Quadratic Equations Class 10 OP Malhotra Exe-5A ICSE Maths Solutions Ch-5. We Provide Step by Step Solutions / Answer of Exe-5A Questions of S Chand OP Malhotra Maths . Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Quadratic Equations Class 10 OP Malhotra Exe-5A ICSE Maths Solutions Ch-5
Board | ICSE |
Publications | S Chand |
Subject | Maths |
Class | 10th |
Chapter-5 | Quadratic Equations |
Writer | OP Malhotra |
Exe-5A | Solving by Factorisation |
Edition | 2024-2025 |
Quadratic Equations by Factorisation
- Step 1: Consider the quadratic equation ax2 + bx + c = 0
- Step 2: Now, find two numbers such that their product is equal to ac and sum / difference equals to b = ac,
- Step 3: Now, split the middle B term using these two numbers,
- Step 4: Take the common factors out and simplify
Exercise- 5A
Solve :
Que-1: (i) (x – 3)(x + 7) = 0
(ii) (x – 3)(x + 7) = 0
Sol: (i) (x – 3)(x + 7) = 0
⇒ x – 3 = 0 or x + 7 = 0
⇒ x = 3 or x = -7.
(ii) (3x + 4)(2x – 11) = 0
⇒ 3x + 4 = 0 or 2x – 11 = 0
⇒ x = -4/3 or x = 11/2.
Que-2: x2 = 4x
Sol: x2 = 4x
⇒ (x2 – 4x) = 0
⇒ x(x – 4) = 0
⇒ x = 0 or x – 4 = 0
⇒ x = 0 or x = 4
Que-3: ((1x/3) – 1)((1x/2) + 7) = 0
Sol: ((1x/3) – 1)((1x/2) + 7) = 0
⇒ x/3 – 1 = 0 or x/2 + 7 = 0
⇒ x/3 = 1 or x/2 = -7
⇒ x = 3 or x = -14
Que-4: (x2 – 5x) / 2 = 0
Sol: (x2 – 5x) / 2 = 0
⇒ (x2 – 5x) = 0
⇒ x(x – 5) = 0
⇒ x = 0 or x – 5 = 0
⇒ x = 0 or 5
Que-5: x2 – 3x – 10 = 0
Sol: x2 – 3x – 10 = 0
⇒ x2 – 5x + 2x – 10 = 0
⇒ x(x – 5) + 2(x – 5) = 0
⇒ (x – 5)(x + 2) = 0
⇒ x = 5 or -2
Que-6: x2 + x – 12 = 0
Sol: x2 + x – 12 = 0
⇒ x2 + 4x – 3x – 12 = 0
⇒ x(x + 4) – 3(x + 4) = 0
⇒ (x + 4)(x – 3) = 0
⇒ x = -4 or 3
Que-7: 2(x2 + 1) = 5x
Sol: 2(x2 + 1) = 5x
⇒ 2x2 – 5x + 2 = 0
⇒ 2x2 – 4x – x + 2 = 0
⇒ 2x(x – 2) – 1(x – 2) = 0
⇒ (x – 2)(2x – 1) = 0
⇒ x = 2 or 1/2
Que-8: x(2x + 5) = 3
Sol: 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
⇒ x + 3 = 0 or 2x – 1 = 0
⇒ x = -3 or 1/2
Que-9: 4x2 – 3x – 1 = 0
Sol: 4x2 – 3x – 1 = 0
⇒ 4x2 – 4x + x – 1 = 0
⇒ 4x(x – 1) + 1(x – 1) = 0
⇒ (x – 1)(4x + 1) = 0
⇒ x – 1 = 0 or 4x + 1 = 0
⇒ x = 1 or -1/4
Que-10: 6x2 – 13x + 5 = 0
Sol: 6x2 – 13x + 5 = 0
⇒ 6x2 – 10x – 3x + 5 = 0
⇒ 2x(3x – 5) – 1(3x – 5) = 0
⇒ (3x – 5)(2x – 1) = 0
⇒ 3x – 5 = 0 or 2x – 1 = 0
⇒ x = 5/3 or 1/2
Que-11: 3x2 – 5x – 12 = 0
Sol: 3x2 – 5x – 12 = 0
⇒ 3x2 – 9x + 4x – 12 = 0
⇒ 3x(x – 3) + 4(x – 3) = 0
⇒ (x – 3)(3x + 4) = 0
⇒ x – 3 = 0 or 3x + 4 = 0
⇒ x = 3 or -4/3
Que-12: 2x2 – 11x + 5 = 0
Sol: 2x2 – 11x + 5 = 0
⇒ 2x2 – 10x – x + 5 = 0
⇒ 2x(x – 5) – 1(x – 5) = 0
⇒ (x – 5)(2x – 1) = 0
⇒ x – 5 = 0 or 2x – 1 = 0
⇒ x = 5 or 1/2
Que-13: x/2 + 6/x = 4
Sol: (x/2) + (6/x) = 4
⇒ (x2 + 12)/2x = 4
⇒ x2 – 8x + 12 = 0
⇒ x2 – 6x – 2x + 12 = 0
⇒ x(x – 6) – 2(x – 6) = 0
⇒ (x – 6)(x – 2) = 0
⇒ x – 6 = 0 or x – 2 = 0
⇒ x = 6 or 2.
