RS Aggarwal Class 10 Solutions Chapter 10 - Quadratic Equations Ex 10B (10.2)

RS Aggarwal Class 10 Chapter 10 - Quadratic Equations Ex 10B (10.2) Solutions Free PDF

The RS Aggarwal Solutions for Class 10 Chapter 10 are considered extremely helpful in solving difficult questions which may be asked in the exam. These solutions help in simplify all your maths doubts and complete the maths syllabus of Class 10 in a proper time. By practicing the questions from the RS Aggarwal Maths textbook you can enhance your problem-solving skills. Students can refer to these solutions whenever they get stuck or have any doubt.

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Q1: 2x2 + ax – a2 = 0

We write, ax = 2ax – ax as 2x2 × (-a2) = -2a2x2 = 2ax × (-ax)

2x2 + ax – a2 = 0

2x2 + 2ax – ax – a2 = 0

2x(x + a) – a(x + a) =0

(x + a)(2x – a) = 0

x+a=0 or 2x-a=0

x = -a or x =\(\frac{a}{2}\)

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

Q2: 4x² + 4bx – (a² – b²) = 0

We write, 4bx = 2(a + b)x – 2(a – b)x as

4x2 ([- (a2 -b2)] = -4(a2 -b2) x2= 2(a + b)x ( [-2(a – b)x]

4x2 + 4bx – (a2 – b2) = 0

4x2 + 2(a + b)x – 2(a – b)x – (a – b)(a + b) = 0

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

[2x + (a + b)][2x – (a – b)] = 0

2x + (a + b) = 0 or 2x – (a – b) = 0

Hence, and are the roots of the given equation.

Q3: 4x² -4a²x + (a4 –b4) = 0

We write, -4a2x = -2(a2 b2)x – 2 (a2 – b2)x as

4x2 ( a4-b4)=4(a4-b4)x2 = [-2(a2 +b2)]x ( [-2(a2 -b2)]x

4x2 -4a2x + (a4 –b4) = 0

4x2 – 2 (a2 + b2)x – 2 (a2 – b2)x+ (a2 – b2) (a2 +b2) = 0

2x [2x – (a2+b2)] – (a2 – b2) [2x- (a2 + b2)] = 0

[2x – (a2 + b2)] [2x- (a2 – b2)] = 0

2x – (a2 +b2) = 0 or 2x – (a2 -b2) = 0

x=\(\frac{a^{2} +b^{2}}{2}\) or x=\(\frac{ a^{2} -b^{2}}{2}\) –*

Hence,* are the roots of the given equation.

Q4: x2 +5x – (a2 +a-6) =0

We write, 5x = (a + 3)x – (a – 2)x as

x2 ([- (a2 + a – 6)] = – (a2 + a – 6)x2 = (a + 3)x (-(a – 2)x])

.’. x2 +5x – (a2 +a-6) =0

x2 + (a + 3)x – (a – 2)x – (a + 3)(a – 2) = 0

x[x + (a + 3)] – (a – 2)[x + (a + 3)] = 0

[x + (a + 3)][x – (a – 2)] =0

x+(a+3)=0 or x – (a -2)=0

x=-(a+3) or x=a-2

Hence, -(a + 3) and (a – 2) are the roots of the given equation.

Q5: x² – 2ax – (4b² – a²) = 0

We write, -2ax = (2b – a)x – (2b+ a)x as

x2 (- (4b2 – a2)1 = – (4b2 a2) x2=(2b – a)x [- (2b + a)x]

.’. X2 – 2ax – (4b2 – a2) = 0

x2 + (2b – a)x – (2b+ a)x – (2b – a)(2b+ a) = 0

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

[x + (2b – a)][x – (2b + a)] =0

.x+(2b-a)=0 or x-(2b+a)=0

x=-(2b-a) or x=2b+a x=a-2b or x=a+2b

Hence, a – 2b and a + 2b are the roots of the given equation.

