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

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**ICSE Class 10 Maths syllabus**## S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Exercise 5C

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

2x^{2} + x – 3 = 0

**S Chand ICSE Maths Solutions: **

Given that 2x^{2} + x – 3 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 2, b = 1, c = -3

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 1 or -3/2.

**Ques No 2**

6x^{2} + 7x – 20 = 0

**Solutions:**

Given that 6x^{2} + 7x – 20 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 6, b = 7, c = -20

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 4/3 or -5/2.

**Ques No 3**

9x^{2} + 6x = 35

**Solutions:**

Given that 9x^{2} + 6x = 35 ⇒ 9x^{2} + 6x – 35 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 9, b = 6, c = -35

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 5/3 or -7/3.

**Ques No 4**

3x^{2} + 7x – 6 = 0

**Solutions:**

Given that 3x^{2} + 7x – 6 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 3, b = 7, c = -6

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 2/3 or -3.

**Ques No 5**

x^{2} – 66x + 189 = 0

**Solutions:**

Given that x^{2} – 66x + 189 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 1, b = -66, c = 189

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 63 or 3.

**Ques No 6**

√3x^{2} + 11x + 6√3 = 0

**Solutions:**

Given that √3x^{2} + 11x + 6√3 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = √3, b = 11, c = 6√3

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = -2√3/3 or -3√3.

**Ques No 7**

36x^{2} + 23 = 60x

**Solutions:**

Given that 36x^{2} + 23 = 60x ⇒ 36x^{2} – 60x + 23 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 36, b = -60, c = 23

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = (5 + √2)/6 or (5 – √2)/6.

**Ques No 8**

x^{2} – 2x + 5 = 0

**Solutions:**

Given that x^{2} – 2x + 5 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 1, b = -2, c = 5

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = (1 + 2√(-1)) or (1 – 2√(-1)).

**Ques No 9**

3x^{2} – 17x + 25 = 0

**Solutions:**

Given that 3x^{2} – 17x + 25 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 3, b = -17, c = 25

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = (17 + √(-11))/6 or (17 – √(-11))/6.

**Ques No 10**

15x^{2} – 28 = x

**Solutions:**

Given that 15x^{2} – 28 = x ⇒ 15x^{2} – x – 28 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 15, b = -1, c = -28

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 7/5 or -4/3.

**Ques No 11**

x^{2} + 3x – 3 = 0, giving your answer correct to two decimal places.

**Solutions:**

Given that x^{2} + 3x – 3 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 1, b = 3, c = -3

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 0.79 or -3.79.

**Ques No 12**

(2/3)x = (-1/6)x^{2} – 1/3, giving your answer correct to two decimal places.

**Solutions:**

Given that (2/3)x = (-1/6)x^{2} – 1/3 ⇒ x^{2} + 4x + 2 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 1, b = 4, c = 2

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 3.41 or -0.59.

**Ques No 13**

x^{2} + 6x – 10 = 0

**Solutions:**

Given that x^{2} + 6x – 10 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 1, b = 6, c = -10

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = -3-√19 or -3+√19.

**Ques No 14**

(x^{2} + 8)/11 = 5x – x^{2} – 5

**Solutions:**

Given that (x^{2} + 8)/11 = 5x – x^{2} – 5 ⇒ 12x^{2} – 55x + 63 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 12, b = -55, c = 63

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 7/3 or 9/4.

**Ques No 15**

y – 3/y = 1/2

**Solutions:**

Given that y – 3/y = 1/2 ⇒ 2y^{2} – y – 6 = 0.

Then, comparing this quadratic equation with ay^{2} + by + c = 0, we get a = 2, b = -1, c = -6

Putting the values of a, b and c quadratic formula y = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are y = 2 or -3/2.

**Ques No 16**

2x + 4/x = 9

**Solutions:**

Given that 2x + 4/x = 9 ⇒ 2x^{2} – 9x + 4 = 0.

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 2, b = -9, c = 4

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 4 or 1/2.

**Ques No 17**

x/(x + 1) + (x + 1)/x = 34/15, x ≠ 0, x ≠ -1.

