Quadratic Equations Class 10 OP Malhotra Exe-5C ICSE Maths Solutions Ch-5. We Provide Step by Step Solutions / Answer of Exe-5C 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-5C ICSE Maths Solutions
Board | ICSE |
Publications | S Chand |
Subject | Maths |
Class | 10th |
Chapter-5 | Quadratic Equations |
Writer | OP Malhotra |
Exe-5C | Quadratic Equations Using Formula |
Edition | 2024-2025 |
How to Solve Quadratic Equations Using Formula
The quadratic equation formula to solve the equation ax2 + bx + c = 0 is x = [-b ± √(b2 – 4ac)]/2a. Here we obtain the two values of x, by applying the plus and minus symbols in this formula. Hence the two possible values of x are [-b + √(b2 – 4ac)]/2a, and [-b – √(b2 – 4ac)]/2a
Exe-5C
( Quadratic Equations Class 10 OP Malhotra Exe-5C ICSE Maths Solutions Ch-5 )
Using Quadratic formula, find the roots of the following equations :
Que-1: 2x2 + x – 3 = 0
Sol: Given that 2x2 + x – 3 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 2, b = 1, c = -3
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-1±√(1²-4×2×(-3))]/2×2
x = [-1±√(1+24)]/4
x = [-1±√25]/4 = [-1±5]/4
x = [-1+5]/4 or [-1-5]/4
x = 1 or -3/2
Que-2: 6x2 + 7x – 20 = 0
Sol: Given that 6x2 + 7x – 20 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 6, b = 7, c = -20
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-7±√(7²-4×6×(-20))]/2×6
x = [-7±√(49+480)]/12
x = [-7±√529]/12 = [-7±23]/12
x = [-7+23]/12 or [-7-23]/12
x = 16/12 or -30/12
x = 4/3 = 1*(1/3) or -5/2 = -2*(1/2).
Que-3: 9x2 + 6x = 35
Sol: Given that 9x2 + 6x = 35 ⇒ 9x2 + 6x – 35 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 9, b = 6, c = -35
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-6±√(6²-4×9×(-35))]/2×9
x = [-6±√(36+1260)]/18
x = [-6±√1296]/18 = [-6±36]/18
x = [-6+36]/18 or [-6-36]/18
x = 30/18 or -42/18
x = 5/3 = 1*(2/3) or -7/3 = -2*(1/3).
Que-4: 3x2 + 7x – 6 = 0
Sol: Given that 3x2 + 7x – 6 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 3, b = 7, c = -6
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-7±√(7²-4×3×(-6))]/2×3
x = [-7±√(49+72)]/6
x = [-7±√121]/6 = [-7±11]/6
x = [-7+11]/6 or [-7-11]/6
x = 4/6 or -18/6
x = 2/3 or -3.
Que-5: x2 – 66x + 189 = 0
Sol: Given that x2 – 66x + 189 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 1, b = -66, c = 189
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(-66)±√((-66)²-4×1×189)]/2×1
x = [66±√(4356-756)]/2
x = [66±√3600]/2 = [66±60]/2
x = [66+60]/2 or [66-60]/2
x = 126/2 or 6/2
x = 63 or 3.
Que-6: √3x2 + 11x + 6√3 = 0
Sol: Given that √3x2 + 11x + 6√3 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = √3, b = 11, c = 6√3
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(11)±√((11)²-4×√3×6√3)]/2×√3
x = [-11±√(121-72)]/2√3
x = [-11±√49]/2√3 = [-11±7]/2√3
x = [-11+7]/2√3 or [-11-7]/2√3
x = -4/2√3 or -18/2√3
x = -2√3/3 or -3√3.
Que-7: 36x2 + 23 = 60x
Sol: Given that 36x2 + 23 = 60x ⇒ 36x2 – 60x + 23 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 36, b = -60, c = 23
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(-60)±√((-60)²-4×36×23)]/2×36
x = [60±√(3600-3312)]/72
x = [60±√288]/72 = [60±12√2]/72
x = [5±√2]/6
x = (5+√2)/6 or (5-√2)/6.
Que-8: x2 – 2x + 5 = 0
Sol: Given that x2 – 2x + 5 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 1, b = -2, c = 5
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(-2)±√((-2)²-4×1×5)]/2×1
x = [2±√(4-20)]/2
x = [2±√-16]/2 = [2±4√-1]/2
x = (1±2√-1)
x = 1+2√-1 or 1-2√-1.
Que-9: 3x2 – 17x + 25 = 0
Sol: Given that 3x2 – 17x + 25 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 3, b = -17, c = 25
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(-17)±√((-17)²-4×3×25)]/2×3
x = [17±√(289-300)]/6
x = [17±√-11]/6
x = (17+√-11)/6 or (17-√-11)/6
Que-10: 15x2 – 28 = x
Sol: Given that 15x2 – 28 = x ⇒ 15x2 – x – 28 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 15, b = -1, c = -28
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(-1)±√((-1)²-4×15×(-28))]/2×15
x = [1±√(1+1680)]/30
x = [1±√1681]/30 = [1±41]/30
x = [1+41]/30 or [1-41]/30
x = 42/30 or -40/30
x = 7/5 or -4/3.
