RS Aggarwal Class 10 Solutions Chapter 10 - Quadratic Equations Ex 10C ( 10.3)

RS Aggarwal Class 10 Chapter 10 - Quadratic Equations Ex 10C ( 10.3) Solutions Free PDF

For students the best way to score good marks in the exam by practicing more and more important questions on a regular basis. It helps in improving your maths solving skills. The RS Aggarwal Maths textbook comprises of a greater number of questions for a certain topic and these solutions are based on helping students to solve questions easily of Class 10 level. Students of Class 10 can score good marks by referring to these RS Aggarwal Maths solutions to solve difficult maths problems.

The RD Sharma Maths textbook comprises of good questions for you to practice to improve your level of preparation. Using these solutions, you can greatly enhance your problem-solving skills. The solutions are designed based on the recent syllabus of the CBSE and comply fully with the CCE guidelines.

Download PDF of RS Aggarwal Class 10 Solutions Chapter 10– Quadratic Equations Ex 10C (10.3)

Q1 : \( \frac{x – 4}{x – 5} + \frac{x – 6}{x – 7} = 3 \frac{1}{3} , x \neq 5, 7 \)

\( \frac{x – 4}{x – 5} + \frac{x – 6}{x – 7} = 3 \frac{1}{3} , x \neq 5, 7 \)

\( \Rightarrow \frac{(x – 4)(x – 7) + (x – 5)(x – 6)}{(x – 5)(x – 7)} = \frac{10}{3} \)

\( \Rightarrow \frac{x^{2} – 11x + 28 + x^{2} – 11x + 30}{x^{2} – 12x + 35} = \frac{10}{3} \)

\( \Rightarrow \frac{2x^{2} – 22x + 58}{x^{2} – 12x + 35} = \frac{10}{3} \)

\( \Rightarrow \frac{x^{2} – 11x + 29}{x^{2} – 12x + 35} = \frac{5}{3} \)

\( \Rightarrow \) 3x2 – 33x + 87 = 5x2 – 60x + 175

\( \Rightarrow \) 2x2 – 16x – 11x + 88 = 0

\( \Rightarrow 2x^{2} – 16x – 11x + 88 = 0 \)

\( \Rightarrow \) 2x(x – 8) – 11( x – 8) = 0

\( \Rightarrow \) (x – 8)(2x – 11) = 0

\( \Rightarrow \) x – 8 = 0 or 2x – 11 = 0

\( \Rightarrow x = 8 or x = \frac{11}{2} \)

Hence, 8 and (11/2) are the roots of the equation.

Q2: \( \frac{x – 1}{x – 2} + \frac{x – 3}{x – 4} = 3 \frac{1}{3}, x \neq 2, 4 \)

\( \frac{x – 1}{x – 2} + \frac{x – 3}{x – 4} = 3 \frac{1}{3}, x \neq 2, 4 \)

\( \Rightarrow \frac{(x – 1)(x – 4) + (x – 2)(x – 3)}{(x – 2)(x – 4)} = \frac{10}{3} \)

\( \Rightarrow \frac{x^{2} – 5x + 4 + x^{2} – 5x + 6}{x^{2} – 6x + 8} = \frac{10}{3} \)

\( \Rightarrow \frac{2x^{2} – 10x + 10}{x^{2} – 6x + 8} \frac{10}{3} \)

\( \Rightarrow \frac{x^{2} – 5x + 5}{x^{2} – 6x + 8} = \frac{5}{3} \)

\( \Rightarrow \) 3x2 – 15x + 15 = 5x2 – 30x + 40

\( \Rightarrow \) 2x2 – 15x + 25 = 0

\( \Rightarrow \) 2x2 – 10x – 5x + 25 = 0

\( \Rightarrow \) 2x(x – 5) – 5(x – 5) = 0

\( \Rightarrow \) x – 5 = 0 or 2x – 5 = 0

\( \Rightarrow x = 5 \;or \;x = \frac{5}{2} \)

Hence, 5 and (5/2) are the roots of the given equation.

Q3: \( \frac{1}{x – 2} + \frac{2}{x – 1} = \frac{6}{x} \)

Given:

\( \frac{1}{x – 2} + \frac{2}{x – 1} = \frac{6}{x} \)

\( \Rightarrow \frac{(x – 1) + 2(x – 2)}{(x – 1)(x – 2)} = \frac{6}{2} \)

\( \Rightarrow \frac{3x – 5}{x^{2} – 3x + 2} = \frac{6}{x} \)

\( \Rightarrow \) 3x2 – 5x = 6x2 – 18x + 12 [On cross multiplication]

\( \Rightarrow \) 3x2 – 13x + 12 = 0

\( \Rightarrow \) 3x2 – (9 + 4)x + 12 = 0

\( \Rightarrow \) 3x2 – 9x – 4x + 12 = 0

\( \Rightarrow \) 3x(x – 3) – 4(x – 3) = 0

\( \Rightarrow \) (x – 3)(3x – 4) = 0

\( \Rightarrow x= \frac{4}{3} or x = 3 \)

Hence, the roots of the equation are (4/3) and 3.

