In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 10 Quadratic Equations 10 D Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 10 Quadratic Equations 10 D Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations 10 D Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 10 |

Chapter Name | Quadratic Equations |

Exercise | 10 D |

Category | RS Aggarwal Solutions |

**RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations 10D**

**Question 1.****Solution:**

**Question 2.****Solution:**

**Question 3.****Solution:**

x² + px – q² = 0

Comparing it with ax² + bx + c = 0

a = 1, b = p, c = -q²

Discriminant (D) = b² – 4ac

= (p)² – 4 x 1 x (-q²)

= p² + 4 q²

p and q are of two powers

p² + 4q² is always greater than 0

The roots are real for all real values of p and q.

**Question 4.****Solution:**

The quadratic equation is 3x² + 2kx + 27 = 0

Comparing it with ax² + bx + c = 0

a = 3, b = 2k, c = 27

Discriminant (D) = b² – 4ac

= (2k)² – 4 x 3 x 27

= (2k)² – 324

Roots are real and equal

(2k)² – 324 = 0

⇒ (2k)² – (18)² = 0

⇒ (k)² – (9)² = 0

⇒ (k + 9) (k – 9) = 0

Either k + 9 = 0, then k = -9

or k – 9 = 0, then k = 9

Hence, k = 9, -9

**Question 5.****Solution:**

The quadratic equation is

kx (x – 2√5) x + 10 = 0

kx² – 2√5 kx + 10 = 0

Comparing it with ax² + bx + c = 0

a = k, b = -2√5 k, c = 10

D = b² – 4ac = (-2 k)² – 4 x k x 10 = 20k² – 40k

Roots are real and equal.

D = 0

20k² – 40k = 0

⇒ k² – 2k = 0

⇒ k (k – 2) = 0

Either, k = 0 or k – 2 = 0, then k = 2

k = 0, k = 2

**Question 6.****Solution:**

The quadratic equation is 4x² + px + 3 = 0

Comparing it with ax² + bx + c = 0

a = 4, b = p, c = 3

D = b² – 4ac = p² – 4 x 4 x 3 = p²- 48

Roots are real and equal.

D = 0

⇒ p² – 48 = 0

⇒ p² = 48 = (±4√3)²

⇒ P = ± 4√3

P = 4√3, p = -4√3

**Question 7.****Solution:**

The quadratic equation is 9x² – 3kx + k = 0

Comparing it with ax? + bx + c = 0

a = 9, b = -3k, c = k

D = b² – 4ac = (-3k)² – 4 x 9 x k = 9k² – 36k

Roots are real and equal.

D = 0

9k² – 36k = 0

⇒ 9k (k – 4) = 0

Either, k = 0 or k – 4 = 0, then k = 4

The value of k is non-zero.

k = 4

**Question 8.****Solution:**

(i) The equation is (3k + 1) x² + 2(k + 1) x + 1 = 0

Comparing it with ax² + bx + c = 0

a = (3k + 1), b = 2(k + 1), c = 1

D = b² – 4 ac

= [2(k + 1)]² – 4(3k + 1) x 1

= 4k² + 4 + 8k – 12k – 4

= 4k² – 4k

= 4k (k – 1)

Roots are real and equal.

Either, k = 0 or k – 1 = 0, then k = 1

k = 0, k = 1

(ii) x² + k(2x + k – 1) + 2 = 0

⇒ x² + 2kx + (k² – k + 2) = 0

Here, a = 1, b = 2k, c = (k² – k + 2)

Discriminant (D) = b² – 4ac = (2k)² – 4 x 1 x (k² – k + 2)

= 4k² – 4k² + 4k – 8

= 4k – 8

Roots are real and equal.