Que-11: x2 + 3x – 3 = 0, giving your answer correct to two decimal places.
Sol: Given that x2 + 3x – 3 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 1, b = 3, c = -3
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(3)±√((3)²-4×1×(-3))]/2×1
x = [-3±√(9+12)]/2
x = [-3±√21]/2 = [-3±4.58]/2
x = [-3+4.58]/2 or [-3-4.58]/2
x = 1.58/2 or -7.58/2
x = 0.79 or -3.79.
Que-12: (2/3)x = (-1/6)x2 – 1/3, giving your answer correct to two decimal places.
Sol: Given that (2/3)x = (-1/6)x2 – 1/3 ⇒ x2 + 4x + 2 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 1, b = 4, c = 2
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(4)±√((4)²-4×1×(2))]/2×1
x = [-4±√(16-8)]/2
x = [-4±√8]/2 = [-4±2.82]/2
x = [-4+2.82]/2 or [-4-2.82]/2
x = -1.18/2 or -6.82/2
x = -0.59 or -3.41.
Que-13: x2 + 6x – 10 = 0
Sol: Given that x2 + 6x – 10 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 1, b = 6, c = -10
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(6)±√((6)²-4×1×(-10))]/2×1
x = [-6±√(36+40)]/2
x = [-6±√76]/2 = [-6±2√2]/2
x = -3±√19.
Que-14: (x2 + 8)/11 = 5x – x2 – 5
Sol: Given that (x2 + 8)/11 = 5x – x2 – 5 ⇒ 12x2 – 55x + 63 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 12, b = -55, c = 63
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(-55)±√((-55)²-4×12×(-63))]/2×12
x = [55±√(3025-3024)]/24
x = [55±√1]/24 = [55±1]/24
x = [55+1]/24 or [55-1]/24
x = 56/24 or 54/24
x = 7/3 or 9/4.
Que-15: y – 3/y = 1/2
Sol: Given that y – 3/y = 1/2 ⇒ 2y2 – y – 6 = 0.
Then, comparing this quadratic equation with ay2 + by + c = 0, we get a = 2, b = -1, c = -6
Putting the values of a, b and c quadratic formula y = [- b ± √(b2 – 4ac)]/2a.
x = [-(-1)±√((-1)²-4×2×(-6))]/2×2
x = [1±√(1+48)]/4
x = [1±√49]/4 = [1±7]/4
x = [1+7]/4 or [1-7]/4
x = 8/4 or -6/4
x = 2 or -3/2.
Que-16: 2x + 4/x = 9
Sol: Given that 2x + 4/x = 9 ⇒ 2x2 – 9x + 4 = 0.
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 2, b = -9, c = 4
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(-9)±√((-9)²-4×2×(4))]/2×2
x = [9±√(81-32)]/4
x = [9±√49]/4 = [9±7]/4
x = [9+7]/4 or [9-7]/4
x = 16/4 or -2/4
x = 4 or -1/2.
Que-17: [x/(x + 1)] + [(x + 1)/x] = 34/15, x ≠ 0, x ≠ -1.
Sol: Given that x/(x + 1) + (x + 1)/x = 34/15
Put y = x/(x + 1), then
⇒ y + 1/y = 34/15
⇒ 15y2 – 34y + 15 = 0
Then, comparing this quadratic equation with ay2 + by + c = 0, we get a = 15, b = -34, c = 15
Putting the values of a, b and c quadratic formula y = [- b ± √(b2 – 4ac)]/2a.
y = [-(-34)±√((-34)²-4×15×(15))]/2×15
y = [34±√(1156-900)]/30
y = [34±√256]/30 = [34±16]/30
y = [34+16]/30 or [34-16]/30
y = 50/30 or 18/30
y = 5/3 or 3/5
If y = 5/3, then
x/(x + 1) = 5/3
⇒ 3x = 5x + 5
⇒ 2x = -5
⇒ x = -5/2
If y = 3/5, then
x/(x + 1) = 3/5
⇒ 5x = 3x + 3
⇒ 2x = 3
⇒ x = 3/2
Thus, the roos of the given equation are x = -5/2 or 3/2.
