CBSE Class 10 Mathematics Quadratic Equation Worksheet Set 06

CBSE Class 10 Quadratic Equation (13). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Very Short Answer Type Questions

Question Q1. Show that x=-3 is the solution of equation x²+6x+9=0 .
Answer: To show that \( x = -3 \) is a solution (or root) of the quadratic equation \( x^2 + 6x + 9 = 0 \), we substitute \( x = -3 \) into the left-hand side (L.H.S.) of the equation:
\[ \text{L.H.S.} = (-3)^2 + 6(-3) + 9 \]
\[ \text{L.H.S.} = 9 - 18 + 9 \]
\[ \text{L.H.S.} = 18 - 18 = 0 \]
Since \( \text{L.H.S.} = \text{R.H.S.} \), \( x = -3 \) satisfies the equation.
Hence, \( x = -3 \) is a solution of the equation.
In simple words: We plug -3 into the x-positions of the equation. The calculation simplifies to zero, which matches the right side and proves it is a valid solution.

Exam Tip: Always state "L.H.S. = R.H.S." explicitly at the end of a "show that" or verification question to conclude the proof formally.

Question Q2. For what value of k are the roots of quadratic equation 3x²+2kx+27=0 real and equal .
Answer: The given quadratic equation is \( 3x^2 + 2kx + 27 = 0 \).
Comparing this with the standard form \( ax^2 + bx + c = 0 \), we get:
\( a = 3, b = 2k, c = 27 \)

For a quadratic equation to have real and equal roots, its discriminant (\( D \)) must be equal to zero:
\[ D = b^2 - 4ac = 0 \]
Substitute the values into the formula:
\[ (2k)^2 - 4(3)(27) = 0 \]
\[ 4k^2 - 324 = 0 \]
\[ 4k^2 = 324 \]
\[ k^2 = 81 \]
\[ k = \pm \sqrt{81} = \pm 9 \]
Therefore, the roots of the equation are real and equal when \( k = \pm 9 \).
In simple words: For the roots to be real and equal, the discriminant (\( b^2 - 4ac \)) must be zero. Solving this condition for the given equation yields \( k = 9 \) or \( k = -9 \).

Exam Tip: Do not forget to write both the positive and negative signs (\( \pm \)) when taking the square root of a variable.

Question Q3. Find the discriminant of the quadratic equation 3√3 x+10 x+√3=0
Answer: The given quadratic equation is \( 3\sqrt{3}x^2 + 10x + \sqrt{3} = 0 \).
Comparing this with the standard form \( ax^2 + bx + c = 0 \), we have:
\( a = 3\sqrt{3}, b = 10, c = \sqrt{3} \)

The discriminant (\( D \)) of a quadratic equation is defined as:
\[ D = b^2 - 4ac \]
Substitute the coefficients:
\[ D = 10^2 - 4(3\sqrt{3})(\sqrt{3}) \]
\[ D = 100 - 12(3) \]
\[ D = 100 - 36 = 64 \]
Therefore, the discriminant of the quadratic equation is 64.
In simple words: We find the discriminant using the formula \( b^2 - 4ac \). Plunging the values from our equation into this formula gives a final value of 64.

Exam Tip: Be careful when multiplying radical terms; remember that \( \sqrt{3} \times \sqrt{3} = 3 \).

Question Q4. Write the nature of roots of quadratic equation 4x²+4√3 x+3=0
Answer: The given quadratic equation is \( 4x^2 + 4\sqrt{3}x + 3 = 0 \).
Comparing with \( ax^2 + bx + c = 0 \), we identify the coefficients:
\( a = 4, b = 4\sqrt{3}, c = 3 \)

To find the nature of the roots, we calculate the discriminant (\( D \)):
\[ D = b^2 - 4ac \]
\[ D = (4\sqrt{3})^2 - 4(4)(3) \]
\[ D = (16 \times 3) - 48 \]
\[ D = 48 - 48 = 0 \]
Since the discriminant \( D = 0 \), the quadratic equation has real and equal roots.
In simple words: We calculate the discriminant, which turns out to be exactly zero. This tells us that the roots of the equation are real and identical to each other.

Exam Tip: There are three possible natures of roots depending on \( D \): \( D > 0 \) (real and distinct), \( D = 0 \) (real and equal), and \( D < 0 \) (no real roots).

Question Q5. Which of the following are quadratic equations- a) x³-x=x²+2 b) √x +4 =(x +1) c) (x+1)(x²-2)=(x+3)³
Answer: A quadratic equation is a polynomial equation of degree 2, which can be written in the standard form \( ax^2 + bx + c = 0 \) (where \( a \neq 0 \)).

(a) For the equation \( x^3 - x = x^2 + 2 \):
The highest power of the variable \( x \) is 3. Since the degree is 3, this is a cubic equation, not a quadratic equation.

(b) For the equation \( \sqrt{x} + 4 = x + 1 \):
The term \( \sqrt{x} = x^{1/2} \) has a fractional exponent. A quadratic equation must have only non-negative integer powers of \( x \), so this is not a quadratic equation.

(c) For the equation \( (x + 1)(x^2 - 2) = (x + 3)^3 \):
Expanding both sides:
L.H.S.: \( x^3 + x^2 - 2x - 2 \)
R.H.S.: \( x^3 + 9x^2 + 27x + 27 \)
Equating them:
\( x^3 + x^2 - 2x - 2 = x^3 + 9x^2 + 27x + 27 \)
Subtract \( x^3 \) from both sides:
\( 8x^2 + 29x + 29 = 0 \)
Since the \( x^3 \) terms cancel out, the remaining polynomial is of degree 2. This is a quadratic equation.
In simple words: A quadratic equation must have \( x^2 \) as its highest power. Only equation (c) simplifies to a degree of 2 because the \( x^3 \) terms on both sides cancel out.

Exam Tip: Always fully expand and simplify the expressions on both sides of the equation before concluding its degree.

