OP Malhotra Class 10 Maths Solutions Chapter 5 Quadratic Equations Ex 5(c)

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S Chand Class 10 ICSE Maths Solutions Chapter 5 Quadratic Equations Ex 5(c)

Using quadratic formula, find the roots of the following equations:

Question 1.
2x² + x – 3 = 0
Solution:
2x² + x – 3 = 0
Here a = 2, b = 1, c = – 3
∴ D = b² 4ac = (1)² – 4 x 2 x (- 3) = 1 + 24 = 25
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-1 \pm \sqrt{25}}{2 \times 2}\)
= \(\frac{-1 \pm 5}{4}\)
∴ x₁ = \(\frac{-1+5}{4}=\frac{4}{4}\) = 1
x₂ = \(\frac{-1-5}{4}=\frac{-6}{4}=\frac{-3}{2}\)
∴ x = 1, \(\frac { -3 }{ 2 }\)

Question 2.
6x² + 7x – 20 = 0
Solution:
6x² + 7x – 20 = 0
Here a = 6, b = 1, c = – 20
∴ D = b² – 4ac = (7)² – 4 x 6 x (- 20)
= 49 + 480 = 529
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-7 \pm \sqrt{529}}{2 \times 6}\)
= \(\frac{-7 \pm 23}{12}\)
∴ x₁ = \(\frac { -7+23 }{ 12 }\) = \(\frac { 16 }{ 12 }\) = \(\frac { 4 }{ 3 }\)
x₂ = \(\frac { -7-23 }{ 12 }\) = \(\frac { -30 }{ 12 }\) = \(\frac { -5 }{ 2 }\)
x = \(\frac { 4 }{ 3 }\), \(\frac { -5 }{ 2 }\) or 1\(\frac { 1 }{ 3 }\), – 2\(\frac { 1 }{ 2 }\)

Question 3.
9x² + 6x = 35
Solution:
9x² + 6x = 35
⇒ 9x² + 6x – 35 = 0
Here a = 9, b = 6, c = – 35
D = b² – 4ac = (6)² – 4 x 9 x (- 35)
= 36 + 1260 = 1296
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-7 \pm \sqrt{529}}{2 \times 6}\)
= \(\frac{-6 \pm 36}{18}\)
∴ x₁ = \(\frac{-6+36}{18}=\frac{30}{18}=\frac{5}{3}=1 \frac{2}{3}\)
x₂ = \(\frac{-6-36}{18}=\frac{-42}{18}=\frac{-7}{3}=-2 \frac{1}{3}\)
∴ x = \(1 \frac{2}{3},-2 \frac{1}{3}\)

Question 4.
3x² + 7x – 6 = 0
Solution:
3x² + 7x – 6 = 0
Here a = 3, b = 7, c = – 6
∴ D = b² – 4ac = (7)² – 4 x 3 x (- 6) = 49 + 72 = 121
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-7 \pm \sqrt{121}}{2 \times 3}\)
= \(\frac{-7 \pm 11}{6}\)
∴ x₁ = \(\frac{-7+11}{6}=\frac{4}{6}=\frac{2}{3}\)
x₂ = \(\frac{-7-11}{6}=\frac{-18}{6}\)
∴ x = \(\frac{2}{3}\), – 3

Question 5.
x² – 66x + 189 = 0
Solution:
x² – 66x + 189 = 0
Here a = 1, b = – 66, c = 189
∴ D = b² – 4ac = (- 6)² – 4 x 9 x (- 35) = 36 + 1260 = 1296
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-66) \pm \sqrt{3600}}{2 \times 1}\)
= \(\frac{66 \pm 60}{2}\)
∴ x₁ = \(\frac{66+60}{2}=\frac{126}{2}\) = 63
x₂ = \(\frac{66-60}{2}=\frac{6}{2}\) = 3
∴ x = 63, 3

Question 6.
\(\sqrt{3}\)x² + 11x + 6\(\sqrt{3}\) = 0
Solution:
\(\sqrt{3}\)x² + 11x + 6\(\sqrt{3}\) = 0
Here a = \(\sqrt{3}\), b = 11, c = 6\(\sqrt{3}\)
∴ D = b² – 4ac = (11)² – 4 x \(\sqrt{3}\) x 6\(\sqrt{3}\)
= 121 – 72 = 49
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-11 \pm \sqrt{49}}{2 \times \sqrt{3}}\)
= \(\frac{-11 \pm 7}{2 \sqrt{3}}\)
∴ x₁ = \(\frac{-11+7}{2 \sqrt{3}}=\frac{-4}{2 \sqrt{3}}=\frac{-2}{\sqrt{3}}=\frac{-2 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)
= \(\frac{-11-7}{2 \sqrt{3}}=\frac{-18}{2 \sqrt{3}}=\frac{-18 \times \sqrt{3}}{2 \sqrt{3} \times \sqrt{3}}\)
= \(\frac{-18 \sqrt{3}}{6}\) = – 3\(\sqrt{3}\)
∴ x = \(\frac{-2 \sqrt{3}}{3}\), 3\(\sqrt{m}\)

