here, \(\alpha=3, \beta=5\). Sum of roots = \(3+5=8\). Product of roots = \(3 \times 5 = 15\). So, the equation is \(x^2 - 8x + 15 = 0\).

🎯 Exam Tip: Remember the relationship between roots and coefficients: sum of roots \( = -\frac{b}{a}\) and product of roots \( = \frac{c}{a}\) for \(ax^2 + bx + c = 0\).

(vi) Out of the following equations, find the equation having the sum of its roots -5.
(A) \(3x^2 - 15x + 3 = 0\)
(B) \(x^2 - 5x + 3 = 0\)
(C) \(x^2 + 3x - 5 = 0\)
(D) \(3x^2 + 15x + 3 = 0\)
Answer: (D) \(3x^2 + 15x + 3 = 0\)
In simple words: The sum of roots for \(ax^2 + bx + c = 0\) is \(-\frac{b}{a}\). For option (D), \(a=3, b=15\), so sum of roots is \(-\frac{15}{3} = -5\).

🎯 Exam Tip: Always identify \(a\) and \(b\) correctly from the standard form \(ax^2 + bx + c = 0\) to find the sum of roots using the formula \(-\frac{b}{a}\).

(vii) \(\sqrt{5}m^2 - \sqrt{5} m + \sqrt{5} =0\) which of the following statement is true for this given equation?
(A) Real and unequal roots
(B) Real and equal roots
(C) Roots are not real
(D) Three roots
Answer: (C) Roots are not real
In simple words: Calculate the discriminant \(b^2 - 4ac\). For \(a=\sqrt{5}, b=-\sqrt{5}, c=\sqrt{5}\), the discriminant is \((-\sqrt{5})^2 - 4(\sqrt{5})(\sqrt{5}) = 5 - 4(5) = 5 - 20 = -15\). Since the discriminant is negative, the roots are not real.

🎯 Exam Tip: A negative discriminant (\(\Delta < 0\)) indicates that the roots of the quadratic equation are not real (they are complex conjugates).

(viii) One of the roots of equation \(x^2 + mx - 5 = 0\) is 2; find m.
(A) -2
(B) \(-\frac{1}{2}\)
(C) \(\frac{1}{2}\)
(D) 2
Answer: (C) \(\frac{1}{2}\)
In simple words: If 2 is a root, substituting \(x=2\) into the equation should satisfy it. So, \((2)^2 + m(2) - 5 = 0 \implies 4 + 2m - 5 = 0 \implies 2m - 1 = 0 \implies 2m = 1 \implies m = \frac{1}{2}\).

🎯 Exam Tip: If a number is a root of an equation, it means that number satisfies the equation when substituted for the variable. This is a direct substitution problem.

Question 2. Which of the following equations is quadratic
(i) \(x^2 + 2x + 11 = 0\)
(ii) \(x^2 - 2x + 5 = x^2\)
(iii) \((x + 2)^2 = 2x^2\)
Solution:
(i) The given equation is
\(x^2 + 2x + 11 = 0\)
Here, \(x\) is the only variable and maximum index of the variable is 2.
\(a = 1, b = 2, c = 11\) are real numbers and
\(a \neq 0\).
The given equation is a quadratic equation.

(ii) The given equation is
\(x^2 - 2x + 5 = x^2\)
\(\implies x^2 - x^2 - 2x + 5 = 0\)
\(\implies -2x + 5 = 0\)
Here, \(x\) is the only variable and maximum index of the variable is not 2.
\(\implies\) The given equation is not a quadratic equation.

(iii) The given equation is
\((x + 2)^2 = 2x^2\)
\(\implies x^2 + 4x + 4 = 2x^2\)
\(\implies 2x^2 - x^2 - 4x - 4 = 0\)
\(\implies x^2 - 4x - 4 = 0\)
Here, \(x\) is the only variable and maximum index of the variable is 2.
\(a = 1, b = -4, c = -4\) are real numbers and
\(a \neq 0\).
\(\implies\) The given equation is a quadratic equation.
In simple words: A quadratic equation has the highest power of the variable as 2. Equations (i) and (iii) satisfy this condition after simplification, while equation (ii) simplifies to a linear equation.

🎯 Exam Tip: For each equation, expand and simplify all terms, then arrange them in the standard form \(ax^2 + bx + c = 0\) to correctly identify if it's quadratic. Ensure \(a \neq 0\).