Short Answer Type Questions-I

Question Q6. Solve for x: a) x²-2(a²+b²)x+ (a²- b²)²=0 b) 4/x -3 = 5/2x +3 , x = 0 , -3/2 c) √2 x²+7 x + 5√2 =0 d) (a + b)²x²+8 (a²- b² ) x+16(a-b)²=0 e) \( \frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \), \( a \neq 0 \), \( b \neq 0 \), \( x \neq 0 \)
Answer: Let's solve each equation step-by-step:

(a) \( x^2 - 2(a^2 + b^2)x + (a^2 - b^2)^2 = 0 \)
Using the quadratic formula where \( A = 1, B = -2(a^2 + b^2), C = (a^2 - b^2)^2 \):
The discriminant is:
\[ D = B^2 - 4AC = 4(a^2 + b^2)^2 - 4(a^2 - b^2)^2 \]
\[ D = 4[(a^2 + b^2)^2 - (a^2 - b^2)^2] = 4[4a^2b^2] = 16a^2b^2 \]
Now, finding the roots:
\[ x = \frac{-B \pm \sqrt{D}}{2A} = \frac{2(a^2 + b^2) \pm 4ab}{2} = (a^2 + b^2) \pm 2ab \]
This gives two roots:
- \( x_1 = a^2 + b^2 + 2ab = (a + b)^2 \)
- \( x_2 = a^2 + b^2 - 2ab = (a - b)^2 \)

(b) \( \frac{4}{x} - 3 = \frac{5}{2x + 3} \)
Multiply both sides to obtain a single denominator:
\[ \frac{4 - 3x}{x} = \frac{5}{2x + 3} \]
Cross-multiplying:
\[ (4 - 3x)(2x + 3) = 5x \implies 8x + 12 - 6x^2 - 9x = 5x \]
\[ -6x^2 - x + 12 = 5x \implies 6x^2 + 6x - 12 = 0 \]
Divide by 6:
\[ x^2 + x - 2 = 0 \implies (x + 2)(x - 1) = 0 \implies x = 1 \text{ or } x = -2 \]

(c) \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
We split the middle term:
The product of the outer terms is \( \sqrt{2} \times 5\sqrt{2} = 10 \). We can write \( 7x = 5x + 2x \):
\[ \sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0 \]
\[ \sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0 \]
\[ (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \]
This gives:
- \( x = -\frac{5}{\sqrt{2}} \)
- \( x = -\sqrt{2} \)

(d) \( (a + b)^2x^2 + 8(a^2 - b^2)x + 16(a-b)^2 = 0 \)
We rewrite the terms to identify a perfect square:
Let \( (a+b)x = X \) and \( 4(a-b) = Y \). The equation matches \( X^2 + 2XY + Y^2 = 0 \):
\[ [(a+b)x + 4(a-b)]^2 = 0 \implies (a+b)x + 4(a-b) = 0 \]
\[ x = -\frac{4(a-b)}{a+b} = \frac{4(b-a)}{a+b} \]
Since it is a perfect square, both roots are equal to \( \frac{4(b-a)}{a+b} \).

(e) \( \frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \)
Subtract \( \frac{1}{x} \) from both sides:
\[ \frac{1}{a+b+x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \implies \frac{x - (a+b+x)}{x(a+b+x)} = \frac{a+b}{ab} \]
\[ \frac{-(a+b)}{x(a+b+x)} = \frac{a+b}{ab} \implies \frac{-1}{ax+bx+x^2} = \frac{1}{ab} \]
\[ -ab = x^2 + ax + bx \implies x^2 + ax + bx + ab = 0 \]
\[ x(x+a) + b(x+a) = 0 \implies (x+a)(x+b) = 0 \]
This yields the roots:
- \( x = -a \)
- \( x = -b \)
In simple words: We solve each sub-question using algebraic identities, factorizing, or standard quadratic formulas to find the values of x.

Exam Tip: For part (e), moving the \( \frac{1}{x} \) term to the left-hand side before taking a common denominator is a crucial step that makes the factorization much simpler.

Question Q7. If one root of equation 3 x²- kx -2 =0 is 2, find the value of k.Also find the other root.
Answer: Since \( x = 2 \) is a root of the quadratic equation \( 3x^2 - kx - 2 = 0 \), it must satisfy the equation.
Substituting \( x = 2 \):
\[ 3(2)^2 - k(2) - 2 = 0 \]
\[ 3(4) - 2k - 2 = 0 \]
\[ 12 - 2k - 2 = 0 \]
\[ 10 - 2k = 0 \implies 2k = 10 \implies k = 5 \]

Now substitute \( k = 5 \) back into the original quadratic equation:
\[ 3x^2 - 5x - 2 = 0 \]
We factorize by splitting the middle term:
\[ 3x^2 - 6x + x - 2 = 0 \]
\[ 3x(x - 2) + 1(x - 2) = 0 \]
\[ (3x + 1)(x - 2) = 0 \]
This gives the roots:
- \( x = 2 \) (the given root)
- \( x = -\frac{1}{3} \) (the other root)

Therefore, \( k = 5 \) and the other root is \( -\frac{1}{3} \).
In simple words: We substitute 2 into the equation to solve for \( k \), which gives 5. Then, factorizing the resulting equation \( 3x^2 - 5x - 2 = 0 \) gives the second root as \( -1/3 \).

Exam Tip: You can also find the second root quickly by using the product of roots formula: \( \alpha \beta = \frac{c}{a} = -\frac{2}{3} \). Since \( \alpha = 2 \), \( 2\beta = -\frac{2}{3} \implies \beta = -\frac{1}{3} \).

Question Q8. If -5 is a root of the quadratic equation 2x²+px -15 =0 and the quadratic equation p(x²+x) + k =0 has equal roots,find the value of k.
Answer: Since \( x = -5 \) is a root of \( 2x^2 + px - 15 = 0 \), it satisfies the equation:
\[ 2(-5)^2 + p(-5) - 15 = 0 \]
\[ 2(25) - 5p - 15 = 0 \]
\[ 50 - 5p - 15 = 0 \]
\[ 35 - 5p = 0 \implies 5p = 35 \implies p = 7 \]

Now substitute \( p = 7 \) into the second quadratic equation:
\[ p(x^2 + x) + k = 0 \]
\[ 7(x^2 + x) + k = 0 \]
\[ 7x^2 + 7x + k = 0 \]

This equation has equal roots, meaning its discriminant (\( D \)) must be zero. Here, \( a = 7, b = 7, c = k \):
\[ D = b^2 - 4ac = 0 \]
\[ 7^2 - 4(7)(k) = 0 \]
\[ 49 - 28k = 0 \]
\[ 28k = 49 \]
\[ k = \frac{49}{28} = \frac{7}{4} \]
Therefore, the value of \( k \) is \( \frac{7}{4} \).
In simple words: We plug in the root -5 to find \( p = 7 \). Substituting this into the second equation and setting the discriminant to zero allows us to solve for \( k = 7/4 \).

Exam Tip: In questions with two connected equations, find the numerical value from the first equation completely before starting on the second equation.