Q6: x2 – (2b – 1)x + (b² – b – 20) = 0

We write, (2b – 1)x = -(b- 5)x – (b + 4)x as

x2 (b2 – b – 20) = (b2 – b – 20) x2 = [-(b – 5)x] [-(b + 4)x]

x2 – (2b – 1)x + (b2 – b – 20) = 0

x2– (b – 5)x – (b+ 4)x + (b – 5)(b+ 4) =0

x[x – (b – 5)] – (b + 4)[x – (b – 5)] = 0

x – (b – 5)][x – (b+ 4)] = 0

x – (b – 5) = 0 or x – (b + 4) = 0

x=b-5 or x=b+4

Hence, b – 5 and b+ 4 are the roots of the given equation.

Q7: x2 + 6x – (a² + 2a – 8)= 0

We write, 6x = (a +4)x – (a – 2)x as

x2 [-(a2 + 2a – 8)] = -(a2 + 2a – 8)x2 =(a + 4)x [- (a – 2)x]

x2 + 6x – (a2 + 2a – 8)= 0

x2 + (a+ 4)x – (a – 2)x – (a + 4)(a – 2) = 0

x[x + (a+ 4)] – (a – 2)[x + (a +4)] = 0

[x+ (a + 4)][x – (a – 2)] = 0

x + (a + 4) =0 or x- (a- 2)= 0

x = -(a + 4) or x = a – 2

Hence, -(a + 4) and (a – 2) are the roots of the given equation.

Q8: abx2 + (b2 – ac)x – bc = 0

Given:

abx2 + (b2 – ac)x – bc = 0

abx2 + b2x – acx – bx = 0

bx(ax + b) – c(ax + b) = 0

(bx – c)(ax + b) = 0

bx – c = 0 or ax + b = 0

x= \(\frac{c}{b}\) or \(\frac{-b}{a}\)

Q9: x2 – 4ax – b2 + 4a2 = 0

We write, -4ax = -(b + 2a)x+ (b – 2a)x as

x2 x (-b2 +4a2)= (-b2+4a2)x2= -(b+2a)x (b- 2a)x

x2 – 4ax – b2 + 4a2 = 0

x2 – (b + 2a)x + (b – 2a)x – (b – 2a)(b + 2a) = 0

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

[x-(b+2a)][x+(b-2a)] =0

x-(b+2a)=0 or x+(b-2a)=0

x=2a+b or x=-(b-2a)

x=2a+b or x=2a-b

Hence, (2a+ b) and (2a- b) are the roots of the given equation.

Q10: 4x2 – 2 (a2 ± b2 )x + a2b2 = 0

Given :

4x2 – 2 (a2 ± b2 )x + a2b2 = 0

4x2 – 2a2x – 2b2x + a2b2 = 0

2x (2x – a2) – b2 (2x – a2) = 0

(2x – b2) (2x – a2) = 0

2x – b2 = 0 or 2x – a2=0

x=\(\frac{ b^{2}}{2}\) or x=\(\frac{ a^{2}}{2}\)

Hence, the roots of the equation.

Q11: 12abx2 – 9a2x + 8b2x – 6ab = 0

Given : 12abx2 – (9a2– 8b2)x – 6ab = 0

12abx2 – 9a2x + 8b2x – 6ab = 0

3ax(4bx – 3a) + 2b(4bx – 3a) = 0

(3ax + 2b)(4bx – 3a) = 0

3ax + 2b = 0 or 4bx – 3a = 0

X=\(\frac{ -2b}{3a}\) or x= \(\frac{ 3a}{4b}\)

Hence, the roots of the equation

Q12: a2b2x2 + b2x – a2x – 1 = 0

Given :

a2b2x2 + b2x – a2x – 1 = 0

b2x(a2x + 1) – 1(a2x + 1) =0

(b2x – 1) (a2x + 1) = 0

(b2x – 1) = 0 or (a2x + 1) = 0

x=\(\frac{ 1}{b^{2}}\) or x=\(\frac{-1}{ a^{2}}\)