**Solutions:**

Given that x/(x + 1) + (x + 1)/x = 34/15

Put y = x/(x + 1), then

⇒ y + 1/y = 34/15

⇒ 15y^{2} – 34y + 15 = 0

Then, comparing this quadratic equation with ay^{2} + by + c = 0, we get a = 15, b = -34, c = 15

Putting the values of a, b and c quadratic formula y = [- b ± √(b^{2 }– 4ac)]/2a.

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

**Ques No 18**

2x/(x – 4) + (2x – 5)/(x – 3) = 8 1/3

**Solutions:**

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[2x^{2} – 6x + 2x^{2} – 13x + 20] = 25(x – 4)(x – 3)

⇒ 3[4x^{2} – 19x + 20] = 25(x^{2} – 7x + 12)

⇒ 12x^{2} – 57x + 60 = 25x^{2} – 175x + 300

⇒ 13x^{2} – 118x + 240 = 0

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 13, b = -118, c = 240

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 6 or 40/13.

**Ques No 19**

(x + 6)/(x + 7) – (x + 1)/(x + 2) = 1/(3x + 1)

**Solutions:**

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)(x^{2} + 8x + 12 – x^{2} – 8x – 7) = (x^{2} + 9x + 14)

⇒ 5(3x + 1) = (x^{2} + 9x + 14)

⇒ 15x + 5 = x^{2} + 9x + 14

⇒ x^{2} – 6x + 9 = 0

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 1, b = -6, c = 9

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 3 or 3.

**Ques No 20**

(x + 1)/(2x + 5) = (x + 3)/(3x + 4)

**Solutions:**

Given that (x + 1)/(2x + 5) = (x + 3)/(3x + 4)

⇒ (x + 1)(3x + 4) = (x + 3)(2x + 5)

⇒ 3x^{2} + 7x + 4 = 2x^{2} + 11x + 15

⇒ x^{2} – 4x – 11 = 0

Then, comparing this quadratic equation with ax^{2} + bx + c = 0, we get a = 1, b = -4, c = -11

Putting the values of a, b and c quadratic formula x = [- b ± √(b^{2 }– 4ac)]/2a.

Thus, the roos of the given equation are x = 2 + √15 or 2 – √15.

**Ques No 21(i)**

Solve using quadratic formula : a^{2}x^{2} – 3abx + 2b^{2} = 0

**Solutions:**

Given that a^{2}x^{2} – 3abx + 2b^{2} = 0

Then, comparing this quadratic equation with Ax^{2} + Bx + C = 0, we get A = a^{2}, B = -3ab, C = 2b^{2}

Putting the values of A, B and C quadratic formula x = [- B ± √(B^{2 }– 4AC)]/2A.

Thus, the roos of the given equation are x = 2b/a or b/a.

**Ques No 21(ii)**

Solve using quadratic formula : x^{2} – x – a(a + 1) = 0

**Solutions:**

Given that x^{2} – x – a(a + 1) = 0

Then, comparing this quadratic equation with Ax^{2} + 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 ± √(B^{2 }– 4AC)]/2A.

Thus, the roos of the given equation are x = -a or (a + 1).

**S Chand ICSE Solutions Class 10 Math Quadratic Equations Exercise 5C: Ques No 21(iii)**

Solve using quadratic formula : 10x^{2} + 3bx + a^{2} – 7ax – b^{2} = 0

**S Chand ICSE Maths Solutions:**

Given that 10x^{2} + 3bx + a^{2} – 7ax – b^{2} = 0

⇒ 10x^{2} + (3b – 7a)x + (a^{2} – b^{2}) = 0

Then, comparing this quadratic equation with Ax^{2} + Bx + C = 0, we get A = 10, B = (3b – 7a), C = (a^{2 }– b^{2}).

Putting the values of A, B and C quadratic formula x = [- B ± √(B^{2 }– 4AC)]/2A.

Thus, the roos of the given equation are x = (a – b)/2 or (a + b)/5.

S Chand ICSE Class 10 Maths Chapterwise Notes & Solutions |
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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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