Question Q9. Find the value of k for which the roots of quadratic equation are equal (k-4)x² + 2(k-4)x+2=0
Answer: The given quadratic equation is \( (k - 4)x^2 + 2(k - 4)x + 2 = 0 \).
Comparing this with \( ax^2 + bx + c = 0 \), we have:
- \( a = k - 4 \)
- \( b = 2(k - 4) \)
- \( c = 2 \)

For the roots to be equal, the discriminant (\( D \)) must be equal to zero:
\[ D = b^2 - 4ac = 0 \]
\[ [2(k - 4)]^2 - 4(k - 4)(2) = 0 \]
\[ 4(k - 4)^2 - 8(k - 4) = 0 \]
Factor out \( 4(k - 4) \):
\[ 4(k - 4)[(k - 4) - 2] = 0 \]
\[ 4(k - 4)(k - 6) = 0 \]
This gives:
- \( k = 4 \) or \( k = 6 \)

However, if \( k = 4 \), the coefficient of \( x^2 \) becomes \( a = 4 - 4 = 0 \), which means the equation is no longer quadratic. Thus, \( k = 4 \) is rejected.
Therefore, the only valid value of \( k \) is 6.
In simple words: Setting the discriminant to zero gives \( k = 4 \) or \( k = 6 \). We reject \( k = 4 \) because it would make the \( x^2 \) coefficient zero, so \( k = 6 \) is the final answer.

Exam Tip: Always check if your calculated values of the variable make the leading coefficient (\( a \)) equal to zero, as this is a common trick used to catch students out.

Question Q10. Find the value of p so that the quadratic equation px(x-3)+9=0 has two equal roots.
Answer: First, expand the given equation \( px(x - 3) + 9 = 0 \) into standard form:
\[ px^2 - 3px + 9 = 0 \]
Here, \( a = p \), \( b = -3p \), and \( c = 9 \).

Since the quadratic equation has two equal roots, its discriminant (\( D \)) must be equal to zero:
\[ D = b^2 - 4ac = 0 \]
\[ (-3p)^2 - 4(p)(9) = 0 \]
\[ 9p^2 - 36p = 0 \]
Factor out \( 9p \):
\[ 9p(p - 4) = 0 \]
This gives:
- \( p = 0 \) or \( p = 4 \)

For the equation to remain quadratic, the leading coefficient \( a = p \) cannot be zero. Thus, \( p = 0 \) is rejected.
Therefore, the required value of \( p \) is 4.
In simple words: Expanding the equation and setting the discriminant to zero yields \( p = 0 \) or \( p = 4 \). Since \( p = 0 \) would destroy the quadratic term, the correct answer is \( p = 4 \).

Exam Tip: Remember to always state why a solution is rejected (e.g., "p cannot be 0 because the equation must be quadratic").

Question Q11. Find the value of k for which the equation x²+kx+64=0 and x²-8x+k=0 will both have Real roots.
Answer: For a quadratic equation to have real roots, its discriminant (\( D \)) must be greater than or equal to zero (\( D \geq 0 \)).

1. **For the first equation \( x^2 + kx + 64 = 0 \):**
- \( a = 1, b = k, c = 64 \)
\[ D_1 = b^2 - 4ac \geq 0 \implies k^2 - 4(1)(64) \geq 0 \]
\[ k^2 - 256 \geq 0 \implies k^2 \geq 256 \implies k \geq 16 \text{ or } k \leq -16 \tag{Condition 1} \]

2. **For the second equation \( x^2 - 8x + k = 0 \):**
- \( a = 1, b = -8, c = k \)
\[ D_2 = b^2 - 4ac \geq 0 \implies (-8)^2 - 4(1)(k) \geq 0 \]
\[ 64 - 4k \geq 0 \implies 4k \leq 64 \implies k \leq 16 \tag{Condition 2} \]

For both equations to have real roots simultaneously, we find the intersection of Condition 1 and Condition 2.
The only value of \( k \) that satisfies both \( (k \geq 16 \text{ or } k \leq -16) \) and \( k \leq 16 \) is:
- \( k = 16 \) or \( k \leq -16 \).
Therefore, \( k = 16 \) is the common value of \( k \) (or \( k \leq -16 \)).
In simple words: Setting \( D \geq 0 \) for both equations yields \( k^2 \geq 256 \) (which means \( k \geq 16 \) or \( k \leq -16 \)) and \( k \leq 16 \). Combining these conditions gives the solution \( k = 16 \) (or \( k \leq -16 \)).

Exam Tip: Be very precise with inequality signs (\( \geq \) vs \( \leq \)) when combining multiple conditions.

Question Q12. In the following determine the set of values of p for which the given equation has real roots- a) 2x²+px+3=0 b) px²+4x+1=0
Answer: For a quadratic equation to have real roots, its discriminant (\( D \)) must satisfy the inequality \( D \geq 0 \).

(a) For the equation \( 2x^2 + px + 3 = 0 \):
- \( a = 2, b = p, c = 3 \)
\[ D = b^2 - 4ac \geq 0 \implies p^2 - 4(2)(3) \geq 0 \]
\[ p^2 - 24 \geq 0 \implies p^2 \geq 24 \]
Taking the square root on both sides:
\[ p \geq 2\sqrt{6} \text{ or } p \leq -2\sqrt{6} \]

(b) For the equation \( px^2 + 4x + 1 = 0 \):
- \( a = p, b = 4, c = 1 \)
\[ D = b^2 - 4ac \geq 0 \implies 4^2 - 4(p)(1) \geq 0 \]
\[ 16 - 4p \geq 0 \implies 4p \leq 16 \implies p \leq 4 \]
Also, for the equation to be quadratic, the leading coefficient \( a = p \) cannot be zero:
- \( p \neq 0 \)
Thus, the set of values is \( p \leq 4 \) and \( p \neq 0 \).
In simple words: To ensure real roots, we solve \( b^2 - 4ac \geq 0 \). For part (a), this gives \( p \geq 2\sqrt{6} \) or \( p \leq -2\sqrt{6} \). For part (b), it gives \( p \leq 4 \) (with \( p \neq 0 \)).

Exam Tip: Remember to write down \( p \neq 0 \) for the leading coefficient in part (b), as quadratic terms can never have a zero coefficient.

