Interactive OP Malhotra Class 10 Solutions Chapter 5 Quadratic Equations Ex 5(a) engage students in active learning and exploration.
S Chand Class 10 ICSE Maths Solutions Chapter 5 Quadratic Equations Ex 5(a)
Question 1.
(i) (x – 3)(x + 7) = 0
(ii) (3x + 4) (2x – 11) = 0
Solution:
(i) (x – 3) (x + 7) = 0
Either x – 3 = 0, then x = 3
or x + 7 = 0, then x = – 7
∴ x = 3, – 7
(ii) (3x + 4)(2x – 11) = 0
Either 3x + 4 = 0, then 3x = – 4 ⇒ x = \(\frac { -4 }{ 3 }\)
or 2x – 11 = 0, then 2x = 11 ⇒ x = \(\frac { 11 }{ 2 }\)
∴ x = \(\frac { -4 }{ 3 }\), \(\frac { 11 }{ 2 }\)
Question 2.
x² = 4x
Solution:
x² = 4x
⇒ x² – 4x = 0
⇒ x (x – 4) = 0
Either x = 0
or x – 4 = 0, then x = 4
∴ x = 0, 4
Question 3.
\(\left(\frac{1}{3} x-1\right)\left(\frac{1}{2} x+7\right)\) = 0
Solution:
\(\left(\frac{1}{3} x-1\right)\left(\frac{1}{2} x+7\right)\) = 0
Either \(\frac { 1 }{ 3 }\)x – 1 = 0, then \(\frac { 1 }{ 3 }\)x = 1 ⇒ x = 3
or \(\frac { 1 }{ 2 }\)x + 7 = 0, then \(\frac { 1 }{ 2 }\)x = – 7 ⇒ x = – 7 x 2
= – 14
∴ x = 3, – 14
Question 4.
\(\frac{x^2-5 x}{2}\) = 0
Solution:
\(\frac{x^2-5 x}{2}\) = 0
⇒ x² – 5x = 0
⇒ x (x – 5) = 0
Either x = 0
or x – 5 = 0, then x = 5
x = 0, 5
Question 5.
x² – 3x – 10 = 0
Solution:
x² – 3x – 10 = 0
⇒ x² – 5x + 2x – 10 = 0
⇒ x (x – 5) + 2 (x – 5) = 0
⇒ (x – 5) (x + 2) = 0
Either x – 5 = 0, then x = 5
or x + 2 = 0, then x = – 2
∴ x = 5, -2
Question 6.
x² + x – 12 = 0
Solution:
x² + x – 12 = 0
⇒ x² + 4x – 3x – 12 = 0
⇒ x (x + 4) – 3(x + 4) = 0
⇒ (x + 4) (x – 3) = 0
Either x + 4 = 0, then x = – 4
or x – 3 = 0, then x = 3
∴ x = – 4, 3
Question 7.
2 (x² + 1) = 5x (ICSE)
Solution:
2 (x² + 1) = 5x
⇒ 2x² + 2 – 5x = 0
⇒ 2x² – 5x + 2 = 0
⇒ 2x² – 4x – x + 2 = 0
⇒ 2x (x – 2) – 1 (x – 2) = 0
⇒ (x – 2) (2x – 1) = 0
Either x – 2 = 0, then x = 2
or 2x – 1 = 0, then 2x = 1 ⇒ x = \(\frac { 1 }{ 2 }\)
∴ x = 2, \(\frac { 1 }{ 2 }\)
Question 8.
x (2x + 5) = 3 (ICSE)
Solution:
x (2x + 5) = 3
⇒ 2x² + 5x – 3 = 0
⇒ 2x² + 6x – x – 3 = 0
⇒ 2x (x + 3) – 1 (x + 3) = 0
⇒ (x + 3) (2x – 1) = 0
Either x + 3 = 0, then x = -3
or 2x – 1 = 0, then 2x = 1 ⇒ x = \(\frac { 1 }{ 2 }\)
∴ x = – 3, \(\frac { 1 }{ 2 }\)
Question 9.
4x² – 3x – 1 = 0 (ICSE)
Solution:
4x² – 3x – 1 = 0
⇒ 4x² – 4x + x – 1 = 0
⇒ 4x (x – 1) + 1 (x – 1) = 0
⇒ (x – 1)(4x + 1) = 0
Either x – 1 = 0, then x = 1
or 4x + 1 = 0, then 4x = -1
⇒ x = \(\frac { -1 }{ 4 }\)
∴ x = 1, \(\frac { -1 }{ 4 }\)
Question 10.