Question 3. Find the value of discriminant for each of the following equations.
(i) \(2y^2 - y + 2 = 0\)
(ii) \(5m^2 - m = 0\)
(iii) \(\sqrt{5} x^2 - x - \sqrt{5} = 0\)
Solution:
(i) \(2y^2 - y + 2 = 0\)
Comparing the above equation with
\(ay^2 + by + c = 0\), we get
\(a = 2, b = -1, c = 2\)
\(\implies b^2 - 4ac = (-1)^2 - 4 \times 2 \times 2\)
\( = 1 - 16\)
\(\implies b^2 - 4ac = -15\)

(ii) \(5m^2 - m = 0\)
\(\implies 5m^2 - m + 0 = 0\)
Comparing the above equation with
\(am^2 + bm + c = 0\), we get
\(a = 5, b = -1, c = 0\)
\(\implies b^2 - 4ac = (-1)^2 - 4 \times 5 \times 0\)
\( = 1 - 0\)
\(\implies b^2 - 4ac = 1\)

(iii) \(\sqrt{5}x^2 - x - \sqrt{5} = 0\)
Comparing the above equation with
\(ax^2 + bx + c = 0\), we get
\(a = \sqrt{5}, b = -1, c = -\sqrt{5}\)
\(\implies b^2 - 4ac = (-1)^2 - 4 \times \sqrt{5} \times (-\sqrt{5})\)
\( = 1 + 20\)
\(\implies b^2 - 4ac = 21\)
In simple words: The discriminant (\(\Delta\)) is calculated using the formula \(b^2 - 4ac\). For each equation, identify the coefficients \(a\), \(b\), and \(c\), then substitute these values into the formula.

🎯 Exam Tip: Pay close attention to the signs of \(b\) and \(c\) when substituting into the discriminant formula. Mistakes often occur with negative numbers or square roots.

Question 4. One of the roots of quadratic equation \(2x^2 + kx - 2 = 0\) is - 2, find k.
Solution:
-2 is one of the roots of the equation
\(2x^2 + kx - 2 = 0\).
\(\implies\) Putting \(x = -2\) in the given equation, we get
\(2(-2)^2 + k(-2) - 2 = 0\)
\(\implies 8 - 2k - 2 = 0\)
\(\implies 6 - 2k = 0\)
\(\implies 2k = 6\)
\(\implies k = \frac{6}{2}\)
\(\implies k = 3\)
In simple words: Since -2 is a root, substitute -2 for \(x\) in the equation. This will create a simple linear equation in \(k\), which can then be solved.

🎯 Exam Tip: When a root is given, direct substitution is the most straightforward method to find an unknown coefficient. Be careful with calculations involving negative numbers and squares.

Question 5. Two roots of quadratic equations are given; frame the equation.
(i) 10 and -10
(ii) \(1-3\sqrt{5}\) and \(1 + 3\sqrt{5}\)
(iii) 0 and 7
Solution:
(i) Let \(\alpha = 10\) and \(\beta = -10\)
\(\implies \alpha + \beta = 10 - 10 = 0\)
and \(\alpha \times \beta = 10 \times -10 = -100\)
\(\implies\) The required quadratic equation is
\(x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
\(\implies x^2 - 0x + (-100) = 0\)
\(\implies x^2 - 100 = 0\)

(ii) Let \(\alpha = 1 - 3 \sqrt{5}\) and \(\beta = 1 + 3\sqrt{5}\)
\(\alpha + \beta = (1 - 3\sqrt{5}) + (1 + 3\sqrt{5}) = 2\)
and \(\alpha \times \beta = (1 - 3\sqrt{5}) (1 + 3\sqrt{5})\)
\( = (1)^2 - (3\sqrt{5})^2\)
\( = 1 - 45\)
\( = -44\)
\(\implies\) The required quadratic equation is
\(x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
\(\implies x^2 - 2x - 44 = 0\)

(iii) Let \(\alpha = 0\) and \(\beta = 7\)
\(\implies \alpha + \beta = 0 + 7 = 7\)
and \(\alpha \times \beta = 0 \times 7 = 0\)
\(\implies\) The required quadratic equation is
\(x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
\(\implies x^2 - 7x + 0 = 0\)
\(\implies x^2 - 7x = 0\)
In simple words: To form a quadratic equation from its roots \(\alpha\) and \(\beta\), use the formula \(x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0\). First, calculate \(\alpha + \beta\) and \(\alpha\beta\), then substitute these values into the formula.