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

Q13: \( \frac{x}{x – 1} + \frac{x – 1}{x} = 4 \frac{1}{4} , x \neq 0, 1 \)

\( \frac{x}{x – 1} + \frac{x – 1}{x} = 4 \frac{1}{4} , x \neq 0, 1 \)

\( \Rightarrow \frac{x^{2} + (x – 1)^{2}}{x(x – 1)} = \frac{17}{4} \)

\( \Rightarrow \frac{x^{2} + x^{2} – 2x + 1}{x^{2} – x} = \frac{17}{4} \)

\( \Rightarrow \frac{2x^{2} – 2x + 1}{x^{2} – x} = \frac{17}{4} \)

\( \Rightarrow \) 8x2 – 8x + 4 = 17x2 – 17x

\( \Rightarrow \) 9x2 – 9x – 4 = 0

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

\( \Rightarrow\) 3x(3x – 4) + 1(3x – 4) = 0

\(\Rightarrow\) (3x – 4)(3x + 1) = 0

\( \Rightarrow \) 3x – 4 = 0 or 3x + 1 = 0

\( \Rightarrow x = \frac{4}{3} or x = \frac{-1}{3} \)

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

Q14: 9x2 — 9(a+ b)x+ (2a2 + 5ab+ 2b2) = 0

We write, —9(a+ b)x = —3(2a + b)x – 3(a+ 2b)x as

9x2 x (2a2 +5ab+ 2b2) = 9(2a2 + 5ab+2b2)x2 = [-3(2a + b)x] x [-3(a +2b)x]

Therefore, 9x2 — 9(a+ b)x+ (2a2 + 5ab+ 2b2) = 0

\(\Rightarrow\) 9x2 — 3(2a + b)x — 3(a + 2b)x + (2a+ b)(a + 2b) = 0

\(\Rightarrow\) 3x[3x — (2a+ b)] – (a +2b)[3x — (2a+ b)] = 0

\(\Rightarrow\) [3x — (2a + b)] [3x — (a + 2b)] = 0

\(\Rightarrow\) 3x — (2a+ b) = 0 or 3x — (a + 2b) = 0

\(\Rightarrow\) X=\(\frac{2a+b}{3}\) or x=\(\frac{a+2b}{3}\)

Hence, \(\frac{2a+b}{3}\) and \(\frac{a+2b}{3}\) are the roots of the given equation.

Q15: \(\frac{16}{x}\)-1= \(\frac{15}{x+1}\),x \(\neq\)0, —1

\(\Rightarrow\) \(\frac{16}{x}\)\(\frac{15}{x+1}\)=1

\(\Rightarrow\) \(\frac{16x+16-15x}{x(x+1)}\)=1

\(\Rightarrow\)

\(\frac{x+16}{x^{2}+x}\)=1

\(\Rightarrow\) x2 + x = x + 16 (Cross multiplication)

\(\Rightarrow\) x2 — 16 = 0

\(\Rightarrow\) (x + 4)(x — 4) = 0

\(\Rightarrow\) x+4=0 or x-4=0

\(\Rightarrow\) x = —4 or x = 4

Hence, -4 and 4 are the roots of the given equation.

Q16: \(\frac{4}{x}\)-3=\(\frac{5}{2x+3}\),x\(\neq\)0,- \(\frac{3}{2}\)

\(\frac{4}{x}\)-3=\(\frac{5}{2x+3}\),x\(\neq\)0,- \(\frac{3}{2}\)

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

\(\Rightarrow\) \(\frac{8x+12-5x}{x(2x+3)}\)=3

\(\Rightarrow\)  \(\frac{3x+12}{2x^{2}+3x}\)=3

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

\(\Rightarrow\) 2x2+3x = x+4 (Cross multiplication)

\(\Rightarrow\) 2x2 + 2x — 4 = 0

\(\Rightarrow\) x2+x —2=0

\(\Rightarrow\) x2+2x-x-2=0

\(\Rightarrow\) x(x + 2) — 1(x + 2) = 0

\(\Rightarrow\) (x + 2)(x — 1)=0

\(\Rightarrow\) x+2=0 or x-1=0

\(\Rightarrow\) x=-2 or x=1

Hence, -2 and 1 are the roots of the given equation.