Que-14: 10x – (1/x) = 3
Sol: 10x – (1/x) = 3
⇒ 10x2 – 1 = 3x
⇒ 10x2 – 3x – 1 = 0
⇒ 10x2 – 5x + 2x – 1 = 0
⇒ 5x(2x – 1) + 1(2x – 1) = 0
⇒ (2x – 1)(5x + 1) = 0
⇒ 2x – 1 = 0 or 5x + 1 = 0
⇒ x = 1/2 or -1/5
Que-15: 9x + (1/x) = 6
Sol: 9x + 1/x = 6
⇒ 9x2 + 1 = 6x
⇒ 9x2 – 6x + 1 = 0
⇒ (3x)2 – 2(3x)(1) + 1 = 0
⇒ (3x – 1)2 = 0
⇒ (3x – 1)(3x – 1) = 0
⇒ (3x – 1) = 0 or (3x – 1) = 0
⇒ x = 1/3 or 1/3
Que-16: (x/5) + [28/(x + 2)] = 5
Sol: [x(x + 2) + 140] / 5(x + 2) = 5
⇒ x2 + 2x + 140 = 25(x + 2)
⇒ x2 + 2x + 140 = 25x + 50
⇒ x2 + 2x – 25x + 140 – 50 = 0
⇒ x2 – 23x + 90 = 0
⇒ x2 – 18x – 5x + 90 = 0
⇒ x(x – 18) – 5(x – 18) = 0
⇒ (x – 18)(x – 5) = 0
⇒ (x – 18) = 0 or (x – 5) = 0
⇒ x = 18 or x = 5
⇒ x = 18 or 5
Que-17: [x/(x – 1)] + [(x – 1)/x] = 2*(1/2)
Sol: [x/(x – 1)] + [(x – 1)/x] = 2*(1/2)
⇒ [x2 + (x – 1)2] / x(x – 1) = 5/2
⇒ 2[x2 + x2 – 2x + 1] = 5x(x – 1)
⇒ 4x2 – 4x + 2 = 5x2 – 5x
⇒ x2 – x – 2 = 0
⇒ x2 – 2x + x – 2 = 0
⇒ x(x – 2) + 1(x – 2) = 0
⇒ (x + 1)(x – 2) = 0
⇒ (x + 1) = 0 or (x – 2) = 0
⇒ x = -1 or 2
Que-18: (x/a) – [(a + b)/x] = b(a + b)/ax
Sol: (x/a) – [(a + b)/x] = [b(a + b)/ax]
⇒ [x2 – a(a + b)]/ax = b(a + b)/ax
⇒ x2 – a(a + b) = b(a + b)
⇒ x2 – a(a + b) – b(a + b) = 0
⇒ x2 – (a + b)2 = 0
⇒ [x + (a + b)][x – (a + b)] = 0
⇒ [x + (a + b)] = 0 or [x – (a + b)] = 0
⇒ x = (a + b) or -(a + b)
Que-19: [1/(a + b + x)] – (1/x) = (1/a) + (1/b)
Sol: [1/(a + b + x)] – (1/x) = (1/a) + (1/b)
⇒ (x – a – b – x)/x(a + b + x) = (a + b)/ab
⇒ -(a + b)/x(a + b + x) = (a + b)/ab
⇒ -1/x(a + b + x) = 1/ab
⇒ -x(a + b + x) = ab
⇒ -x2 – ax – bx – ab = 0
⇒ x2 + ax + bx + ab = 0
⇒ x(x + a) + b(x + a) = 0
⇒ (x + a)(x + b) = 0
⇒ (x + a) = 0 or (x + b) = 0
⇒ x = -a or -b
Que-20: [(x + 3)/(x – 2)] – [(1 – x)/x] = 17/4, x ≠ 0, 2.