🎯 Exam Tip: For roots involving surds (like \(\sqrt{5}\)), remember the difference of squares formula \((a-b)(a+b) = a^2 - b^2\) for simplifying the product of roots. Also, ensure you handle signs correctly for the sum of roots.

Question 6. Determine the nature of roots for each of the quadratic equation.
(i) \(3x^2 - 5x + 7 = 0\)
(ii) \(\sqrt{3} x^2 + \sqrt{2}x - 2\sqrt{3} = 0\)
(iii) \(m^2 - 2m + 1 = 0\)
Solution:
(i) \(3x^2 - 5x + 7 = 0\)
Comparing the above equation with
\(ax^2 + bx + c = 0\), we get
\(a = 3, b = -5, c = 7\)
\(\implies \Delta = b^2 - 4ac\)
\( = (-5)^2 - 4 \times 3 \times 7\)
\( = 25 - 84\)
\(\implies \Delta = -59\)
\(\implies \Delta < 0\)
\(\implies\) Roots of the given quadratic equation are not real.

(ii) \(\sqrt{3} x^2 + \sqrt{2} x - 2 \sqrt{3} = 0\)
Comparing the above equation with
\(ax^2 + bx + c = 0\), we get
\(a = \sqrt{3}, b = \sqrt{2}, c = -2\sqrt{3}\)
\(\implies \Delta = b^2 - 4ac\)
\( = (\sqrt{2})^2 - 4 \times \sqrt{3} \times (-2\sqrt{3})\)
\( = 2 + 24\)
\(\implies \Delta = 26\)
\(\implies \Delta > 0\)
\(\implies\) Roots of the given quadratic equation are real and unequal.

(iii) \(m^2 - 2m + 1 = 0\)
Comparing the above equation with
\(am^2 + bm + c = 0\), we get
\(a = 1, b = -2, c = 1\)
\(\implies \Delta = b^2 - 4ac\)
\( = (-2)^2 - 4 \times 1 \times 1\)
\( = 4 - 4\)
\(\implies \Delta = 0\)
\(\implies\) Roots of the given quadratic equation are real and equal
In simple words: The nature of roots is determined by the discriminant, \(\Delta = b^2 - 4ac\). If \(\Delta < 0\), roots are not real. If \(\Delta > 0\), roots are real and unequal. If \(\Delta = 0\), roots are real and equal.

🎯 Exam Tip: Systematically calculate the discriminant for each equation. Remember the three cases for the value of \(\Delta\) and their corresponding nature of roots. Double-check calculations, especially with square roots and negative numbers.

Question 7. Solve the following quadratic equations.
(i) \(\frac{1}{x+5} = \frac{1}{x^2}\)
(ii) \(x^2 - \frac{3x}{10} - \frac{1}{10} = 0\)
(iii) \((2x + 3)^2 = 25\)
(iv) \(m^2 + 5m + 5 = 0\)
(v) \(5m^2 + 2m+1 = 0\)
(vi) \(x^2 - 4x - 3 = 0\)
Solution:
(i) \(\frac{1}{x+5} = \frac{1}{x^2}\)
\(\implies x^2 = x+5\)
\(\implies x^2 - x - 5 = 0\)
Comparing the above equation with
\(ax^2 + bx + c = 0\), we get
\(a = 1, b = -1, c = -5\)
\(b^2 - 4ac = (-1)^2 - 4 \times 1 \times -5\)
\( = 1 + 20 = 21\)
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\( = \frac{-(-1) \pm \sqrt{21}}{2(1)}\)
\(x = \frac{1 \pm \sqrt{21}}{2}\)
\(x = \frac{1 + \sqrt{21}}{2}\) or \(x = \frac{1 - \sqrt{21}}{2}\)
\(\implies\) The roots of the given quadratic equation are \(\frac{1 + \sqrt{21}}{2}\) and \(\frac{1 - \sqrt{21}}{2}\).
In simple words: First, cross-multiply to remove fractions and rearrange the equation into standard form \(ax^2 + bx + c = 0\). Then, use the quadratic formula to find the roots, as it's often the most reliable method for such equations.