Short Answer Type Questions-II

Question Q13. If the roots of the equation (b-c)x²+(c-a)x+(a-b)=0 are equal then prove that 2b=a+c.
Answer: Comparing the equation \( (b-c)x^2 + (c-a)x + (a-b) = 0 \) with the standard form \( Ax^2 + Bx + C = 0 \), we have:
- \( A = b - c \)
- \( B = c - a \)
- \( C = a - b \)

Notice that the sum of the coefficients is:
\[ A + B + C = (b - c) + (c - a) + (a - b) = 0 \]
This implies that \( x = 1 \) is always a root of this equation.
Since we are given that the roots of the equation are equal, both roots must be equal to 1.
The product of the roots is:
\[ \text{Product of roots} = 1 \times 1 = \frac{C}{A} \implies 1 = \frac{a - b}{b - c} \]
\[ \implies b - c = a - b \implies 2b = a + c \]
Hence, proven.
In simple words: Since the sum of the coefficients is zero, 1 is a root of the equation. Since the roots are equal, both roots must be 1. Equating their product \( \frac{a-b}{b-c} \) to 1 proves \( 2b = a+c \).

Exam Tip: The "sum of coefficients is zero" trick is a beautiful and elegant shortcut that saves you from doing highly complex discriminant expansions.

Question Q14. Find the values of a and b if the sum and product of roots of the equation 4ax²+4bx+3=0 are 1/2 and 3/16 .
Answer: The given quadratic equation is \( 4ax^2 + 4bx + 3 = 0 \).
We are given:
- Sum of roots = \( \frac{1}{2} \)
- Product of roots = \( \frac{3}{16} \)

The formulas for sum and product of roots of \( Ax^2 + Bx + C = 0 \) are:
- \( \text{Sum of roots} = -\frac{B}{A} \)
- \( \text{Product of roots} = \frac{C}{A} \)

Substituting the coefficients \( A = 4a, B = 4b, C = 3 \):

1. **For the product of roots:**
\[ \frac{3}{4a} = \frac{3}{16} \implies 4a = 16 \implies a = 4 \]

2. **For the sum of roots:**
\[ -\frac{4b}{4a} = \frac{1}{2} \implies -\frac{b}{a} = \frac{1}{2} \]
Substitute \( a = 4 \):
\[ -\frac{b}{4} = \frac{1}{2} \implies b = -2 \]
Therefore, \( a = 4 \) and \( b = -2 \).
In simple words: The product formula \( \frac{c}{a} = \frac{3}{4a} = \frac{3}{16} \) immediately gives \( a = 4 \). Plugging this into the sum formula \( -\frac{b}{a} = \frac{1}{2} \) gives \( b = -2 \).

Exam Tip: Always solve the product of roots equation first in this setup, as the constant term \( C \) is fully known, which lets you solve directly for \( a \).

Question Q15. If the roots of the equation (c²-ab)x²-2(a²-bc)x+b²-ac=0 Are equal prove that either a=0 or a³+b³+c³=3abc
Answer: For the roots of the equation \( (c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0 \) to be equal, the discriminant (\( D \)) must be zero.
Here, \( A = c^2 - ab \), \( B = -2(a^2 - bc) \), and \( C = b^2 - ac \).
\[ D = B^2 - 4AC = 0 \]
\[ [-2(a^2 - bc)]^2 - 4(c^2 - ab)(b^2 - ac) = 0 \]
\[ 4(a^2 - bc)^2 - 4(c^2 - ab)(b^2 - ac) = 0 \]
Divide both sides by 4 and expand the terms:
\[ (a^4 - 2a^2bc + b^2c^2) - (b^2c^2 - ac^3 - ab^3 + a^2bc) = 0 \]
\[ a^4 - 2a^2bc + b^2c^2 - b^2c^2 + ac^3 + ab^3 - a^2bc = 0 \]
Group and simplify the remaining terms:
\[ a^4 + ab^3 + ac^3 - 3a^2bc = 0 \]
Factor out \( a \):
\[ a(a^3 + b^3 + c^3 - 3abc) = 0 \]
This implies:
- \( a = 0 \), or
- \( a^3 + b^3 + c^3 - 3abc = 0 \implies a^3 + b^3 + c^3 = 3abc \)
Hence, proven.
In simple words: Setting the discriminant to zero and expanding the algebraic terms leads to \( a(a^3 + b^3 + c^3 - 3abc) = 0 \). This directly proves that either \( a = 0 \) or \( a^3 + b^3 + c^3 = 3abc \).

Exam Tip: Take your time when expanding terms like \( (c^2 - ab)(b^2 - ac) \) to ensure no negative signs are dropped.

Question Q16. If the roots of the equation (1+m²)x +2mcx+(c²-a²)=0 Prove that c²=a²(1+m²)
Answer: Though the question contains a typo in the first term, we treat the equation in its mathematically correct form \( (1+m^2)x^2 + 2mcx + (c^2 - a^2) = 0 \).
Since the roots of this quadratic equation are equal, the discriminant (\( D \)) must be equal to zero.
Here, \( A = 1+m^2 \), \( B = 2mc \), and \( C = c^2 - a^2 \).
\[ D = B^2 - 4AC = 0 \]
\[ (2mc)^2 - 4(1+m^2)(c^2 - a^2) = 0 \]
\[ 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0 \]
Divide both sides by 4:
\[ m^2c^2 - (c^2 - a^2 + m^2c^2 - m^2a^2) = 0 \]
\[ m^2c^2 - c^2 + a^2 - m^2c^2 + m^2a^2 = 0 \]
The \( m^2c^2 \) terms cancel out:
\[ -c^2 + a^2 + m^2a^2 = 0 \implies c^2 = a^2(1 + m^2) \]
Hence, proven.
In simple words: Setting the discriminant of \( (1+m^2)x^2 + 2mcx + (c^2-a^2) = 0 \) to zero simplifies to \( c^2 = a^2 + m^2a^2 \). Factoring out \( a^2 \) gives the required proof \( c^2 = a^2(1+m^2) \).

Exam Tip: This is a very common coordinate geometry result for the tangency of a line to a circle; memorizing the algebraic steps is highly recommended.

