RS Aggarwal Solutions Class 10 Ex 10B

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.


Practise This Question

‘s’ is called irrational if it cannot be written in the form of _____ where p and q are integers and ______