🎯 Exam Tip: When solving fractional equations, ensure you eliminate denominators correctly. The quadratic formula is applicable to all quadratic equations, making it a powerful tool, especially when factorization is not obvious.

(ii) \(x^2 - \frac{3x}{10} - \frac{1}{10} = 0\)
\(\implies 10x^2 - 3x - 1 = 0\) ...[Multiplying both sides by 10]
\(\implies 10x^2 - 5x + 2x - 1 = 0\)
\(\implies 5x(2x - 1) + 1(2x - 1) = 0\)
\(\implies (2x - 1)(5x + 1) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(\implies 2x - 1 = 0\) or \(5x + 1 = 0\)
\(\implies 2x = 1\) or \(5x = -1\)
\(\implies x = \frac{1}{2}\) or \(x = -\frac{1}{5}\)
\(\implies\) The roots of the given quadratic equation are \(\frac{1}{2}\) and \(-\frac{1}{5}\).
In simple words: Eliminate fractions by multiplying the entire equation by the LCD. Then, factorize the quadratic expression by splitting the middle term. Set each factor to zero to find the roots.

🎯 Exam Tip: Clearing denominators is crucial. Factorization by splitting the middle term is an efficient method when applicable. Practice identifying two numbers whose product is \(ac\) and sum is \(b\).

(iii) \((2x + 3)^2 = 25\)
\(\implies (2x + 3)^2 - 25 = 0\)
\(\implies (2x + 3)^2 - (5)^2 = 0\)
\(\implies (2x + 3 - 5) (2x + 3 + 5) = 0\) ..... [ \((a^2 - b^2) = (a - b) (a + b)\) ]
\(\implies (2x - 2) (2x + 8) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(\implies 2x - 2 = 0\) or \(2x + 8 = 0\)
\(\implies 2x = 2\) or \(2x = -8\)
\(\implies x = \frac{2}{2}\) or \(x = \frac{-8}{2}\)
\(\implies x = 1\) or \(x = -4\)
\(\implies\) The roots of the given quadratic equation are 1 and -4.
In simple words: Move the constant to the left side and use the difference of squares formula \((a^2 - b^2) = (a - b)(a + b)\) to factorize the equation. Then, set each factor to zero to find the roots.

🎯 Exam Tip: Recognizing the difference of squares is a quick way to solve equations in this format. Alternatively, taking the square root of both sides also works, remembering to include both positive and negative roots: \((2x+3) = \pm 5\).

(iv) \(m^2 + 5m + 5 = 0\)
Comparing the above equation with
\(am^2 + bm + c = 0\), we get
\(a = 1, b = 5, c = 5\)
\(\implies b^2 - 4ac = (5)^2 - 4 \times 1 \times 5\)
\( = 25 - 20 = 5\)
\(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\( = \frac{-5 \pm \sqrt{5}}{2(1)}\)
\(\implies m = \frac{-5 \pm \sqrt{5}}{2}\)
\(m = \frac{-5 + \sqrt{5}}{2}\) or \(m = \frac{-5 - \sqrt{5}}{2}\)
\(\implies\) The roots of the given quadratic equation are \(\frac{-5 + \sqrt{5}}{2}\) and \(\frac{-5 - \sqrt{5}}{2}\).
In simple words: This equation cannot be easily factored, so use the quadratic formula. Identify \(a, b, c\), calculate the discriminant, and substitute the values into the formula to find the two roots.

🎯 Exam Tip: The quadratic formula is a universal method for finding roots. Ensure precise substitution and calculation of the discriminant, especially when it's not a perfect square.

(v) \(5m^2 + 2m+1 = 0\)
Comparing the above equation with
\(am^2 + bm + c = 0\), we get
\(a = 5, b = 2, c = 1\)
\(\implies b^2 - 4ac = (2)^2 - 4 \times 5 \times 1\)
\( = 4 - 20\)
\( = -16\)
\(\implies b^2 - 4ac < 0\)
\(\implies\) Roots of the given quadratic equation are not real.
In simple words: Calculate the discriminant. Since the discriminant is negative (-16), the equation has no real roots; its roots are complex.

🎯 Exam Tip: Always check the discriminant first. If \(\Delta < 0\), you don't need to proceed with the quadratic formula to find the roots, as they are not real numbers.