Que-18: [2x/(x – 4)] + [(2x – 5)/(x – 3)] = 8*(1/3)
Sol: Given that 2x/(x – 4) + (2x – 5)/(x – 3) = 8 1/3
⇒ [2x(x – 3) + (2x – 5)(x – 4)]/(x – 4)(x – 3) = 25/3
⇒ 3[2x2 – 6x + 2x2 – 13x + 20] = 25(x – 4)(x – 3)
⇒ 3[4x2 – 19x + 20] = 25(x2 – 7x + 12)
⇒ 12x2 – 57x + 60 = 25x2 – 175x + 300
⇒ 13x2 – 118x + 240 = 0
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 13, b = -118, c = 240
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(-118)±√((-118)²-4×13×(240))]/2×13
x = [118±√(13924-12480)]/26
x = [118±√1444]/26 = [118±38]/26
x = [118+38]/26 or [118-38]/26
x = 156/26 or 80/26
x = 6 or 40/13.
Thus, the roos of the given equation are x = 6 or 40/13.
Que-19: [(x + 6)/(x + 7)] – [(x + 1)/(x + 2)] = [1/(3x + 1)]
Sol: Given that (x + 6)/(x + 7) – (x + 1)/(x + 2) = 1/(3x + 1)
⇒ [(x + 6)(x + 2) – (x + 1)(x + 7)]/(x + 7)(x + 2) = 1/(3x + 1)
⇒ (3x + 1)(x2 + 8x + 12 – x2 – 8x – 7) = (x2 + 9x + 14)
⇒ 5(3x + 1) = (x2 + 9x + 14)
⇒ 15x + 5 = x2 + 9x + 14
⇒ x2 – 6x + 9 = 0
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 1, b = -6, c = 9
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(-6)±√((-6)²-4×1×(9))]/2×1
x = [6±√(36-36)]/2
x = [6±√0]/2 = [6±0]/2
x = 3, 3.
Que-20: (x + 1)/(2x + 5) = (x + 3)/(3x + 4)
Sol: Given that (x + 1)/(2x + 5) = (x + 3)/(3x + 4)
⇒ (x + 1)(3x + 4) = (x + 3)(2x + 5)
⇒ 3x2 + 7x + 4 = 2x2 + 11x + 15
⇒ x2 – 4x – 11 = 0
Then, comparing this quadratic equation with ax2 + bx + c = 0, we get a = 1, b = -4, c = -11
Putting the values of a, b and c quadratic formula x = [- b ± √(b2 – 4ac)]/2a.
x = [-(-4)±√((-4)²-4×1×(-11))]/2×1
x = [4±√(16+44)]/2
x = [4±√60]/2 = [4±2√15]/2
x = 2±√15.
Que-21: Solve using quadratic formula :
(i) a2x2 – 3abx + 2b2 = 0
(ii) x2 – x – a(a + 1) = 0
(iii) 10x2 + 3bx + a2 – 7ax – b2 = 0
Sol: (i) Given that a2x2 – 3abx + 2b2 = 0
Then, comparing this quadratic equation with Ax2 + Bx + C = 0, we get A = a2, B = -3ab, C = 2b2
Putting the values of A, B and C quadratic formula x = [- B ± √(B2 – 4AC)]/2A.
= [-(-3ab)±√((-3ab)²-4×a²×(2b²))]/2×a²
x = [3ab±√(9a²b²-8a²b²)]/2a²
x = [3ab±√a²b²]/26 = [3ab±ab]/2a²
x = [3ab+ab]/2a² or [3ab-ab]/2a²
x = 4ab/2a² or 2ab/2a²
x = 2b/a or b/a.
(ii) Given that x2 – x – a(a + 1) = 0
Then, comparing this quadratic equation with Ax2 + Bx + C = 0, we get A = 1, B = -1, C = -a(a + 1)
Putting the values of A, B and C quadratic formula x = [- B ± √(B2 – 4AC)]/2A.
x = [-(-1)±√((-1)²-4×1×-a(a+1))]/2×1
x = [1±√(1+4a+4a²)]/2
x = [1±√(2a+1)²]/2 = [1±(2a+1)]/2
x = [1+2a+1]/2 or [1-2a-1]/2
x = (2a+2)/2 or -2a/2
x = a+1 or -a.
(iii) Given that 10x2 + 3bx + a2 – 7ax – b2 = 0
⇒ 10x2 + (3b – 7a)x + (a2 – b2) = 0
Then, comparing this quadratic equation with Ax2 + Bx + C = 0, we get A = 10, B = (3b – 7a), C = (a2 – b2).
Putting the values of A, B and C quadratic formula x = [- B ± √(B2 – 4AC)]/2A.
x = [-(3b-7a)±√((3b-7a)²-4×10×(a²+b²))]/2×10
x = [-3b+7a±√(9b²+49a²-42ab-40a²+40b²)]/20
x = [-3b+7a±√(9a²+49b²-42ab)]/20
= [-3b+7a±√(3a-7b)²]/20
x = [-3b+7a±(3a-7b)]/20
x = [-3b+7a+3a-7b]/20 or [-3b+7a-3a+7b]/20
x = (10a-10b)/20 or (4a+4b)/20
x = (a-b)/2 or (a+b)/5.
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