Solution:
6x² – 13x + 5 = 0
⇒ 6x² – 3x – 10x + 5 = 0
⇒ 3x (2x – 1) – 5 (2x – 1) = 0
⇒ (2x – 1) (3x – 5) = 0
Either 2x – 1 = 0, then 2x = 1 ⇒ x = \(\frac { 1 }{ 2 }\)
or 3x – 5 = 0, then 3x = 5 ⇒ x = \(\frac { 5 }{ 3 }\)
∴ x = \(\frac { 1 }{ 2 }\), \(\frac { 5 }{ 3 }\)
Question 11.
3x² – 5x – 12 = 0
Solution:
3x² – 5x – 12 = 0
⇒ 3x² – 9x + 4x – 12 = 0
⇒ 3x (x – 3) + 4 (x – 3) = 0
⇒ (x – 3) (3x + 4) = 0
Either x – 3 = 0, then x = 3
or 3x + 4 = 0, then 3x = – 4 ⇒ x = \(\frac { – 4 }{ 3 }\)
∴ x = 3, \(\frac { -4 }{ 3 }\)
Question 12.
2x² – 11x + 5 = 0
Solution:
2x² – 11x + 5 = 0
⇒ 2x² – 10x – x + 5 = 0
⇒ 2x (x – 5) – 1 (x – 5) = 0
⇒ (x – 5) (2x – 1) = 0
Either x – 5 = 0, then x = 5
or 2x – 1 = 0. then 2x = 1 ⇒ x = \(\frac { 1 }{ 2 }\)
∴ x = 5, \(\frac { 1 }{ 2 }\)
Question 13.
\(\frac { x }{ 2 }\) + \(\frac { 6 }{ x }\) = 4
Solution:
\(\frac { x }{ 2 }\) + \(\frac { 6 }{ x }\) = 4
⇒ \(\frac{x^2+12}{2 x}\) = 4 ⇒ x² + 12 = 4 x 2x
⇒ x² + 12 = 8x
⇒ x² – 8x + 12 = 0
⇒ x² – 2x – 6x + 12 = 0
⇒ (x – 2) – 6 (x – 2) = 0
⇒ (x – 2) (x – 6) = 0
Either x – 2 = 0, then x = 2
or x – 6 = 0, then x = 6
∴ x = 2, 6
Question 14.
10x – \(\frac { 1 }{ x }\) = 3
Solution:
10x – \(\frac { 1 }{ x }\) = 3
⇒ \(\frac{10 x^2-1}{x}\) = 3 ⇒ 10²x – 1 = 3x
⇒ 10x² – 3x – 1 = 0
⇒ 10x² – 5x + 2x – 1 = 0
⇒ 5x (2x – 1) + 1 (2x – 1) = 0
⇒ (2x- 1) (5x + 1) = 0
Either 2x – 1 = 0, then 2x = 1 ⇒ x = \(\frac { 1 }{ 2 }\)
or 5x + 1 = 0, then 5x = – 1 ⇒ x = – \(\frac { 1 }{ 5 }\)
∴ x = \(\frac { 1 }{ 2 }\), – \(\frac { 1 }{ 5 }\)
Question 15.
9x + \(\frac { 1 }{ x }\) = 6
Solution:
9x + \(\frac { 1 }{ x }\) = 6 ⇒ \(\frac{9 x^2+1}{x}\)
⇒ 9x² +1 = 6x
⇒ 9x² – 6x + 1 = 0
⇒ 9x² – 3x – 3x + 1 = 0
⇒ 3x (3x – 1) – 1 (3x – 1) = 0
⇒ (3x – 1) (3x – 1) = 0
Either 3x – 1 = 0, then 3x = 1 ⇒ x = \(\frac { 1 }{ 3 }\)
or 3x – 1 = 0, then 3x = 1 ⇒ x = \(\frac { 1 }{ 3 }\)
∴ x = \(\frac { 1 }{ 3 }\), \(\frac { 1 }{ 3 }\)
Question 16.