(vi) \(x^2 - 4x - 3 = 0\)
Comparing the above equation with
\(ax^2 + bx + c = 0\), we get
\(a = 1, b = -4, c = -3\)
\(\implies b^2 - 4ac = (-4)^2 - 4 \times 1 \times -3\)
\( = 16 + 12\)
\( = 28\)
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\( = \frac{-(-4) \pm \sqrt{28}}{2(1)}\)
\( = \frac{4 \pm \sqrt{4 \times 7}}{2}\)
\( = \frac{4 \pm 2\sqrt{7}}{2}\)
\( = \frac{2(2 \pm \sqrt{7})}{2}\)
\(\implies x = 2 \pm \sqrt{7}\)
\(\implies x = 2 + \sqrt{7}\) or \(x = 2 - \sqrt{7}\)
\(\implies\) The roots of the given quadratic equation are \(2 + \sqrt{7}\) and \(2 - \sqrt{7}\).
In simple words: Use the quadratic formula to solve this equation. Identify \(a, b, c\), calculate the discriminant, and simplify the square root term to find the exact values of the two real roots.

🎯 Exam Tip: Simplify the square root in the quadratic formula as much as possible, for example, \(\sqrt{28}\) simplifies to \(2\sqrt{7}\). This makes the final answer cleaner and often required.

Question 8. Find m, if \((m - 12) x^2 + 2(m - 12) x + 2 = 0\) has real and equal roots.
Solution:
\((m - 12) x^2 + 2(m - 12)x + 2 = 0\)
Comparing the above equation with
\(ax^2 + bx + c = 0\), we get
\(a = m - 12, b = 2(m - 12), c = 2\)
\(\implies \Delta = b^2 - 4ac\)
\( = [2(m -12)]^2 - 4 \times (m - 12) \times 2\)
\( = 4(m - 12)^2 - 8(m - 12)\)
\( = 4(m - 12) [(m - 12) - 2]\)
\(\implies \Delta = 4(m - 12) (m - 14)\)
Since, the roots are real and equal.
\(\implies \Delta = 0\)
\(\implies 4(m - 12) (m - 14) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(\implies m - 12 = 0\) or \(m - 14 = 0\)
\(\implies m = 12\) or \(m = 14\)
But, if \(m = 12\), then quadratic coefficient becomes zero.
\(\implies m \neq 12\)
So \(m = 14\)
In simple words: For real and equal roots, the discriminant must be zero. Calculate the discriminant in terms of \(m\), set it to zero, and solve for \(m\). Remember that the coefficient of \(x^2\) cannot be zero for a quadratic equation.

🎯 Exam Tip: Always consider the condition that for a quadratic equation, the coefficient of the \(x^2\) term (\(a\)) cannot be zero. If a value of \(m\) makes \(a=0\), discard that solution.

Question 9. The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.
Solution:
Let \(\alpha\) and \(\beta\) be the roots of the quadratic equation.
According to the given conditions,
\(\alpha + \beta = 5\) and \(\alpha^3 + \beta^3 = 35\)
Now, \((\alpha + \beta)^3 = \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3\)
\(\implies (\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha\beta (\alpha + \beta)\)
\(\implies (5)^3 = 35 + 3\alpha\beta(5)\)
\(\implies 125 = 35 + 15\alpha\beta\)
\(\implies 125 - 35 = 15\alpha\beta\)
\(\implies 15\alpha\beta = 90\)
\(\implies \alpha\beta = \frac{90}{15}\)
\(\implies \alpha\beta = 6\)
\(\implies\) The required quadratic equation is
\(x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
\(\implies x^2 - 5x + 6 = 0\)
In simple words: Use the given sum of roots and the identity \((\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta)\) to find the product of roots \(\alpha\beta\). Once both the sum and product of roots are known, form the quadratic equation using the standard formula.

🎯 Exam Tip: Be familiar with algebraic identities like the sum of cubes factorization. This allows you to find \(\alpha\beta\) from the given information, which is then used to construct the quadratic equation.