Question Q17. Solve for x: a) \( \frac{2}{x+1} + \frac{3}{2(x-2)} = \frac{23}{5x} \), \( x \neq 0, -1, 2 \) b) \( \frac{x+1}{x-1} + \frac{x-2}{x+2} = 3 \), \( x \neq 1, -2 \)
Answer:
(a) \( \frac{2}{x+1} + \frac{3}{2(x-2)} = \frac{23}{5x} \)
Take a common denominator on the left-hand side:
\[ \frac{2[2(x-2)] + 3(x+1)}{2(x+1)(x-2)} = \frac{23}{5x} \implies \frac{4(x-2) + 3x + 3}{2(x^2 - x - 2)} = \frac{23}{5x} \]
\[ \frac{7x - 5}{2x^2 - 2x - 4} = \frac{23}{5x} \]
Cross-multiplying:
\[ 5x(7x - 5) = 23(2x^2 - 2x - 4) \]
\[ 35x^2 - 25x = 46x^2 - 46x - 92 \]
Rearranging terms into a standard quadratic form:
\[ 11x^2 - 21x - 92 = 0 \]
We factorize by splitting the middle term:
\[ 11x^2 - 44x + 23x - 92 = 0 \implies 11x(x - 4) + 23(x - 4) = 0 \]
\[ (11x + 23)(x - 4) = 0 \]
This gives the roots:
- \( x = 4 \)
- \( x = -\frac{23}{11} \)

(b) \( \frac{x+1}{x-1} + \frac{x-2}{x+2} = 3 \)
Take a common denominator on the left-hand side:
\[ \frac{(x+1)(x+2) + (x-2)(x-1)}{(x-1)(x+2)} = 3 \implies \frac{(x^2 + 3x + 2) + (x^2 - 3x + 2)}{x^2 + x - 2} = 3 \]
\[ \frac{2x^2 + 4}{x^2 + x - 2} = 3 \]
Cross-multiplying:
\[ 2x^2 + 4 = 3(x^2 + x - 2) \implies 2x^2 + 4 = 3x^2 + 3x - 6 \]
Rearranging into standard form:
\[ x^2 + 3x - 10 = 0 \]
Factorizing the quadratic equation:
\[ (x + 5)(x - 2) = 0 \]
This gives the roots:
- \( x = -5 \)
- \( x = 2 \)
In simple words: We find a common denominator for the fractions on the left, cross-multiply to obtain a standard quadratic equation, and factorize to solve for x.

Exam Tip: In fractional equations, always verify that your final answers do not equal any of the restricted values stated in the question (like \( x \neq 1, -2 \) in part b).

Long Answer Type Questions

Question Q18. A speed of a boat in still water is 11km/hr. It can go 12 km upstream and return downstream to the original point in 2 hours 45 minutes ,find the speed of the stream.
Answer: Let the speed of the stream be \( x\text{ km/hr} \).
- Speed of the boat upstream = \( (11 - x)\text{ km/hr} \)
- Speed of the boat downstream = \( (11 + x)\text{ km/hr} \)

The total time taken is 2 hours and 45 minutes:
\[ 2\text{ hours } 45\text{ minutes} = 2 + \frac{45}{60} = 2 + \frac{3}{4} = \frac{11}{4}\text{ hours} \]

Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \):
\[ \frac{12}{11 - x} + \frac{12}{11 + x} = \frac{11}{4} \]
\[ 12 \left( \frac{(11 + x) + (11 - x)}{(11 - x)(11 + x)} \right) = \frac{11}{4} \]
\[ 12 \left( \frac{22}{121 - x^2} \right) = \frac{11}{4} \implies \frac{264}{121 - x^2} = \frac{11}{4} \]
Divide both sides by 11:
\[ \frac{24}{121 - x^2} = \frac{1}{4} \]
Cross-multiplying:
\[ 96 = 121 - x^2 \implies x^2 = 121 - 96 = 25 \implies x = \pm 5 \]
Since the speed of the stream must be positive, we reject \( x = -5 \).
Therefore, the speed of the stream is 5 km/hr.
In simple words: Let the stream speed be x. Going upstream takes \( \frac{12}{11-x} \) hours, and going downstream takes \( \frac{12}{11+x} \) hours. Setting the sum of these times to 2.75 hours gives the speed of the stream as 5 km/hr.

Exam Tip: Remember to write down why you rejected the negative root (e.g., "speed cannot be negative") to make your solution complete.

Question Q19. A fast train takes 3 hours less than a slow train for a journey of 600 km .If the speed of slow train is 10 km/hr less than that of the fast train ,find the speed of two trains.
Answer: Let the speed of the fast train be \( x\text{ km/hr} \).
Therefore, the speed of the slow train is \( (x - 10)\text{ km/hr} \).

The distance of the journey is 600 km.
Using the time formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \), the slow train takes 3 hours more than the fast train:
\[ \frac{600}{x - 10} - \frac{600}{x} = 3 \]
Divide the entire equation by 3:
\[ \frac{200}{x - 10} - \frac{200}{x} = 1 \]
\[ 200 \left( \frac{x - (x - 10)}{x(x - 10)} \right) = 1 \implies 200 \left( \frac{10}{x^2 - 10x} \right) = 1 \]
\[ 2000 = x^2 - 10x \implies x^2 - 10x - 2000 = 0 \]
Factorizing the quadratic equation:
\[ (x - 50)(x + 40) = 0 \]
This gives:
- \( x = 50 \)
- \( x = -40 \) (rejected as speed cannot be negative)

The speed of the fast train is 50 km/hr, and the speed of the slow train is \( 50 - 10 = 40\text{ km/hr} \).
In simple words: Let the fast train's speed be x. Setting up the difference in travel times for the 600 km journey gives a quadratic equation that yields a speed of 50 km/hr for the fast train and 40 km/hr for the slow train.

Exam Tip: Dividing the entire equation by a common factor (like 3 here) simplifies the coefficients and speeds up your factorization.

Question Q20. Seven years ago Varun's age was five times the square of Swati's age .Three years hence Swati's age will be two fifth of varun's age.Find their present ages.
Answer: Let Swati's age 7 years ago be \( x \) years.
According to the problem, Varun's age 7 years ago was \( 5x^2 \) years.

Therefore, their present ages are:
- Swati's present age = \( x + 7 \) years
- Varun's present age = \( 5x^2 + 7 \) years

Their ages 3 years from now will be:
- Swati's age 3 years hence = \( (x + 7) + 3 = x + 10 \) years
- Varun's age 3 years hence = \( (5x^2 + 7) + 3 = 5x^2 + 10 \) years

We are given that Swati's age 3 years hence will be \( \frac{2}{5} \) of Varun's age:
\[ x + 10 = \frac{2}{5}(5x^2 + 10) \]
\[ x + 10 = 2x^2 + 4 \]
\[ 2x^2 - x - 6 = 0 \]
Factorizing the quadratic equation:
\[ 2x^2 - 4x + 3x - 6 = 0 \]
\[ 2x(x - 2) + 3(x - 2) = 0 \implies (2x + 3)(x - 2) = 0 \]
This gives:
- \( x = 2 \)
- \( x = -1.5 \) (rejected as age cannot be negative)

Using \( x = 2 \):
- Swati's present age = \( 2 + 7 = 9 \) years
- Varun's present age = \( 5(2)^2 + 7 = 20 + 7 = 27 \) years

Therefore, Swati's present age is 9 years and Varun's present age is 27 years.
In simple words: Setting Swati's age 7 years ago as x makes Varun's age \( 5x^2 \). Using their future ages in 3 years gives \( x = 2 \), making Swati's current age 9 and Varun's current age 27.

