Quadratic Equations Class 10 OP Malhotra Exe-5B ICSE Maths Ch-5. We Provide Step by Step Solutions / Answer of Exe-5B 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-5B ICSE Maths Ch-5
Board | ICSE |
Publications | S Chand |
Subject | Maths |
Class | 10th |
Chapter-5 | Quadratic Equations |
Writer | OP Malhotra |
Exe-5B | Equations Reducible to Quadratic |
Edition | 2024-2025 |
Equations Reducible to Quadratic
the equations that are not quadratic but are reducible to quadratic equations. These equations can be easily solved if we convert them to a quadratic form which is otherwise difficult to solve
The primary method is to check if the equations are reduceable to quadratic form and then use the substitution method to convert them into quadratic equations
Exe-5B
(Quadratic Equations Class 10 OP Malhotra ICSE Maths Ch-5)
Solve the following equations by reducing them to quadratic equations :
Que-1: x4 + 5x2 – 36 = 0
Sol: Given that x4 + 5x2 – 36 = 0.
Put y = x2.
⇒ y2 + 5y – 36 = 0
⇒ y2 + 9y – 4y – 36 = 0
⇒ y(y + 9) – 4(y + 9) = 0
⇒ (y + 9)(y – 4) = 0
⇒ (y + 9) = 0 or (y – 4) = 0
⇒ y = -9 or 4
If y = -9, then x2 = -9 ⇒ x is NOT real.
If y = 4, then x2 = 4 ⇒ x = ±2.
Thus, x = 2 or -2
Que-2: x4 – 25x2 + 144 = 0
Sol: Given that x4 – 25x2 + 144 = 0.
Put y = x2.
⇒ y2 – 25y + 144 = 0
⇒ y2 – 16y – 9y + 144 = 0
⇒ y(y – 16) – 9(y – 16) = 0
⇒ (y – 9)(y – 16) = 0
⇒ (y – 9) = 0 or (y – 16) = 0
⇒ y = 9 or 16
If y = 9, then x2 = 9 ⇒ x = ±3.
If y = 16, then x2 = 16 ⇒ x = ±4.
Thus, x = -4, -3, 3, 4.
Que-3: (x2 + x)2 – (x2 + x) – 2 = 0
Sol: Given that (x2 + x)2 – (x2 + x) – 2 = 0.
Put y = (x2 + x).
⇒ y2 – y – 2 = 0
⇒ y2 – 2y + y – 2 = 0
⇒ y(y – 2) + 1(y – 2) = 0
⇒ (y + 1)(y – 2) = 0
⇒ (y + 1) = 0 or (y – 2) = 0
⇒ y = -1 or 2
If y = -1, then x2 + x = -1 ⇒ x2 + x + 1 = 0
Since its discriminant D = 1 – 4 = -3 < 0, there is NO real values of x.
If y = 2, then
x2 + x = 2
⇒ 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-4: [(x – 2)/(x + 2)]2 – 4[(x – 2)/(x + 2)] + 3 = 0, x ≠ 2
Sol: Given that [(x – 2)/(x + 2)]2 – 4[(x – 2)/(x + 2)] + 3 = 0.
Put y = [(x – 2)/(x + 2)].
⇒ y2 – 4y + 3 = 0
⇒ y2 – 3y – y + 3 = 0
⇒ y(y – 3) – 1(y – 3) = 0
⇒ (y – 1)(y – 3) = 0
⇒ (y – 1) = 0 or (y – 3) = 0
⇒ y = 1 or 3
If y = 1, then [(x – 2)/(x + 2)] = 1
⇒ x – 2 = x + 2
⇒ – 2 = 2
⇒ There is no value of x.
If y = 3, then [(x – 2)/(x + 2)] = 3.
⇒ x – 2 = 3(x + 2)
⇒ x – 2 = 3x + 6
⇒ x – 3x = 2 + 6
⇒ -2x = 8
⇒ x = -4
Thus, x = -4
Que-5: 4[(7x – 1)/x]2 – 8[(7x – 1)/x] + 3 = 0
Sol: Given that 4[(7x – 1)/x]2 – 8[(7x – 1)/x] + 3 = 0.
Put y = [(7x – 1)/x].