Question 10. Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation
\(2x^2 + 2(p + q)x + p^2 + q^2 = 0\).
Solution:
The given quadratic equation is
\(2x^2 + 2(p+q)x + p^2 + q^2 = 0\)
Here, \(a = 2, b = 2(p+q), c = p^2 + q^2\)
\(\alpha + \beta = -\frac{b}{a} = -\frac{2(p+q)}{2} = -(p+q)\)
and \(\alpha\beta = \frac{c}{a} = \frac{p^2 + q^2}{2}\)
\((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\)
\((\alpha - \beta)^2 = [-(p + q)]^2 - 4 \times \frac{p^2 + q^2}{2}\)
\( = (p+q)^2 - 2(p^2 + q^2)\)
\( = p^2 + 2pq + q^2 - 2p^2 - 2q^2\)
\( = -p^2 + 2pq - q^2\)
\( = -(p^2 - 2pq + q^2)\)
\( = -(p - q)^2\)
...[ \((a^2-2ab+b^2) = (a - b)^2\) ]
According to the given condition,
Roots of the required quadratic equation are
\((\alpha + \beta)^2\) and \((\alpha - \beta)^2\)
Now, Sum of the roots
\( = (\alpha + \beta)^2 + (\alpha - \beta)^2\)
\( = [-(p+q)]^2 - (p-q)^2\)
\( = (p+q)^2 - (p-q)^2\)
\( = p^2 + 2pq + q^2 - (p^2 - 2pq + q^2)\)
.....[ \((a+b)^2 = a^2 + 2ab + b^2\) and \((a - b)^2 = a^2 - 2ab + b^2\) ]
\( = p^2 + 2pq + q^2 - p^2 + 2pq - q^2\)
\( = 4pq\)
Product of the roots
\( = (\alpha + \beta)^2 (\alpha - \beta)^2\)
\( = [-(p+q)]^2 [-(p - q)^2]\)
\( = (p+q)^2 (p-q)^2\)
\( = [(p+q) (p-q)]^2 = (p^2 - q^2)^2\)
The required quadratic equation is
\(x^2 - [(\alpha + \beta)^2 + (\alpha-\beta)^2]x + [(\alpha + \beta)^2 (\alpha - \beta)^2] = 0\)
\(x^2 - (4pq)x + (p^2 - q^2)^2 = 0\)
In simple words: First, find the sum and product of roots (\(\alpha+\beta\) and \(\alpha\beta\)) for the given equation using coefficient relationships. Then, calculate \((\alpha - \beta)^2\). The new roots are \((\alpha + \beta)^2\) and \((\alpha - \beta)^2\). Use these new roots to find their sum and product, and finally, construct the new quadratic equation.

🎯 Exam Tip: This problem involves nested use of root properties. Keep track of the original roots' sum and product, then use them to calculate the sum and product of the *new* roots. Careful algebraic simplification is key.

Question 11. Mukund possesses Rs. 50 more than what Sagar possesses. The product of the amount they have is 15,000. Find the amount each one has.
Solution:
Let the amount Sagar possesses be Rs. \(x\).
\(\implies\) the amount Mukund possesses = Rs. \((x + 50)\)
According to the given condition,
\(x(x + 50) = 15000\)
\(\implies x^2 + 50x - 15000 = 0\)
\(\implies x^2 + 150x - 100x - 15000 = 0\)
\(\implies x(x + 150) - 100(x + 150) = 0\)
\(\implies (x + 150)(x - 100) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(\implies x + 150 = 0\) or \(x - 100 = 0\)
\(\implies x = -150\) or \(x = 100\)
But, amount cannot be negative.
\(\implies x = 100\) and \(x + 50 = 100 + 50 = 150\)
\(\implies\) The amount possessed by Sagar and Mukund are Rs. 100 and Rs. 150 respectively.
In simple words: Set up an equation based on the given information: Sagar's amount is \(x\), Mukund's is \(x+50\), and their product is 15000. Solve the resulting quadratic equation for \(x\), discarding any negative solutions as amount cannot be negative.

🎯 Exam Tip: In word problems, define variables clearly. When solving for real-world quantities like money or time, ensure your answer makes physical sense (e.g., discard negative values if not applicable).