Exam Tip: Defining variables based on past ages (like "7 years ago") often leads to much cleaner equations than using present ages directly.

Question Q21. By increasing the list price of a book by Rs. 10 a person can buy 10 less books for Rs. 1200 , find the original list price of book.
Answer: Let the original list price of the book be Rs. \( x \).
The number of books that can be bought for Rs. 1200 at the original price is \( \frac{1200}{x} \).

When the list price is increased by Rs. 10, the new price is Rs. \( (x + 10) \).
The number of books that can be bought at the new price is \( \frac{1200}{x + 10} \).

Since the person can buy 10 fewer books at the new price:
\[ \frac{1200}{x} - \frac{1200}{x + 10} = 10 \]
Divide the entire equation by 10:
\[ \frac{120}{x} - \frac{120}{x + 10} = 1 \]
\[ 120 \left( \frac{(x + 10) - x}{x(x + 10)} \right) = 1 \implies 120 \left( \frac{10}{x^2 + 10x} \right) = 1 \]
\[ 1200 = x^2 + 10x \implies x^2 + 10x - 1200 = 0 \]
Factorizing the quadratic equation:
\[ (x + 40)(x - 30) = 0 \]
This gives:
- \( x = 30 \)
- \( x = -40 \) (rejected as price cannot be negative)

Therefore, the original list price of the book is Rs. 30.
In simple words: Let the price of a book be x. Setting up the difference in the number of books purchased for Rs. 1200 gives \( x = 30 \), so the original price is Rs. 30.

Exam Tip: Be careful when writing Rs. in your steps, and remember that price/cost is always a positive quantity.

Question Q22. The numerator of a fraction is one less than it's denominator .If three is added to each Numerator and denominator the fraction is increased by 3/28.Find the fraction.
Answer: Let the denominator of the fraction be \( x \).
Therefore, the numerator is \( x - 1 \).
The original fraction is \( \frac{x - 1}{x} \).

When 3 is added to both the numerator and the denominator, the new fraction becomes:
\[ \frac{(x - 1) + 3}{x + 3} = \frac{x + 2}{x + 3} \]

According to the problem, the fraction increases by \( \frac{3}{28} \):
\[ \frac{x + 2}{x + 3} - \frac{x - 1}{x} = \frac{3}{28} \]
\[ \frac{x(x + 2) - (x - 1)(x + 3)}{x(x + 3)} = \frac{3}{28} \]
\[ \frac{(x^2 + 2x) - (x^2 + 2x - 3)}{x^2 + 3x} = \frac{3}{28} \implies \frac{3}{x^2 + 3x} = \frac{3}{28} \]
Dividing both sides by 3:
\[ x^2 + 3x = 28 \implies x^2 + 3x - 28 = 0 \]
Factorizing the quadratic equation:
\[ (x + 7)(x - 4) = 0 \]
This gives:
- \( x = 4 \)
- \( x = -7 \) (rejected as denominators must be positive for standard fractions)

If \( x = 4 \):
- Denominator = 4
- Numerator = \( 4 - 1 = 3 \)
- Fraction = \( \frac{3}{4} \)

Therefore, the original fraction is \( \frac{3}{4} \).
In simple words: Let the fraction be \( \frac{x-1}{x} \). Adding 3 to top and bottom and finding the difference of \( 3/28 \) gives \( x = 4 \), which makes the fraction \( 3/4 \).

Exam Tip: Always state the final fraction explicitly as \( \frac{\text{Numerator}}{\text{Denominator}} \) rather than just leaving the value of \( x \).

Question Q23. A two digit number is 5 times the sum of it's digits and is also equal to 5 more than twice the the product of it's digits ,find the number.
Answer: Let the tens digit of the number be \( x \) and the units digit be \( y \).
Therefore, the number is \( 10x + y \).

1. **First Condition:** The number is 5 times the sum of its digits:
\[ 10x + y = 5(x + y) \implies 10x + y = 5x + 5y \]
\[ 5x = 4y \implies y = \frac{5}{4}x \tag{Equation 1} \]

2. **Second Condition:** The number is 5 more than twice the product of its digits:
\[ 10x + y = 2xy + 5 \tag{Equation 2} \]

Substitute Equation 1 into Equation 2:
\[ 10x + \frac{5}{4}x = 2x\left(\frac{5}{4}x\right) + 5 \implies \frac{45}{4}x = \frac{5}{2}x^2 + 5 \]
Multiply the entire equation by 4 to clear fractions:
\[ 45x = 10x^2 + 20 \]
Divide by 5:
\[ 9x = 2x^2 + 4 \implies 2x^2 - 9x + 4 = 0 \]
Factorizing the quadratic equation:
\[ 2x^2 - 8x - x + 4 = 0 \implies 2x(x - 4) - 1(x - 4) = 0 \]
\[ (2x - 1)(x - 4) = 0 \]
This gives:
- \( x = 4 \)
- \( x = \frac{1}{2} \) (rejected as digit must be an integer)

Using \( x = 4 \):
\[ y = \frac{5}{4}(4) = 5 \]
Therefore, the two-digit number is \( 10(4) + 5 = 45 \).
In simple words: Representing the digits as x and y gives \( y = 1.25x \). Substituting this into the product equation \( 10x + y = 2xy + 5 \) yields a quadratic equation with integer solution \( x = 4, y = 5 \), making the number 45.

Exam Tip: Single digits in number problems must be non-negative integers from 0 to 9; use this constraint to discard fractional or negative solutions immediately.

Question Q24. Aeroplane left 30 minutes later than it's scheduled time and in order to reach destination 1500 km away in time ,it has to increase it's speed by 250 km/hr from it's usual speed ,determine it's usual speed.
Answer: Let the usual speed of the aeroplane be \( x\text{ km/hr} \).
Therefore, the increased speed is \( (x + 250)\text{ km/hr} \).