D = 0

⇒ 4k – 8 = 0

⇒ k = 2

Hence, k = 2

**Question 9.****Solution:**

**Question 10.****Solution:**

The given quadratic equation is

(p + 1) x² – 6(p + 1) x + 3(p + 9) = 0, p ≠ -1

**Question 11.****Solution:**

-5 is a root of 2x² + px – 15 = 0

x = -5 will satisfy it

Now, substituting the value of x = -5

⇒ 2(-5)² + p(-5) – 15 = 0

⇒ 50 – 5p – 15 = 0

⇒ 35 – 5p = 0

⇒ 5p = 35

⇒ P = 7

In quadratic equation p(x² + x) + k = 0

⇒ 7 (x² + x) + k = 0 (p = 7)

⇒ 7x² + 7x + k = 0

Comparing it with ax² + bx + c = 0

a = 7, b = 7, c = k

D = b² – 4ac = (7)² – 4 x 7 x k

= 49 – 28k

Roots are real and equal.

49 – 28k = 0

⇒ 28k = 49

k = 4928 = 74

**Question 12.****Solution:**

3 is a root of equation x² – x + k = 0

It will satisfy it

Now, substituting the value of x = 3 in it

(3)² – (3) + k = 0

⇒ 9 – 3 + k = 0

⇒ 6 + k = 0

⇒ k = -6

Now in the equation, x² + k (2x + k + 2) + p = 0

x² + (-6)(2x – 6 + 2) + p = 0

⇒ x² – 12x + 36 – 12 + p = 0

⇒ x² – 12x + (24 + p) = 0

Comparing it with ax² + bx + c = 0

a = 1, b = -12, c = 24 + p

D = b² – 4ac

= (-12)² – 4 x 1 x (24 + p)

= 144 – 96 – 4p = 48 – 4p

Roots are real and equal.

D = 0

48 – 4p = 0

⇒ 4p = 48

⇒ p = 12

Hence, p = 12

**Question 13.****Solution:**

-4 is a root of the equation x² + 2x + 4p = 0

Then it will satisfy the equation

Now, substituting the value of x = -4

(-4)² + 2(-4) + 4p = 0

16 – 8 + 4p = 0

⇒ 8 + 4p = 0

⇒ 4p = -8

⇒ p = -2

In the quadratic equation x² + px (1 + 3k) + 7(3 + 2k) = 0

⇒ x² – 2x (1 + 3k) + 7(3 + 2k) = 0

Comparing it with ax² + bx + c = 0

a = 1, b = -2 (1 + 3k), c = 7 (3 + 2k)

D = b² – 4ac

= [-2(1 + 3k)]² – 4 x 1 x 7(3 + 2k)

= 4(1 + 9k² + 6k) – 28(3 + 2k)

= 4 + 36k² + 24k – 84 – 56k

= 36k² – 32k – 80

Roots are equal.

D = 0

⇒ 36k² – 32k – 80 = 0

⇒ 9k² – 8k – 20 = 0

⇒ 9k² – 18k + 10k – 20 = 0

⇒ 9k (k – 2) + 10(k – 2) = 0

⇒ (k – 2) (9k + 10) = 0

Either, k – 2 = 0, then k = 2

or 9k + 10 = 0, then 9k = -10 ⇒ k = −109

k = 2, k = −109

**Question 14.****Solution:**

**Question 15.****Solution:**

**Question 16.****Solution:**

The quadratic equation is 2x² + px + 8 = 0

Comparing it with ax² + bx + c = 0

a = 2, b = p, c = 8

D = b^{2} – 4ac = p² – 4 x 2 x 8 = p² – 64

Roots are real.

D ≥ 0

p² – 64 ≥ 0

⇒ p² ≥ 64 ≥ (±8)²

p ≥ 8 or p ≤ -8

**Question 17.****Solution:**

Roots are equal

D = 0

⇒ 4(α – 12) (α – 14) = 0

⇒ α – 14 = 0 {(α – 12) ≠ 0}

⇒ α = 14

Hence, α = 14

**Question 18.****Solution:**

9x² + 8kx + 16 = 0

Comparing it with ax^{2} + bx + c = 0

a = 9, b = 8k, c = 16

D = b² – 4ac

= (8k)² – 4 x 9 x 16 = 64k² – 576

Roots are real and equal.

D = 0

64k² – 576 = 0

64k² = 576

⇒ k² = 9 = (±3)²

k = 3, k = -3

**Question 19.****Solution:**

**Question 20.****Solution:**

**Question 21.****Solution:**

**Question 22.****Solution:**

**All Chapter RS Aggarwal Solutions For Class 10 Maths**

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