Question 12. The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.
Solution:
Let the numbers be \(x\) and \(y\) (\(x > y\)).
According to the given condition,
\(x^2 - y^2 = 120\) ...(i)
\(y^2 = 2x\) ...(ii)
Substituting \(y^2 = 2x\) in equation (i), we get
\(x^2 - 2x = 120\)
\(\implies x^2 - 2x - 120 = 0\)
\(\implies x^2 - 12x + 10x - 120 = 0\)
\(\implies x(x - 12) + 10(x - 12) = 0\)
\(\implies (x - 12)(x + 10) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(\implies x - 12 = 0\) or \(x + 10 = 0\)
\(\implies x = 12\) or \(x = -10\)
But \(x \neq -10\)
as, \(y^2 = 2x = 2(-10) = -20\) ...[Since, the square of number cannot be negative]
\(\implies x = 12\)
Smaller number \(= y^2 = 2x\)
\(\implies y^2 = 2 \times 12\)
\(\implies y^2 = 24\)
\(\implies y = \pm \sqrt{24}\) ...[Taking square root of both sides]
\(\implies\) The smaller number is \(\sqrt{24}\) and greater number is 12 or the smaller number is \(-\sqrt{24}\) and greater number is 12.
In simple words: Formulate two equations from the given conditions: \(x^2 - y^2 = 120\) and \(y^2 = 2x\). Substitute the second equation into the first to get a quadratic equation in \(x\). Solve for \(x\), discard invalid solutions, then find \(y\).

🎯 Exam Tip: Carefully set up simultaneous equations from the word problem. When solving for \(y\), remember that the square root can be positive or negative, but ensure \(y^2\) itself is non-negative, and values correspond to the "smaller" and "greater" conditions.

Question 13. Ranjana wants to distribute 540 oranges among some students. If 30 students were more each would get 3 oranges less. Find the number of students.
Solution:
Let the number of students be \(x\).
Total number of oranges = 540
\(\implies\) the number of oranges each student gets = \(\frac{540}{x}\)
If there were 30 more students, the total number of students = \((x + 30)\) and the total number of oranges each student gets
\( = \frac{540}{(x+30)}\)
According to the given condition,
\(\frac{540}{x} - \frac{540}{x+30} = 3\)
\(\implies 540 (\frac{1}{x} - \frac{1}{x+30}) = 3\)
\(\implies 540 (\frac{x+30-x}{x(x+30)}) = 3\)
\(\implies 540 (\frac{30}{x^2+30x}) = 3\)
\(\implies \frac{540 \times 30}{x^2+30x} = 3\)
\(\implies 540 \times 10 = x^2 + 30x\) ...[Dividing both sides by 3 and cross-multiplying]
\(\implies 5400 = x^2 + 30x\)
\(\implies x^2 + 30x - 5400 = 0\)
\(\implies x^2 + 90x - 60x - 5400 = 0\)
\(\implies x(x + 90) - 60(x + 90) = 0\)
\(\implies (x + 90) (x - 60) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(\implies x + 90 = 0\) or \(x - 60 = 0\)
\(\implies x = -90\) or \(x = 60\)
But, number of students cannot be negative,
\(x = 60\)
\(\implies\) The total number of students is 60.
In simple words: Define \(x\) as the initial number of students. Set up an equation representing the orange distribution per student for both scenarios (original \(x\) students and \(x+30\) students), and the difference of 3 oranges. Solve the resulting quadratic equation for \(x\), discarding any negative solutions.

🎯 Exam Tip: Word problems involving quantities like "per person" often lead to fractional equations. Simplify carefully, clear denominators, and ensure your solution fits the context (e.g., number of students cannot be negative). Factorization or the quadratic formula can be used to solve.

Question 14. Mr. Dinesh owns an rectangular agricultural farm at village Talvel. The length of the farm is 10 metre more than twice the breadth. In order to harvest rain water, he dug a square shaped pond inside the farm. The side of pond is \(\frac{1}{3}\) of the breadth of the farm. The area of the farm is 20 times the area of the pond. Find the length and breadth of the farm and side of the pond.
Solution:
Let the breadth of the rectangular farm be x m.
\(\therefore\) Length of rectangular farm = (2x + 10) m
Area of rectangular farm = Length \(\times\) Breadth
= (2x + 10) \(\times\) x
= (2x² + 10x) sq. m
Now, side of square shaped pond = \(\frac{x}{3}\) m
In simple words: This problem involves setting up equations for the dimensions and areas of a rectangular farm and a square pond based on the given relationships. We will use the area relationship to form a quadratic equation and solve for the breadth, then find the length and pond side.

🎯 Exam Tip: Pay close attention to unit consistency (meters, square meters) and correctly formulate the quadratic equation from the given word problem relationships.