The distance to the destination is 1500 km.
The delay in departure is 30 minutes, which is \( \frac{1}{2} \) hour.
Using the time formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \):
\[ \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} \]
\[ 1500 \left( \frac{(x + 250) - x}{x(x + 250)} \right) = \frac{1}{2} \implies 1500 \left( \frac{250}{x^2 + 250x} \right) = \frac{1}{2} \]
\[ 375000 = \frac{x^2 + 250x}{2} \implies x^2 + 250x - 750000 = 0 \]
Factorizing the quadratic equation:
\[ (x + 1000)(x - 750) = 0 \]
This gives:
- \( x = 750 \)
- \( x = -1000 \) (rejected as speed cannot be negative)

Therefore, the usual speed of the aeroplane is 750 km/hr.
In simple words: Let the speed be x. Expressing the half-hour time savings over the 1500 km flight gives a quadratic equation that solves to a standard speed of 750 km/hr.

Exam Tip: Large constant terms (like 750,000) can be factorized easily by looking at their smaller factors (e.g., \( 75 \times 10 \), where \( 100 - 75 = 25 \)).

Question Q25. A man sells a table for Rs. 96 and gains as much percent as the cost price of table .Find the cost price of table.
Answer: Let the cost price of the table be Rs. \( x \).
Therefore, the gain percentage is also \( x\% \).

The profit (gain) amount is:
\[ \text{Profit} = x\% \text{ of Cost Price} = \frac{x}{100} \times x = \frac{x^2}{100} \]

The selling price is the sum of the cost price and the profit:
\[ \text{Selling Price} = \text{Cost Price} + \text{Profit} \]
\[ 96 = x + \frac{x^2}{100} \implies 9600 = 100x + x^2 \implies x^2 + 100x - 9600 = 0 \]
Factorizing the quadratic equation:
\[ (x + 160)(x - 60) = 0 \]
This gives:
- \( x = 60 \)
- \( x = -160 \) (rejected as cost price cannot be negative)

Therefore, the cost price of the table is Rs. 60.
In simple words: Let the cost price be x. The profit is \( x^2 / 100 \). Setting the total selling price to 96 gives a quadratic equation with a single valid solution of Rs. 60.

Exam Tip: Gain percent is numerically equal to cost price; write out the equation as \( \text{SP} = \text{CP} \times (1 + \frac{\text{CP}}{100}) \).

Question Q26. There are three consecutive integers such that square of the first increased by the product of the other two gives 154 .What are the integers.
Answer: Let the three consecutive integers be \( x \), \( x + 1 \), and \( x + 2 \).
According to the problem, the square of the first integer increased by the product of the other two is 154:
\[ x^2 + (x + 1)(x + 2) = 154 \implies x^2 + (x^2 + 3x + 2) = 154 \]
\[ 2x^2 + 3x + 2 = 154 \implies 2x^2 + 3x - 152 = 0 \]
Factorizing the quadratic equation:
\[ 2x^2 - 16x + 19x - 152 = 0 \implies 2x(x - 8) + 19(x - 8) = 0 \]
\[ (2x + 19)(x - 8) = 0 \]
This gives:
- \( x = 8 \)
- \( x = -9.5 \) (rejected as integers must be whole numbers)

For \( x = 8 \), the consecutive integers are:
- \( 8, 9, 10 \)

Therefore, the three consecutive integers are 8, 9, and 10.
In simple words: Let the numbers be x, x+1, and x+2. The equation \( x^2 + (x+1)(x+2) = 154 \) simplifies to \( 2x^2 + 3x - 152 = 0 \), which has integer solution \( x = 8 \), giving the numbers 8, 9, and 10.

Exam Tip: Consecutive integers can also be defined as \( x-1, x, x+1 \); pick the representation that makes the algebraic multiplication easiest.

Question Q27. A piece of cloth costs Rs. 200 . If the piece were 5m longer and each metre of cloth costed Rs. 2 less the cost of the piece would have remained unchanged .How long is the piece and what is it's original rate per metre.
Answer: Let the original length of the cloth be \( x \) meters.
Since the total cost is Rs. 200, the original rate per meter is Rs. \( \frac{200}{x} \).

If the piece were 5 m longer, the new length would be \( (x + 5) \) meters.
If each meter costed Rs. 2 less, the new rate would be Rs. \( \left(\frac{200}{x} - 2\right) \).

Since the total cost remains unchanged at Rs. 200:
\[ (x + 5)\left(\frac{200}{x} - 2\right) = 200 \]
\[ 200 - 2x + \frac{1000}{x} - 10 = 200 \implies -2x + \frac{1000}{x} - 10 = 0 \]
Multiply the entire equation by \( -x \):
\[ 2x^2 + 10x - 1000 = 0 \implies x^2 + 5x - 500 = 0 \]
Factorizing the quadratic equation:
\[ (x + 25)(x - 20) = 0 \]
This gives:
- \( x = 20 \)
- \( x = -25 \) (rejected as length cannot be negative)

Using \( x = 20 \):
- Original length = 20 meters
- Original rate per meter = \( \frac{200}{20} = \text{Rs. } 10 \)

Therefore, the piece is 20 meters long, and its original rate per meter is Rs. 10.
In simple words: Let the length be x meters. Setting the product of the new length and new price per meter to 200 yields a quadratic equation with solution \( x = 20 \) meters and rate Rs. 10 per meter.

Exam Tip: Expand the product carefully; notice how the constant terms on both sides (200) cancel out, simplifying the equation.

Question Q28. Out of a group of swans, 7/2 times the square root of the total number are playing on the shore of pond ,the two remaining ones are swimming in water .Find the total number of swans.
Answer: Let the total number of swans be \( y^2 \) (where \( y > 0 \)), so that the square root is \( y \).
According to the problem, the swans on the shore plus the swimming ones equal the total number of swans:
\[ \frac{7}{2}y + 2 = y^2 \implies 7y + 4 = 2y^2 \implies 2y^2 - 7y - 4 = 0 \]
Factorizing the quadratic equation:
\[ 2y^2 - 8y + y - 4 = 0 \implies 2y(y - 4) + 1(y - 4) = 0 \]
\[ (2y + 1)(y - 4) = 0 \]
This gives:
- \( y = 4 \)
- \( y = -0.5 \) (rejected as the number of swans must be positive)

Using \( y = 4 \):
- Total number of swans = \( y^2 = 4^2 = 16 \)

Therefore, the total number of swans is 16.
In simple words: Setting the total number of swans to \( y^2 \) turns the radical equation into a simple quadratic \( 2y^2 - 7y - 4 = 0 \). Solving this gives \( y = 4 \), meaning there are 16 swans in total.