Q17: \(\frac{3}{x+1}\)\(\frac{1}{2}\) = \(\frac{2}{3x-1}\),x\(\neq\)-1, \(\frac{1}{3}\)

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

\(\Rightarrow\) \(\frac{3}{x+1}\)\(\frac{2}{3x-1}\)= \(\frac{1}{2}\)

\(\Rightarrow\) \(\frac{9x-3-2x-2}{(x+1)(3x-1)}\) = \(\frac{1}{2}\)

\(\Rightarrow\)  \(\frac{6}{x^{2}+4x-5}\)= \(\frac{6}{7}\)

\(\Rightarrow\) 3x2 + 2x — 1 = 14x — 10 (Cross multiplication)

\(\Rightarrow\) 3x2 — 12x + 9 = 0

\(\Rightarrow\) x2 —4x+ 3= 0

\(\Rightarrow\) x2 — 3x—x+ 3 =0

\(\Rightarrow\) x(x — 3) — 1(x — 3) = 0

\(\Rightarrow\) (x — 3)(x — 1) = 0

\(\Rightarrow\) x — 3 = 0 or x — 1 = 0

\(\Rightarrow\) x = 3 or x = 1

Hence, 1 and 3 are the roots of the given equation.

Q18: \(\frac{1}{x-1}\)\(\frac{1}{x+5}\)= \(\frac{6}{7}\),x\(\neq\)1, -5

\(\frac{1}{x-1}\)\(\frac{1}{x+5}\)= \(\frac{6}{7}\),x\(\neq\)1, -5

\(\Rightarrow\) \(\frac{x+5-x+1}{(x-1)(x+5)}\) =\(\frac{6}{7}\)

\(\Rightarrow\)  \(\frac{-(2a+b)}{4x^{2}+4ax+2bx}\)= \(\frac{2a+b}{2ab}\)

\(\Rightarrow\) x2+4x-5= 7

\(\Rightarrow\) x2+4x-12=0

\(\Rightarrow\) x2 +6x-2x-12=0

\(\Rightarrow\) x(x + 6) — 2(x + 6) = 0

\(\Rightarrow\) (X + 6)(X — 2)=0

\(\Rightarrow\) x+6=0 or x-2=0

\(\Rightarrow\) x=-6 or x=2

Hence, -6 and 2 are the roots of the given equation.

Q19: \(\frac{1}{2a+b+2x}\)= \(\frac{1}{2a}\)+ \(\frac{1}{b}\)+ \(\frac{1}{2x}\)

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

\(\Rightarrow\) \(\frac{1}{2a+b+2x}\)\(\frac{1}{2x}\)= \(\frac{1}{2a}\)+ \(\frac{1}{b}\)

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

\(\Rightarrow\)  \(\frac{-(2a+b)}{4x^{2}+4ax+2bx}\)= \(\frac{2a+b}{2ab}\)

\(\Rightarrow\) 4x2 + 4ax + 2bx = —2ab

\(\Rightarrow\) 4x2+4ax+2bx+2ab=0

\(\Rightarrow\) 4x(x + a) + 2b(x + a)= 0

\(\Rightarrow\) (x + a)(4x 2b) = 0

\(\Rightarrow\) X + a=0 or 4x+2b=0

\(\Rightarrow\) x= —a or –\(\frac{b}{2}\)

Hence, —a and –\(\frac{b}{2}\) are the roots of the given equation.