\(\therefore\) Area of square shaped pond = (side)²
= \( (\frac{x}{3})^2 \) m
= \( \frac{x^2}{9} \) m
According to the given condition,
Area of rectangular farm = 20 \(\times\) Area of pond

\( 2x^2 + 10x = 20 \times (\frac{x^2}{9}) \)
\( \implies x^2 + 5x = \frac{10x^2}{9} \) ...[Dividing both sides by 2]
\( \implies 9x^2 + 45x = 10x^2 \) ...[Multiplying both sides by 9]
\( \implies 10x^2 - 9x^2 - 45x = 0 \)
\( \implies x^2 - 45x = 0 \)
\( \implies x(x - 45) = 0 \)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(\therefore\) x = 0 or x - 45 = 0
\(\therefore\) x = 0 or x = 45
But, breadth of the rectangular farm cannot be zero,
\(\therefore\) x = 45
Length of rectangular farm
= 2(45) + 10 = 90 + 10 = 100 m
Side of the pond = \(\frac{x}{3}\) = \(\frac{45}{3}\) = 15 m
\(\therefore\) Length and breadth of the farm and the side of the pond are 100 m, 45 m and 15 m respectively.
In simple words: After setting up equations for the area of the farm and the pond, we found that the area of the farm is 20 times the area of the pond. Solving the resulting quadratic equation \( x^2 - 45x = 0 \) gives \( x = 0 \) or \( x = 45 \). Since breadth cannot be zero, \( x = 45 \) m is the breadth. This leads to a length of 100 m and a pond side of 15 m.

🎯 Exam Tip: Always verify that the physical quantities obtained (like breadth, length, time) are positive and make physical sense in the context of the problem. Negative or zero values for dimensions are usually invalid.

Question 15. A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap take to fill the tank completely?
Solution:
Let the larger tap take x hours to fill the tank completely.
\(\therefore\) Part of tank filled by the larger tap in 1 hour = \(\frac{1}{x}\)
Also, the smaller tap takes (x + 3) hours to fill the tank completely.
\(\therefore\) Part of tank filled by the smaller tap in 1 hour = \(\frac{1}{x+3}\)
\(\therefore\) Part of tank filled by both the taps in 1 hour
= \( (\frac{1}{x} + \frac{1}{x+3}) \)
In simple words: We define the time taken by the larger tap as 'x' hours. This means its filling rate is \( \frac{1}{x} \) per hour. The smaller tap takes \( x+3 \) hours, so its rate is \( \frac{1}{x+3} \) per hour. When both are open, their combined rate is the sum of their individual rates.

🎯 Exam Tip: For 'work and time' problems involving rates, remember that if an entity takes 't' time to complete a task, its rate of work is \( \frac{1}{t} \) per unit of time.

But, the tank gets filled in 2 hours by both the taps.
\(\therefore\) Part of tank filled by both the taps in 1 hour = \(\frac{1}{2}\)
According to the given condition,

\( \frac{1}{x} + \frac{1}{x+3} = \frac{1}{2} \)
\( \implies \frac{x+3+x}{x(x+3)} = \frac{1}{2} \)
\( \implies \frac{2x+3}{x(x+3)} = \frac{1}{2} \)
\(\therefore\) 2(2x + 3) = x(x + 3)
\(\therefore\) \( 4x + 6 = x^2 + 3x \)
\(\therefore\) \( x^2 + 3x - 4x - 6 = 0 \)
\(\therefore\) \( x^2 - x - 6 = 0 \)
\(\therefore\) \( x^2 - 3x + 2x - 6 = 0 \)
\(\therefore\) \( x(x - 3) + 2(x - 3) = 0 \)
\(\therefore\) \( (x - 3)(x + 2) = 0 \)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(\therefore\) x - 3 = 0 or x + 2 = 0
\(\therefore\) x = 3 or x = -2
But, time cannot be negative.
\(\therefore\) x = 3 and x + 3 = 3 + 3 = 6
\(\therefore\) The larger tap takes 3 hours and the smaller tap takes 6 hours to fill the tank completely.
In simple words: The combined rate of the two taps filling the tank is \( \frac{1}{2} \) per hour. We set up the equation \( \frac{1}{x} + \frac{1}{x+3} = \frac{1}{2} \) and solved it to get \( x = 3 \) or \( x = -2 \). Since time cannot be negative, the larger tap takes 3 hours and the smaller tap takes \( 3+3=6 \) hours to fill the tank.

🎯 Exam Tip: Remember to always disregard negative solutions in contexts where quantities like time, distance, or physical dimensions must be positive.