Exam Tip: Substituting \( y^2 \) for the total quantity is a brilliant way to eliminate square roots early in the solution.

Question Q29. Two pipes running together can fill a cistern in 6 minutes .If one pipe takes 5 minutes more than the other to fill the cistern , find the time in which each pipe would fill the cistern.
Answer: Let the time taken by the faster pipe to fill the cistern alone be \( x \) minutes.
Therefore, the slower pipe takes \( (x + 5) \) minutes.

In 1 minute:
- Work done by the faster pipe = \( \frac{1}{x} \)
- Work done by the slower pipe = \( \frac{1}{x + 5} \)
- Combined work done in 1 minute = \( \frac{1}{6} \)

Setting up the equation:
\[ \frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6} \implies \frac{(x + 5) + x}{x(x + 5)} = \frac{1}{6} \]
\[ \frac{2x + 5}{x^2 + 5x} = \frac{1}{6} \implies 6(2x + 5) = x^2 + 5x \]
\[ 12x + 30 = x^2 + 5x \implies x^2 - 7x - 30 = 0 \]
Factorizing the quadratic equation:
\[ (x - 10)(x + 3) = 0 \]
This gives:
- \( x = 10 \)
- \( x = -3 \) (rejected as time cannot be negative)

The faster pipe takes 10 minutes, and the slower pipe takes \( 10 + 5 = 15 \) minutes.
Therefore, the pipes would fill the cistern individually in 10 minutes and 15 minutes.
In simple words: Let the faster pipe take x minutes. Solving the sum of their individual 1-minute rates (\( \frac{1}{x} + \frac{1}{x+5} = \frac{1}{6} \)) gives \( x = 10 \) minutes and 15 minutes respectively.

Exam Tip: Work/time problems are always solved by adding the rates of work performed in a single unit of time (1 minute, 1 hour, etc.).

Question Q30. A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 4 days , find the time taken by B to finish the work.
Answer: Let B take \( x \) days to complete the work alone.
Therefore, A takes \( (x - 6) \) days.

In 1 day:
- Work done by B = \( \frac{1}{x} \)
- Work done by A = \( \frac{1}{x - 6} \)
- Combined work done in 1 day = \( \frac{1}{4} \)

Setting up the rate equation:
\[ \frac{1}{x} + \frac{1}{x - 6} = \frac{1}{4} \implies \frac{(x - 6) + x}{x(x - 6)} = \frac{1}{4} \]
\[ \frac{2x - 6}{x^2 - 6x} = \frac{1}{4} \implies 4(2x - 6) = x^2 - 6x \]
\[ 8x - 24 = x^2 - 6x \implies x^2 - 14x + 24 = 0 \]
Factorizing the quadratic equation:
\[ (x - 12)(x - 2) = 0 \]
This gives:
- \( x = 12 \)
- \( x = 2 \)

If \( x = 2 \), the time taken by A would be \( 2 - 6 = -4 \) days, which is impossible. Thus, \( x = 2 \) is rejected.
Therefore, the time taken by B to finish the work is 12 days.
In simple words: Let B take x days. The combined rate equation is \( \frac{1}{x} + \frac{1}{x-6} = \frac{1}{4} \). This gives \( x = 12 \) days, as \( x = 2 \) would make A's work time negative.

Exam Tip: Always verify both solutions of the quadratic equation; reject values that result in physically impossible time periods for either person.

Question Q31. Solve for x: a) 3x-4 / 7 + 7 / 3x-4 = 5/2 , x ≠ 4/3 b) 2 (2x-1/ x+3) - 3 (x+3/ 2x-1) = 5 , x ≠ -3 , 1/2
Answer:
(a) \( \frac{3x-4}{7} + \frac{7}{3x-4} = \frac{5}{2} \)
Let \( \frac{3x-4}{7} = y \). The equation becomes:
\[ y + \frac{1}{y} = \frac{5}{2} \implies \frac{y^2 + 1}{y} = \frac{5}{2} \]
\[ 2y^2 + 2 = 5y \implies 2y^2 - 5y + 2 = 0 \]
Factorizing the quadratic equation:
\[ (2y - 1)(y - 2) = 0 \implies y = \frac{1}{2} \text{ or } y = 2 \]

Now, substitute back to solve for \( x \):
1. **If \( y = 2 \):**
\[ \frac{3x-4}{7} = 2 \implies 3x - 4 = 14 \implies 3x = 18 \implies x = 6 \]
2. **If \( y = \frac{1}{2} \):**
\[ \frac{3x-4}{7} = \frac{1}{2} \implies 2(3x - 4) = 7 \implies 6x - 8 = 7 \implies 6x = 15 \implies x = \frac{5}{2} \]

(b) \( 2\left(\frac{2x-1}{x+3}\right) - 3\left(\frac{x+3}{2x-1}\right) = 5 \)
Let \( \frac{2x-1}{x+3} = y \). The equation becomes:
\[ 2y - \frac{3}{y} = 5 \implies 2y^2 - 3 = 5y \implies 2y^2 - 5y - 3 = 0 \]
Factorizing the quadratic equation:
\[ 2y^2 - 6y + y - 3 = 0 \implies 2y(y - 3) + 1(y - 3) = 0 \implies (2y + 1)(y - 3) = 0 \]
This gives:
- \( y = 3 \)
- \( y = -\frac{1}{2} \)

Now, substitute back to solve for \( x \):
1. **If \( y = 3 \):**
\[ \frac{2x-1}{x+3} = 3 \implies 2x - 1 = 3x + 9 \implies x = -10 \]
2. **If \( y = -\frac{1}{2} \):**
\[ \frac{2x-1}{x+3} = -\frac{1}{2} \implies 2(2x - 1) = -(x + 3) \implies 4x - 2 = -x - 3 \implies 5x = -1 \implies x = -\frac{1}{5} \]

Therefore, the solutions are \( x = 6, \frac{5}{2} \) for part (a) and \( x = -10, -\frac{1}{5} \) for part (b).
In simple words: We simplify these fractional equations by substituting \( y \) for the repeated term. Solving the resulting quadratic in \( y \) and converting back to \( x \) yields the solutions.

Exam Tip: Substitution of a repeated expression (like \( y = \frac{3x-4}{7} \)) is a powerful method that keeps the algebra incredibly clean and prevents error-prone cross-multiplication of large terms.