⇒ 4y2 – 8y + 3 = 0
⇒ 4y2 – 2y – 6y + 3 = 0
⇒ 2y(2y – 1) – 3(2y – 1) = 0
⇒ (2y – 3)(2y – 1) = 0
⇒ (2y – 3) = 0 or (2y – 1) = 0
⇒ y = 3/2 or 1/2
If y = 3/2, then (7x – 1)/x = 3/2
⇒ 2(7x – 1) = 3x
⇒ 14x – 2 = 3x
⇒ 14x – 3x = 2
⇒ 11x = 2
⇒ x = 2/11
If y = 1/2, then (7x – 1)/x = 1/2
⇒ 2(7x – 1) = x
⇒ 14x – 2 = x
⇒ 14x – x = 2
⇒ x = 2/13
Thus, x = 2/11 or 2/13
Que-6: 4x – 5.2x + 4 = 0
Sol: Given that 4x – 5.2x + 4 = 0 ⇒ (2x)2 – 5.2x + 4 = 0
Put y = 2x ⇒ y2 = 4x.
⇒ y2 – 5y + 4 = 0
⇒ y2 – 4y – y + 4 = 0
⇒ y(y – 4) – 1(y – 4) = 0
⇒ (y – 1)(y – 4) = 0
⇒ (y – 1) = 0 or (y – 4) = 0
⇒ y = 1 or 4
If y = 1, then 2x = 1 ⇒ 2x = 20 ⇒ x = 0.
If y = 4, then 2x = 4 ⇒ 2x = 22 ⇒ x = 2.
Thus, x = 0 or 2.
Que-7: 16.4x+2 – 16.2x+1 + 1 = 0
Sol: Given that 16.4x+2 – 16.2x+1 + 1 = 0
⇒ 16.4x.42 – 16.2x.2 + 1 = 0
⇒ 256(2x)2 – 32(2x) + 1 = 0
Put y = 2x
⇒ 256y2 – 32y + 1 = 0
⇒ (16y – 1)2 = 0
⇒ 16y – 1 = 0
⇒ y = 1/16
If y = 1/16, then 2x = 2-4 ⇒ x = -4.
Thus, x = -4.
Que-8: 34x+1 – 2.32x+2 – 81 = 0
Sol: Given that 34x+1 – 2.32x+2 – 81 = 0
⇒ 34x.31 – 2.32x.32 – 81 = 0
⇒ 3(32x)2 – 18(32x) – 81 = 0
⇒ (32x)2 – 6(32x) – 27 = 0
Put y = 32x
⇒ y2 – 6y – 27 = 0
⇒ y2 – 9y + 3y – 27 = 0
⇒ y(y – 9) + 3(y – 9) = 0
⇒ (y + 3)(y – 9) = 0
⇒ (y + 3) = 0 or (y – 9) = 0
⇒ y = -3 or 9
If y = -3, then 32x = -3 ⇒ There is NO real values of x as ax > 0.
If y = 9, then 32x = 32 ⇒ 2x = 2
Thus, x = 1.
Que-9: [(2x – 3)/(x – 1)] – 4[(x – 1)/(2x – 3)] = 3, x ≠ 1 and 3/2
Sol: Given that [(2x – 3)/(x – 1)] – 4[(x – 1)/(2x – 3)] = 3.
Put y = (2x – 3)/(x – 1)
⇒ y – 4/y = 3
⇒ (y2 – 4)/y = 3
⇒ (y2 – 4) = 3y
⇒ y2 – 3y – 4 = 0
⇒ y2 – 4y + y – 4 = 0
⇒ y(y – 4) + 1(y – 4) = 0
⇒ (y + 1)(y – 4) = 0
⇒ (y + 1) = 0 or (y – 4) = 0
⇒ y = -1 or 4
If y = -1, then (2x – 3)/(x – 1) = -1
⇒ 2x – 3 = -(x – 1)
⇒ 2x – 3 = -x + 1
⇒ 2x + x = 3 + 1
⇒ 3x = 4
⇒ x = 4/3
If y = 4, then (2x – 3)/(x – 1) = 4
⇒ 2x – 3 = 4(x – 1)
⇒ 2x – 3 = 4x – 4
⇒ 2x – 4x = 3 – 4
⇒ -2x = -1
⇒ x = 1/2
Thus, x = 1/2 or 4/3.