Q20: \(\frac{x+3}{x-2}\)\(\frac{1-x}{x}\)= \(\frac{17}{4}\)

Given:

\(\frac{x+3}{x-2}\)\(\frac{1-x}{x}\)= \(\frac{17}{4}\)

\(\Rightarrow\) \(\frac{x(x+3)-(1-x)(x-2)}{(x-2)x}\)= \(\frac{17}{4}\)

\(\Rightarrow\)  \(\frac{x^{2}+3x-(x-2-x^{2}+2x)}{x^{2}-2x}\)= \(\frac{17}{4}\)

\(\Rightarrow\)  \(\frac{x^{2}+3x+x^{2}-3x+2}{x^{2}-2x}\)= \(\frac{17}{4}\)

\(\Rightarrow\)  \(\frac{2x^{2}+2}{x^{2}-2x}\)= \(\frac{17}{4}\)

\(\Rightarrow\) 8x2 + 8 = 17x2 — 34x [On cross multiplying]

\(\Rightarrow\) — 9x2 + 34x + 8 = 0

\(\Rightarrow\) 9x2 — 34x — 8 = 0

\(\Rightarrow\) 9x2 — 36x + 2x — 8 = 0

\(\Rightarrow\) 9x(x — 4) + 2(x — 4) =0

\(\Rightarrow\) (x — 4)(9x + 2) = 0

\(\Rightarrow\) x— 4 = 0 or 9x + 2 = 0

\(\Rightarrow\) x = 4 or x = \(\frac{-2}{9}\)

Hence, the roots of the equation are 4 and \(\frac{-2}{9}\)

Q21.\(\frac{3x-4}{7}\)+ \(\frac{7}{3x-4}\)= \(\frac{5}{2}\),x\(\neq\) \(\frac{4}{3}\)

\(\Rightarrow\)  \(\frac{(3x-4)^{2}+49}{7(3x-4)}\)= \(\frac{5}{2}\)

\(\Rightarrow\)  \(\frac{9x^{2}-24x+16+49}{21x-8}\)= \(\frac{5}{2}\)

\(\Rightarrow\)  \(\frac{9x^{2}-24x+65}{21x-28}\)= \(\frac{5}{2}\)

\(\Rightarrow\) 18x2 — 48x + 130 = 105x — 140

\(\Rightarrow\) 18x2 — 153x + 270 = 0

\(\Rightarrow\) 2x2 — 17x + 30 = 0

\(\Rightarrow\) 2x2 — 12x — 5x + 30 = 0

\(\Rightarrow\) 2x(x — 6) — 5(x — 6) = 0

\(\Rightarrow\) (x — 6)(2x — 5) = 0

\(\Rightarrow\) x-6= 0 or 2x— 5= 0

\(\Rightarrow\) x= 6 or x=\(\frac{5}{2}\)

Hence, 6 and \(\frac{5}{2}\) are the roots of the given equation.

Q22: \( \frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}, x \neq 0 , -1 \)

\( \frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}, x \neq 0 , -1 \)

\( \Rightarrow \frac{x^{2} + (x + 1)^{2}}{x (x + 1)} = \frac{34}{15} \)

\( \Rightarrow \frac{x^{2} + x^{2} + 2x + 1}{x^{2} + x} = \frac{34}{15} \)

\( \Rightarrow \frac{2x^{2} + 2x + 1}{x^{2} + x} = \frac{34}{15} \)

\( \Rightarrow \) 30x2 + 30x + 15 = 34x2 + 34x

\( \Rightarrow \) 4x2 + 4x – 15 = 0

\( \Rightarrow 4x^{2}+ 10x – 6x – 15 = 0 \)

\( \Rightarrow 2x(2x + 5) – 3(2x + 5) = 0 \)

\( \Rightarrow \) (2x + 5)(2x – 3) = 0

\( \Rightarrow \) (2x + 5) = 0 or 2x – 3 = 0

\( \Rightarrow x = \frac{-5}{2} or x = \frac{3}{2} \)

Hence, \( \frac{-5}{2} and \frac{3}{2} \) are the roots of the given equation.

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Which of the following were not used in India in ancient times?