Question 7.
36x² + 23 = 60x
Solution:
36x² + 23 = 60x ⇒ 36x² – 60x + 23 = 0
Here a = 36, b = – 60, c = 23
∴D = b² – 4ac = (- 60)² – 4 x 36 x 23
= 3600 – 3312 = 288
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-60) \pm \sqrt{288}}{2 \times 36}\)
= \(\frac{60 \pm \sqrt{2 \times 144}}{72}=\frac{60 \pm 12 \sqrt{2}}{72}=\frac{5 \pm \sqrt{2}}{6}\)
∴ x = \(\frac{5+\sqrt{2}}{6}, \frac{5-\sqrt{2}}{6}\)

Question 8.
x² – 2x + 5 = 0
Solution:
x² – 2x + 5 = 0
Here a = 1, b = – 2, c = 5
∴ D = b² – 4ac
= (- 2)² – 4 x 1 x 5
= 4 – 20 = – 16
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-2) \pm \sqrt{-16}}{2 \times 1}\)
= \(\frac{2 \pm \sqrt{16 \times(-1)}}{2}=\frac{2 \pm 4 \sqrt{-1}}{2}\)
= 1 ± 2\(\sqrt{-1}\)
∴ x = 1 ± 2\(\sqrt{-1}\)
(But these are not real roots as \(\sqrt{-1}\) is not a real number)

Question 9.
3x² – 17x + 25 = 0
Solution:
3x² – 17x + 25 = 0
Here a = 3, b = – 17, c = 25
∴ D = b² – 4ac = (-17)² – 4 x 3 x 25
= 289 – 300 = – 11
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-17) \pm \sqrt{-11}}{2 \times 3}\)
= \(\frac{17 \pm \sqrt{-11}}{6}\)
∴ x = \(\frac{17+\sqrt{-11}}{6}, \frac{17-\sqrt{-11}}{6}\)
But there are not real roots

Question 10.
15x² – 28 = x
Solution:
15x² – 28 = x ⇒ 15x² – x – 28 = 0
Here a = 15, b = – 1, c = – 28
∴ D = b² – 4ac = (- 1)² – 4 x 15 x (- 28)
= 1 + 1680 = 1681
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-1) \pm \sqrt{1681}}{2 \times 15}\)
= \(\frac{1 \pm 41}{30}\)
x₁ = \(\frac{1+41}{30}=\frac{42}{30}=\frac{7}{5}=1 \frac{2}{5}\)
x₂ = \(\frac{1-41}{30}=\frac{-40}{3 C}=\frac{-4}{3}=-1 \frac{1}{3}\)
∴ x = 1\(\frac { 2 }{ 5 }\), – 1\(\frac { 1 }{ 3 }\)

Question 11.
x² + 3x – 3 = 0, giving your answer to two decimal places.
Solution:
x² + 3x – 3 = 0
Here a = 1, b = 3, c = – 3
∴ D = b² – 4ac = (3)² – 4 x 1 x (- 3)
= 9 + 12 = 21
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-3 \pm \sqrt{21}}{2 \times 1}\)
= \(\frac{-3 \pm 4.58}{2}\)
∴ x₁ = \(\frac{-3+4.58}{2}=\frac{1.58}{2}\) = 0.79
x₂ = \(\frac{-3-4.58}{2}=\frac{-7.58}{2}\) = – 3.79
∴ x = 0.79, – 3.79

Question 12.
\(\frac { 2 }{ 3 }\)x = \(\frac { -1 }{ 6 }\)x² – \(\frac { 1 }{ 3 }\), giving your answer to 2 d.p.
Solution:
\(\frac { 2 }{ 3 }\)x = \(\frac { -1 }{ 6 }\)x² – \(\frac { 1 }{ 3 }\)
⇒ \(\frac { 1 }{ 6 }\) x² + \(\frac { 2 }{ 3 }\)x + \(\frac { 1 }{ 3 }\) = 0
⇒ x² + 4x + 2 = 0
(Multiplying by L.C.M. of 6, 3, 3 = 6)
Here a = 1, b = 4, c = 2
∴ D = b² – 4ac = (4)² – 4 x 1 x 2 = 16 – 8 = 8
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-4 \pm \sqrt{8}}{2 \times 1}\)
= \(\frac{-4 \pm 2 \sqrt{2}}{2}\)
= – 2 ± \(\sqrt{2}\)
= – 2 ± 1.41
∴ x₁ = – 2 + 1.41 = – 0.59
x₂ = – 2 – 1.41 = – 3.41
∴ x = – 0.59, – 3.41