Sol: [(x + 3)/(x – 2)] – [(1 – x)/x] = 17/4
⇒ [x(x + 3) – (x – 2)(1 – x)] / x(x – 2) = 17/4
⇒ [x2 + 3x – x + x2 + 2 – 2x] / (x2 – 2x) = 17/4
⇒ 4[2x2 + 2] = 17(x2 – 2x)
⇒ 8x2 + 8 = 17x2 – 34x
⇒ 17x2 – 8x2 – 34x – 8 = 0
⇒ 9x2 – 34x – 8 = 0
⇒ 9x2 – 36x + 2x – 8 = 0
⇒ 9x(x – 4) + 2(x – 4) = 0
⇒ (x – 4)(9x + 2) = 0
⇒ (x – 4) = 0 or (9x + 2) = 0
⇒ x = 4 or -2/9
Que-21: [2x/(x – 4)] + [(2x – 5)/(x – 3)] = 25/3
Sol: [2x/(x – 4)] + [(2x – 5)/(x – 3)] = 25/3
⇒ [2x(x – 3) + (2x – 5)(x – 4)] / (x – 4)(x – 3) = 25/3
⇒ [2x2 – 6x + 2x2 – 8x – 5x + 20] / (x2 – 7x + 12) = 25/3
⇒ 3[4x2 – 19x + 20] = 25(x2 – 7x + 12)
⇒ 12x2 – 57x + 60 = 25x2 – 175x + 300
⇒ 25x2 – 12x2 – 175x + 57x + 300 – 60 = 0
⇒ 13x2 – 118x + 240 = 0
⇒ 13x2 – 78x – 40x + 240 = 0
⇒ 13x(x – 6) – 40(x – 6) = 0
⇒ (x – 6)(13x – 40) = 0
⇒ (x – 6) = 0 or (13x – 40) = 0
⇒ x = 6 or 40/13
Que-22: [(x + 1)/(x – 1)] – [(x – 1)/(x + 1)] = 5/6, x ≠ 1, -1.
Sol: [(x + 1)/(x – 1)] – [(x – 1)/(x + 1)] = 5/6
⇒ [(x + 1)2 – (x – 1)2] / (x – 1)(x + 1) = 5/6
⇒ 6[x2 + 2x + 1 – x2 +2x – 1] = 5(x2 – 1)
⇒ 6[4x] = 5(x2 – 1)
⇒ 5x2 – 24x – 5 = 0
⇒ 5x2 – 25x + x – 5 = 0
⇒ 5x(x – 5) + 1(x – 5) = 0
⇒ (5x + 1)(x – 5) = 0
⇒ (5x + 1) = 0 or (x – 5) = 0
⇒ x = -1/5 or 5
Que-23: √2x2 – 3x – 2√2 = 0
Sol: √2x2 – 3x – 2√2 = 0
⇒ √2x2 – 4x + x – 2√2 = 0
⇒ √2x(x – 2√2) + 1(x – 2√2) = 0
⇒ (√2x + 1)(x – 2√2) = 0
⇒ (√2x + 1) = 0 or (x – 2√2) = 0
⇒ x = -1/√2 or 2√2
Que-24: a(x2 + 1) – x(a2 + 1) = 0
Sol: a(x2 + 1) – x(a2 + 1) = 0
⇒ ax2 + a – xa2 – x = 0
⇒ ax2 – xa2 – x + a = 0
⇒ ax(x – a) – 1(x – a) = 0
⇒ (ax – 1)(x – a) = 0
⇒ (ax – 1) = 0 or (x – a) = 0
⇒ x = 1/a or a
Que-25: Find the solution set of {x : x2 – 2x – 35 = 0}
Sol: x2 – 2x – 35 = 0
⇒ x2 – 7x + 5x – 35 = 0
⇒ x(x – 7) + 5(x – 7) = 0
⇒ (x + 5)(x – 7) = 0
⇒ (x + 5) = 0 or (x – 7) = 0
⇒ x = -5 or 7
⇒ x ∈ {-5, 7}
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