Q4: \( \frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4} , x \neq -1, -2, -4 \)

\( \frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4} , x \neq -1, -2, -4 \)

\( \Rightarrow \frac{x + 2+ 2x + 2}{(x +1)(x + 2)} = \frac{5}{x + 4} \)

\( \Rightarrow \frac{3x + 4}{x^{2} + 3x + 2} = \frac{5}{x + 4} \)

\( \Rightarrow \) (3x + 4)(x + 4) = 5(x2 + 3x + 2)

\( \Rightarrow \) 3x2 + 16x + 16 = 5x2 + 15x + 10

\( \Rightarrow \) 2x2 – x – 6 = 0

\( \Rightarrow \) 2x2 – 4x + 3x – 6 = 0

\( \Rightarrow \) 2x(x – 2) + 3(x – 2) = 0

\( \Rightarrow \) (x – 2)(2x + 3) = 0

\( \Rightarrow x = 2 or x = \frac{-3}{2} \)

Hence, 2, -(3/2) are the roots of the given equation.

Q5: \( 3 \frac{3x – 1}{2x + 3} – 2\frac{2x + 3}{3x – 1} = 5, x\neq \frac{1}{3}, \frac{-3}{2} \)

\( 3 \frac{3x – 1}{2x + 3} – 2\frac{2x + 3}{3x – 1} = 5, x\neq \frac{1}{3}, \frac{-3}{2} \)

\( \Rightarrow \frac{3(3x – 1)^{2} – 2(2x + 3)^{2}}{(2x + 3)(3x – 1)} = 5 \)

\( \Rightarrow \frac{3(9x^{2} – 6x + 1) – 2 (4x^{2} + 12x + 9}{6x^{2} + 7x – 3} = 5 \)

\( \Rightarrow \frac{27x^{2} – 18x + 3 – 8x^{2} – 24x – 18}{6x^{2} + 7x – 3} = 5 \)

\( \Rightarrow \frac{19x^{2} – 42x – 15}{6x^{2} + 7x – 3} = 5 \)

\( \Rightarrow \) 19x2 – 42x – 15 = 30x2 + 35x – 15

\( \Rightarrow \) 11x2 + 77x = 0

\( \Rightarrow \) 11x(x + 7) = 0

\( \Rightarrow \) x = 0 or x + 7 = 0

\( \Rightarrow \) x = 0 or x = -7

Hence, 0 and -7 are the roots of the given equation.

Q6: \( 3 \frac{7x + 1}{5x – 3} – 4 \frac{5x – 3}{7x + 1} = 11, x \neq \frac{3}{5} , \frac{-7}{7} \)

\( 3 \frac{7x + 1}{5x – 3} – 4 \frac{5x – 3}{7x + 1} = 11, x \neq \frac{3}{5} , \frac{-7}{7} \)

\( \Rightarrow \frac{3(7x + 1)^{2} – 4(5x – 3)^{2}}{(5x – 3)(7x + 1)} = 11 \)

\( \Rightarrow \frac{3(49x^{2} + 14x + 1) – 4(25x^{2} – 30x + 0)}{(5x – 3)(7x + 1)} = 11 \)

\( \Rightarrow \frac{147x^{2} + 42x + 3 – 100x^{2} + 120x – 36}{35x^{2} – 16x – 3} = 11 \)

\( \Rightarrow \frac{47x^{2} + 162x – 33}{35x^{2} – 16x – 3} = 11 \)

\( \Rightarrow \) 47x2 + 162x – 33 = 385x2 – 176x – 33

\( \Rightarrow \) 338x2 – 338x = 0

\( \Rightarrow \) 338x(x – 1) = 0

\( \Rightarrow \) x = 0 or x – 1 = 0

\( \Rightarrow \) x = 0 or x = 1

Hence, 0 and 1 are the roots of the given equation.

Q7: \( \frac{4x – 3}{2x + 1} – 10\frac{(2x + 1)}{(4x – 3)} = 3 \)

\( \frac{4x – 3}{2x + 1} – 10\frac{(2x + 1)}{(4x – 3)} = 3 \)

Putting, \( \frac{4x – 3}{2x + 1} = y \), we get

\( y – \frac{10}{y} = 3 \)

\( \Rightarrow \frac{y^{2} – 10}{y} = 3 \)

\( \Rightarrow \) y2 – 10 = 3y [On cross multiplication]