Que-10: (x + 1/x)2 = 4 + (3/2)(x – 1/x)
Sol: Given that (x + 1/x)2 = 4 + (3/2)(x – 1/x)
⇒ (x – 1/x)2 + 4 = 4 + (3/2)(x – 1/x) [since (a + b)2 = (a – b)2 + 4ab]
⇒ (x – 1/x)2 = (3/2)(x – 1/x)
Put y = (x – 1/x).
⇒ y2 = 3y/2
⇒ y2 – 3y/2 = 0
⇒ y(y – 3/2) = 0
⇒ y = 0 or (y – 3/2) = 0
⇒ y = 0 or 3/2
If y = 0, then (x – 1/x) = 0
⇒ x = 1/x ⇒ x2 = 1 ⇒ x = ±1
If y = 3/2, then (x – 1/x) = 3/2
⇒ (x2 – 1)/x = 3/2
⇒ 2(x2 – 1) = 3x
⇒ 2x2 – 2 = 3x
⇒ 2x2 – 3x – 2 = 0
⇒ 2x2 – 4x + x – 2 = 0
⇒ 2x(x – 2) + 1(x – 2) = 0
⇒ (2x + 1)(x – 2) = 0
⇒ (2x + 1) = 0 or (x – 2) = 0
⇒ x = -1/2 or 2
Thus, x = -1, 1, -1/2 or 2.
Que-11: √(2x + 7) = x + 2
Sol: Given that √(2x + 7) = x + 2. Then, 2x + 7 >= 0 and x + 2 >= 0.
Squaring both sides,
⇒ (2x + 7) = (x + 2)2
⇒ 2x + 7 = x2 + 4x + 4
⇒ x2 + 2x – 3 = 0
⇒ x2 + 3x – x – 3 = 0
⇒ x(x + 3) – 1(x + 3) = 0
⇒ (x – 1)(x + 3) = 0
⇒ (x – 1) = 0 or (x + 3) = 0
⇒ x = 1 or -3
If x = -3, then x + 2 = -3 + 2 = – 1 < 0, so x ≠ -3.
Thus, x = 1
Que-12: 2√(2x + 1) – 2x = 1
Sol: Given that 2√(2x + 1) – 2x = 1
⇒ 2√(2x + 1) = 1 + 2x. Then, 2x + 1 >= 0.
Squaring both sides,
⇒ 4(2x + 1) = (1 + 2x)2
⇒ 8x + 4 = 1 + 4x2 + 4x
⇒ 4x2 – 4x – 3 = 0
⇒ 4x2 – 6x + 2x – 3 = 0
⇒ 2x(x – 3) + 1(2x – 3) = 0
⇒ (2x + 1)(2x – 3) = 0
⇒ (2x + 1) = 0 or (2x – 3) = 0
⇒ x = -1/2 or 3/2 Since at x = -1/2 or 3/2, then value of 1 + 2x = 0, 4.
Thus, x = -1/2 or 3/2
Que-13: √(4x – 3) + √(2x + 3) = 6
Sol: Given that √(4x – 3) + √(2x + 3) = 6
⇒ √(4x – 3) = 6 – √(2x + 3)
Squaring both sides,
⇒ 4x – 3 = [6 – √(2x + 3)]2
⇒ 4x – 3 = 36 + 2x + 3 – 12√(2x + 3)
⇒ 2x – 42 = -12√(2x + 3)
⇒ 6√(2x + 3) = 21 – x, Then, 21 – x >= 0 ⇒ x =< 21.
Squaring both sides,
⇒ 36(2x + 3) = (21 – x)2
⇒ 72x + 108 = 441 + x2 – 42x
⇒ x2 – 114x + 333 = 0
⇒ x2 – 111x – 3x + 333 = 0
⇒ x(x – 111) – 3(x – 111) = 0
⇒ (x – 3)(x – 111) = 0
⇒ (x – 3) = 0 or (x – 111) = 0
⇒ x = 3 or 111
Since x <= 21, thus x = 3
–: Quadratic Equations Class 10 OP Malhotra Exe-5B ICSE Maths Ch-5 Questions :–
Return to : OP Malhotra S Chand Solutions for ICSE Class-10 Maths
Thanks
Please Share with Your Friends