Question 13.
x² + 6x – 10 = 0
Solution:
x² + 6x – 10 = 0
Here a = 1, b = 6, c = – 10
D = b² – 4ac = (6)² – 4 x 1 x (- 10)
= 36 + 40 = 76
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-6 \pm \sqrt{76}}{2 \times 1}\)
= \(\frac{-6 \pm \sqrt{4 \times 19}}{2 \times 1}\)
= \(\frac{-6 \pm 2 \sqrt{19}}{2}\)
= – 3 ± \(\sqrt{-19}\)
∴ x = – 3 + \(\sqrt{19}\), – 3 – \(\sqrt{19}\)

Question 14.
\(\frac{x^2+8}{11}\) = 5x – x² – 5
Solution:
\(\frac{x^2+8}{11}\) = 5x – x² – 5
⇒ x² + 8 = 55x – 11x² – 55
⇒ x² + 8 – 55x + 11x² + 55 = 0
⇒ 12x² – 55x + 63 = 0
Here a = 12, b = – 55, c = 63
∴D = b² – 4ac = (- 55)² – 4 x 12 x 63
= 3025 – 3024 = 1
∴ x₁ = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-55) \pm \sqrt{1}}{2 \times 12}\)
= \(\frac{55 \pm 1}{24}\)
∴ x₁ = \(\frac{55+1}{24}=\frac{56}{24}=\frac{7}{3}=2 \frac{1}{3}\)
x₂ = \(\frac{55-1}{24}=\frac{54}{24}=\frac{9}{4}=2 \frac{1}{4}\)
∴ x = 2\(\frac { 1 }{ 3 }\), 2\(\frac { 1 }{ 4 }\)

Question 15.
y – \(\frac { 3 }{ y }\) = \(\frac { 1 }{ 2 }\)
Solution:
y – \(\frac { 3 }{ y }\) = \(\frac { 1 }{ 2 }\) ⇒ 2y² – 6 = y
⇒ 2y² – y – 6 = 0
Here a = 2, b = – 1, c = – 6
∴D = b² – 4ac = (- 1)² – 4 x 2 x (- 6) = 1 + 48 = 49
∴ y₁ = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(1) \pm \sqrt{49}}{2 \times 2}=\frac{1 \pm 7}{4}\)
y₁ = \(\frac{1+7}{4}=\frac{8}{4}\) = 2
∴ y₁ = \(\frac{1-7}{4}=\frac{-6}{4}=\frac{-3}{2}=-1 \frac{1}{2}\)
∴ y = 2, – 1\(\frac { 1 }{ 2 }\)

Question 16.
2x + \(\frac { 4 }{ x }\) = 9
Solution:
2x + \(\frac { 4 }{ x }\) = 9 ⇒ 2x² + 4 = 9 ⇒ 2x² + 4 = 0
⇒ 2x² – 9x + 4 = 0
Here a = 2, b = – 9, c = 4
∴ D = b² – 4ac = (- 9)² – 4 x 2 x 4
= 81 – 32 = 49
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-9) \pm \sqrt{49}}{2 \times 2}\)
= \(\frac{9 \pm 7}{4}\)
∴ x₁ = \(\frac{9+7}{4}=\frac{16}{4}\) = 4
x₂ = \(\frac{9-7}{4}=\frac{2}{4}=\frac{1}{2}\)
∴ x = 4, \(\frac { 1 }{ 2 }\)

Question 17.
\(\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}\), x ≠ 0, x ≠ 1
Solution:

Question 18.
\(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=8 \frac{1}{3}\)
Solution:
\(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=8 \frac{1}{3}\)
⇒ \(\frac{2 x(x-3)+(x-4)(2 x-5)}{(x-4)(x-3)}=\frac{25}{3}\)
⇒ \(\frac{2 x^2-6 x+2 x^2-5 x-8 x+20}{x^2-3 x-4 x+12}=\frac{25}{3}\)
⇒ \(\frac{4 x^2-19 x+20}{x^2-7 x+12}=\frac{25}{3}\)
⇒ 25x² – 175x + 300 – 12x² – 57x + 60 = 0
(By cross multiplication)
⇒ 25x² – 175x + 300 – 12x² + 57x – 60 = 0
⇒ 13x² – 118x + 240 = 0
Here a = 13, b = – 118, c = 240
D = b² – 4ac = (- 118)² – 4 x 13 x 240
= 13924 – 12480 = 1444
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-118) \pm \sqrt{1444}}{2 \times 13}\)
= \(\frac{118 \pm 38}{26}\)
∴ x₁ = \(\frac{118+38}{26}=\frac{156}{26}\) = 6
x₂ = \(\frac{118-38}{26}=\frac{80}{26}=\frac{40}{13}=3 \frac{1}{13}\)
∴ x = 6, 3\(\frac { 1 }{ 13 }\)

Question 19.
\(\frac{x+6}{x+7}-\frac{x+1}{x+2}=\frac{1}{3 x+1}\)
Solution:

Question 20.
\(\frac{x+1}{2 x+5}=\frac{x+3}{3 x+4}\)
Solution:
\(\frac{x+1}{2 x+5}=\frac{x+3}{3 x+4}\)
⇒ (x + 1) (3x + 4) = (x + 3) (2x + 5)
(By cross multiplication)
⇒ 3x² + 4x + 3x + 4 = 2x² + 5x + 6x + 15
⇒ 3x² + 7x + 4 – 2x² – 11 x – 15 = 0
⇒ x² – 4x – 11 = 0
Here a = 1, b = – 4, c = – 11
∴ D = b² – 4ac = (- 4)² – 4 x 1 x (- 11)
= 16 + 44 = 60
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-4) \pm \sqrt{60}}{2 \times 1}\)
= \(\frac{4 \pm \sqrt{4 \times 15}}{2}=\frac{4 \pm 2 \sqrt{15}}{2}\)
= 2 ± \(\sqrt{15}\)
∴ x = 2 + \(\sqrt{15}\), 2 – \(\sqrt{15}\)

Question 21.
Solve using the quadratic formula :
(i) a²x² – 3abx + 2b² = 0
(ii) x² – x – a(a + 1) = 0
(iii) 10x² + 3bx + a² – 7ax – b² = 0
Solution:
(i) a²x² – 3abx + 2b2 = 0
Here A = a², B = – 3ab, C = 2b²
∴ D = B² – 4AC = (- 3ab)² – 4 x a² x 2b²
= 9a²b² – 8a²b² = a²b²
∴ x = \(\frac{-B \pm \sqrt{B^2-4 A C}}{2 A}\)
= \(\frac{-(-3 a b) \pm \sqrt{a^2 b^2}}{2 a^2}=\frac{3 a b \pm a b}{2 a^2}\)
x₁ = \(\frac{3 a b+a b}{2 a^2}=\frac{4 a b}{2 a^2}=2 \frac{b}{a}\)
x₂ = \(\frac{3 a b-a b}{2 a^2}=\frac{2 a b}{2 a^2}=\frac{b}{a}\)
Hence x = \(\frac { 2b }{ a }\), \(\frac { b }{ a }\)

(ii) x² – x – a(a + 1) = 0
Here A = 1, B = – 1, C = – a (a + 1)
D = B² – 4AC = (- 1)² – 4 x 1 x (- a² – a)
= 1 + 4a² + 4 a = (2a + 1 )²
∴ x = \(\frac{-B \pm \sqrt{B^2-4 A C}}{2 A}\)
= \(\frac{-(-1) \pm \sqrt{(2 a+1)^2}}{2 \times 1}=\frac{1 \pm(2 a+1)}{2}\)
∴ x₁ = \(\frac{1+2 a+1}{2}=\frac{2 a+2}{2}\) = a + 1
x₂\(\frac{1-2 a-1}{2}=\frac{-2 a}{2}\) = – a
∴ x₁ = – a, a + 1

(iii) 10x² + 3bx + a² – 7ax – b² = 0
⇒ 10x² + (3b – 7a) x + (a² – b²) = 0
Here A = 10, B = 3b – 7a, C = a² – b²
∴ D = B² – 4AC = (3b – 7a)² – 4 x 10 x (a² – b²)
= 9b² + 49a² – 42ab – 40 (a² – b²)
= 9b² + 49a² – 42ab – 40a² + 40b²
= 9a² + 49b² – 42ab = (3a)² + (7b)² – 2 x 3 a x 7b = (3a – 7b)²