\( \Rightarrow \) y2 – 3y – 10 = 0

\( \Rightarrow \) y2 – (5 – 2)y – 10 = 0

\( \Rightarrow \) y2 – 5y + 2y – 10 = 0

\( \Rightarrow \) y(y – 5) + 2(y – 5) = 0

\( \Rightarrow \) (y – 5)(y + 2) = 0

\( \Rightarrow \) y – 5 = 0 or y + 2 = 0

\( \Rightarrow \) y = 5 or y = -2

Case 1:

If y = 5, we get

\( \frac{4x – 3}{2x + 1} = 5 \)

\( \Rightarrow \) 4x – 3 = 5(2x + 1) [On cross multiplying]

\( \Rightarrow \) 4x – 3 = 10x + 5

\( \Rightarrow \) -6x = 8

\( \Rightarrow x = \frac{-4}{3} \)

Case 2:

If y = -2, we get

\( \frac{4x – 3}{2x + 1} = -2 \)

\( \Rightarrow \) 4x – 3 = -2(2x + 1)

\( \Rightarrow \) 4x – 3 = -4x – 2

\( \Rightarrow \) 8x = 1

\( \Rightarrow x = \frac{1}{8} \)

Hence, the roots of the equation are \( \frac{-4}{3} and \frac{1}{8} \)

Q8: \( (\frac{x}{x + 1})^{2} – 5(\frac{x}{x + 1}) + 6 = 0 \)

Given:

\( (\frac{x}{x + 1})^{2} – 5(\frac{x}{x + 1}) + 6 = 0 \)

Putting \( \frac{x}{x + 1} = y \) , we get

Y2 – 5y + 6 = 0

\( \Rightarrow \) y2 – 5y + 6 = 0

\( \Rightarrow \) y2 – (3 + 2)y + 6 = 0

\( \Rightarrow \) y2 – 3y – 2y + 6 = 0

\( \Rightarrow \) y(y – 3) – 2( y – 3) = 0

\( \Rightarrow \) (y – 3)(y – 2) = 0

\( \Rightarrow \) y – 3 = 0 or y – 2 = 0

\( \Rightarrow \) y = 3 or y = 2

Case 1:

If y = 3, we get

\( \frac{x}{x + 1} = 3 \)

\( \Rightarrow \) x = 3(x + 1) [On cross multiplication]

\( \Rightarrow \) x = 3x + 3

\( \Rightarrow x = \frac{-3}{2} \)

Case 2:

If y = 2, we get

\( \frac{x}{x + 1} = 2 \)

\( \Rightarrow \) x = 2(x + 1)

\( \Rightarrow \) x = 2x + 2

\( \Rightarrow \) -x = 2

\( \Rightarrow \) x = -2

Hence, the roots of the equation are (-3/2) and -2.

Q9: \( \frac{a}{x – b} + \frac{b}{x – a} = 2 \)

\( \frac{a}{x – b} + \frac{b}{x – a} = 2 \)

\( \Rightarrow [\frac{a}{x – b} – 1 ] + [ \frac{b}{x – a} – 1] = 0 \)

\( \Rightarrow \frac{a – (x – b)}{x – b} + \frac{b – (x – a)}{x – a} = 0 \)

\( \Rightarrow \frac{a – x + b}{x – b} + \frac{a – x + b}{x – a} = 0 \)

\( \Rightarrow (a – x + b) [\frac{1}{x – b} + \frac{1}{x – a} ] = 0 \)

\( \Rightarrow (a – x + b) [ \frac{(x – a) + (x – b)}{(x – b)(x – a)} ] = 0 \)

\( \Rightarrow (a – x + b) [ \frac{2x – (a + b)}{(x – b)(x – a)} ] = 0 \)

\( \Rightarrow (a – x + b) [2x – (a + b)] = 0 \)

\( \Rightarrow \) a – x + b = 0 or 2x – (a + b) = 0

\( \Rightarrow x = a + b\; or\; x = \frac{a + b}{2} \)

Hence, the roots of the equation are \( (a + b) and \frac{a + b}{2} \)

Q11: \( \frac{a}{ax – 1} + \frac{b}{bx – 1} = (a + b) \)

\( \frac{a}{ax – 1} + \frac{b}{bx – 1} = (a + b) \)

\( \Rightarrow [ \frac{a}{ax – 1} – b] + [\frac{b}{bx – 1} – a] = 0 \)

\( \Rightarrow \frac{a – b(ax – 1)}{ax – 1} + \frac{b – a(bx – 1)}{bx – 1} = 0 \)

\( \Rightarrow \frac{a – abx + b}{ax – 1} + \frac{a – abz + b}{bx – 1} = 0 \)

\( \Rightarrow (a – abx + b) [\frac{1}{ax – 1} + \frac{1}{bx – 1}] = 0 \)

\( \Rightarrow (a – abx + b) [\frac{(bx – 1) + (ax – 1)}{(ax – 1)(bx – 1)}] = 0 \)

