S Chand ICSE Maths Solutions Class 10 Chapter 3 Quadratic Equations Exercise 5B

Hi students, Welcome to Amans Maths Blogs (AMB). In this post, you will get S Chand ICSE Maths Solutions Class 10 Chapter 3 Quadratic Equations Exercise 5B. This is the chapter of introduction of ‘Quadratic Equations‘ included in ICSE Class 10 Maths syllabus.

S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Exercise 5B

S Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5B: Ques No 1

x⁴ + 5x² – 36 = 0

S Chand ICSE Maths Solutions:

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

Ques No 2

x⁴ – 25x² + 144 = 0

Solutions:

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.

Ques No 3

(x² + x)² – (x² + x) – 2 = 0

Solutions:

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

Thus, x = 1 or -2

Ques No 4

[(x – 2)/(x + 2)]² – 4[(x – 2)/(x + 2)] + 3 = 0, x ≠ 2 

Solutions:

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

Ques No 5

4[(7x – 1)/x]² – 8[(7x – 1)/x] + 3 = 0 

Solutions:

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

Ques No 6

4^x – 5.2^x + 4 = 0

Solutions:

Given that 4^x – 5.2^x + 4 = 0 ⇒ (2^x)² – 5.2^x + 4 = 0 

Put y = 2^x ⇒ y² = 4^x.

⇒ 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 2^x = 1 ⇒ 2^x = 2⁰ ⇒ x = 0.

If y = 4, then 2^x = 4 ⇒ 2^x = 2² ⇒ x = 2.

Thus, x = 0 or 2.

Ques No 7

16.4^x+2 – 16.2^x+1 + 1 = 0

Solutions:

Given that 16.4^x+2 – 16.2^x+1 + 1 = 0

⇒ 16.4^x.4² – 16.2^x.2 + 1 = 0

⇒ 256(2^x)² – 32(2^x) + 1 = 0

Put y = 2^x

⇒ 256y² – 32y + 1 = 0

⇒ (16y – 1)² = 0

⇒ 16y – 1 = 0

⇒ y = 1/16

If y = 1/16, then 2^x = 2^-4 ⇒ x = -4.

Thus, x = -4.

Ques No 8

3^4x+1 – 2.3^2x+2 – 81 = 0

Solutions:

Given that 3^4x+1 – 2.3^2x+2 – 81 = 0

⇒ 3^4x.3¹ – 2.3^2x.3² – 81 = 0

⇒ 3(3^2x)² – 18(3^2x) – 81 = 0

⇒ (3^2x)² – 6(3^2x) – 27 = 0

Put y = 3^2x

⇒ 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 3^2x = -3 ⇒ There is NO real values of x as a^x > 0.

If y = 9, then 3^2x = 3² ⇒ 2x = 2

Thus, x = 1.

Ques No 9

[(2x – 3)/(x – 1)] – 4[(x – 1)/(2x – 3)] = 3, x ≠ 1 and 3/2

Solutions:

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.

Ques No 10

(x + 1/x)² = 4 + (3/2)(x – 1/x)

Solutions:

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.

Ques No 11

√(2x + 7) = x + 2 

Solutions:

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

Ques No 12

2√(2x + 1) – 2x = 1

Solutions:

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

S Chand ICSE Solutions Class 10 Quadratic Equations Exercise 5B: Ques No 13

√(4x – 3) + √(2x + 3) = 6

S Chand ICSE Maths Solutions:

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

S Chand ICSE Class 10 Maths Chapterwise Notes & Solutions
Chapter 1 : GST Notes & Solutions | Notes | Exercise 1
Chapter 2 : Banking Notes & Solutions | Notes | Exercise 2 | Revision Exercise
Chapter 3 : Shares & Dividends Notes & Solutions | Notes | Exercise 3A | Exercise 3B | Revision Exercise
Chapter 4 : Linear Inequations in One Variable Notes & Solutions | Notes | Exercise 4 | Revision Exercise
Chapter 5 : Quadratic Equations Notes & Solutions | Notes | Exercise 5A | Exercise 5B | Exercise 5C| Exercise 5D| Exercise 5E| Revision Exercise
Chapter 6 : Ratios & Proportions Notes & Solutions | Notes | Exercise 6A | Exercise 6B | Exercise 6C| Revision Exercise

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