\(\frac{x}{5}+\frac{28}{x+2}\) = 5
Solution:
\(\frac{x}{5}+\frac{28}{x+2}\) = 5
⇒ \(\frac{x(x+2)+28 \times 5}{5(x+2)}=5 \Rightarrow \frac{x^2+2 x+140}{5 x+10}\) = 5
⇒ x² + 2x + 140 = 25x + 50
⇒ x² + 2x + 140 – 25x – 50 = 0
⇒ x² – 23x + 90 = 0
⇒ x² – 18x – 5x + 90 = 0
⇒ x(x – 18) – 5 (x – 18) = 0
⇒ (x – 18) (x – 5) = 0
Either x – 18 = 0, then x = 18
or x – 5 = 0, then x = 5
∴ x = 5, 18
Question 17.
\(\frac{x}{x-1}+\frac{x-1}{x}=2 \frac{1}{2}\)
Solution:
Question 18.
\(\frac{x}{a}-\frac{a+b}{x}=\frac{b(a+b)}{a x}\)
Solution:
\(\frac{x}{a}-\frac{a+b}{x}=\frac{b(a+b)}{a x}\)
⇒ \(\frac{x^2-a^2-a b}{a x}=\frac{a b+b^2}{a x}\)
⇒ \(\frac{x^2-a^2-a b=a b+b^2}{a x}\)
⇒ x² – a² – ab – ab – b² = 0
⇒ x² – (a² + 2ab + b²) = 0
⇒ x² – (a + b)² = 0
⇒ {x + a + b} {x – a – b} = 0
Either x + a + b = 0, then x = – a – b = – (a + b)
or x – a – b = 0, then x = a + b
∴ x = (a + b), -(a + b)
Question 19.
\(\frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{1}{b}\)
Solution:
\(\frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{1}{b}\)
⇒ \(\frac{x-a-b-x}{x(a+b+x)}=\frac{b+a}{a b}\)
⇒ \(\frac{-(a+b)}{x(a+b+x)}=\frac{a+b}{a b}\)
⇒ \(\frac{-1}{x(x+a+b)}=\frac{1}{a b}\) {Dividing (a + b)}
⇒ x (x + a + b) = – ab
⇒ x² + (a + b) x + ab = 0
⇒ x² + ax + bx + ab = 0
⇒ x (x + a) + b (x + a) = 0
⇒ (x + a) (x + b) = 0
Either x + a = 0, then x
or x + b = 0, then x = – b
∴ x = – a, – b
Question 20.
\(\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}\), x ≠ 0, x ≠ 2
Solution:
Question 21.
\(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3}\)
Solution:
Question 22.
\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6}\), x ≠ 1, – 1
Solution:
Question 23.
\(\sqrt{2}\)x² – 3x – 2 \(\sqrt{2}\) = 0
Solution:
\(\sqrt{2}\)x² – 3x – 2 \(\sqrt{2}\) = 0
\(\sqrt{2}\) x (-2\(\sqrt{2}\)) = – 4
∴ \(\sqrt{2}\)x² – 4x + x – 2\(\sqrt{2}\) = 0
\(\sqrt{2}\)x(x-2\(\sqrt{2}\)) + 1 (x – 2\(\sqrt{2}\)) = 0
⇒ (x – 2\(\sqrt{2}\)) (\(\sqrt{2}\)x + 1) = 0
Either x – 2\(\sqrt{2}\) = 0, then x = 2\(\sqrt{2}\)
or \(\sqrt{2}\)x + 1 = 0, then \(\sqrt{2}\)x = – 1 ⇒ x = \(\frac{-1}{\sqrt{2}}\)
∴ x = 2V\(\sqrt{2}\), \(\frac{-1}{\sqrt{2}}\)
Question 24.
a (x² + 1) – x (a² + 1) = 0
Solution:
a (x² + 1) – x (a2 + 1) = 0
⇒ ax² + a – a²x -x = 0
⇒ (x – a) (ax – 1) = 0
Either x – a = 0, then x = a
or ax – 1 = 0, then ax = 1 ⇒ x = \(\frac { 1 }{ a }\)
∴ x = a, \(\frac { 1 }{ a }\)
Question 25.
Find the solution set of :
{x : x² – 2x: – 35 = 0}
Solution:
x² – 2x – 35 = 0
⇒ x² – 7x + 5x – 35 = 0
⇒ x (x – 7) + 5 (x – 7) = 0
⇒ (x – 7) (x + 5) = 0
Either x – 7 = 0, then x = 7
or x + 5 = 0, then x = – 5
∴ x = 7, – 5
∴ Solution set = {- 5, 7}