\( \Rightarrow (a – abx + b) [\frac{(a + b)x – 2}{(ax – 1)(bx – 1)}] = 0 \)

\( \Rightarrow (a – abx + b) [(a + b)x – 2] = 0 \)

\( \Rightarrow \) a – abx + b = 0 or (a + b)x – 2 = 0

\( \Rightarrow x = \frac{a + b}{ab} or x = \frac{2}{a + b} \)

Hence, the roots of the equation are, \( \frac{a + b}{ab} and \frac{2}{a + b} \)

Q12: 3(x + 2) + 3-x = 10

3(x + 2) + 3-x = 10

3x . 9 + \( \frac{1}{3^{x}} \) = 10

Let, 3x be equal to y

Hence,

\( 9y + \frac{1}{y} = 10 \)

\( \Rightarrow 9y^{2} + 1 = 10 y \)

\( \Rightarrow 9 y^{2} – 10 + 1 = 0 \)

\( \Rightarrow (y – 1) (9y – 1_ = 0 \)

\( \Rightarrow y – 1 = 0 or 9y – 1 = 0 \)

\( \Rightarrow y = 1 or y = \frac{1}{9} \)

\( \Rightarrow 3^{x} = 1 or 3^{x} = \frac{1}{9} \)

\( \Rightarrow 3^{x} = 3^{0} or 3^{x} = 3^{-2} \)

\( \Rightarrow x = 0 or x = -2 \)

Hence, 0 and -2 are the roots of the given equation.

Q13: 4(x + 1) + 4(1 – x) = 10

Given:

4(x + 1) + 4(1 – x) = 10

\( \Rightarrow 4^{x} . 4 + 4^{1} \frac{1}{4^{x}} = 10 \)

Let 4x be y

Hence,

\( 4y + \frac{4}{y} = 10 \)

\( \Rightarrow 4y^{2} – 10y + 4 = 0 \)

\( \Rightarrow 4y^{2} – 8y – 2y + 4 = 0 \)

\( \Rightarrow 4y(y – 2) – 2(y – 2) = 0 \)

\( \Rightarrow y = 2 or y = \frac{2}{4} = \frac{1}{2} \)

\( \Rightarrow 4^{x} = 2^{2x} = 2^{1} or 2^{2x} = 2^{-1} \)

\( \Rightarrow x = \frac{1}{2} or x = \frac{-1}{2} \)

Hence, \( \frac{1}{2} and \frac{-1}{2} \) are roots of the given equation.

Q14: 22x – 3(2(x + 2)) + 32 = 0

Given:

22x – 3(2(x + 2)) + 32 = 0

\( \Rightarrow \) (2x)2 – 3(2x22) + 32 = 0

Let, 2x be y

Hence,

Y2 – 12y + 32 = 0

\( \Rightarrow \) y2 – 8y – 4y + 32 = 0

\( \Rightarrow \) y (y – 8) – 4(y – 8) = 0

\( \Rightarrow \) (y – 8) = 0 or (y – 4) = 0

\( \Rightarrow \) y = 8 or y = 4

Hence,

2x = 8 or 2x = 4

\( \Rightarrow \) 2x = 23 or 2x = 22

\( \Rightarrow \) x = 2 or 3

Hence, 2 and 3 are the roots of the given equation.

15: 4\(\sqrt{6}\)x^{2}  — 13x — 2\(\sqrt{6}\) = 0  

Given:

4\(\sqrt{6}\)x^{2} — 13x — 2\(\sqrt{6}\) = 0

4\(\sqrt{6}\)x^{2}  — 16x + 3x — 2\(\sqrt{6}\) = 0

4\(\sqrt{2}\)x(\(\sqrt{3}\)x — 2\(\sqrt{2}\)) + \(\sqrt{3}\) (\(\sqrt{3}\)x — 2\(\sqrt{2}\)) = 0

(4\(\sqrt{2}\)x + \(\sqrt{3}\)) (\(\sqrt{3}\)x — 2\(\sqrt{2}\)) = 0

4\(\sqrt{2}\)x + \(\sqrt{3}\) = 0 or \(\sqrt{3}\)x — 2\(\sqrt{2}\) = 0

x = \(\frac{-sqrt{3}}{4[\sqrt{2}}\)= \(\frac{-\sqrt{3}x\sqrt{2}}{4\sqrt{2}x\sqrt{2}}\) = \(\frac{-\sqrt{6}}{8}\) or x = \(\frac{2\sqrt{2}}{ \sqrt{3}}\) = \(\frac{2\sqrt{2}x\sqrt{3}}{\sqrt{3}x\sqrt{3}}\) = \(\frac{2\sqrt{6}}{3}\)

Hence, the roots of the equation are  and \(\frac{-\sqrt{6}}{8}\) and \(\frac{2\sqrt{6}}{3}\).

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The compound which has only primary and tertiary carbons is?