R S Aggarwal Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4D

R S Aggarwal Solutions for Class 10 Maths exercise 4d contains detailed step-by-step explanations to all the problems enlisted under the exercise. This exercise concept helps students to find the nature of roots of a quadratic equation. Students can download free pdf of class 10 Maths Chapter 4 quadratic equations exercise 4d by clicking on the given link below.

Download PDF of R S Aggarwal Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4D

RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D
RS Aggarwal Sol class 10 Maths Chapter 4D

 

Access solution to Maths R S Aggarwal Class 10 Chapter 4 – Quadratic Equations Exercise 4D

Exercise 4A Solutions : 30 Questions (Short Answers)

Exercise 4B Solutions : 16 Questions (Short Answers)

Exercise 4C Solutions : 15 Questions (Short Answers)

Exercise 4D Page No: 199

Question 1:

Find the nature of the roots of the following quadratic equations:

(i) 2x² – 8x + 5 = 0

(ii) 3x² – 2√6 x + 2 = 0

(iii) 5x² – 4x + 1 = 0

(iv) 5x (x – 2) + 6 = 0

(v) 12x² – 4√15 x + 5 = 0

(vi) x² – x + 2 = 0

Solution:

(i)2x² – 8x + 5 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 2, b = – 8, c = 5

Using Discriminant Formula:

D = b^2 – 4ac

= (– 8)^2 – 4.2.5

= 64 – 40

= 24 > 0

Hence the roots of equation are real and unequal.

(ii) 3x² – 2√6 x + 2 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 3, b = – 2√6, c = 2

Using Discriminant Formula:

D = b^2 – 4ac

= (– 2√6)^2 – 4.3.2

=24 – 24

= 0

Roots of equation are real and equal.

(iii) 5x² – 4x + 1 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 5, b = – 4, c = 1

Discriminant:

D = b^2 – 4ac

= (– 4)^2 – 4.5.1

= 16 – 20

= – 4 < 0

Equation has no real roots.

(iv) 5x (x – 2) + 6 = 0

5x^2 – 10x + 6 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 5, b = – 10, c = 6

Discriminant:

D = b^2 – 4ac

= (– 10)2 – 4.5.6

= 100 – 120

= – 20 < 0

Equation has no real roots.

(v) 12x² – 4√15 x + 5 = 0

12x² – 4√15 x + 5 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 12, b = – 4√15, c = 5

Discriminant:

D = b^2 – 4ac

= (– 4√15)^2 – 4.12.5

= 240 – 240

= 0

Equation has real and equal roots.

(vi) x² – x + 2 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 1, b = – 1, c = 2

Discriminant:

D = b^2 – 4ac

= (– 1)^2 – 4.1.2

= 1 – 8

= – 7 < 0

Equation has no real roots.

Question 2:

If a and b are distinct real numbers, show that the quadratic equation 2(a² + b²) x² + 2 (a + b) x + 1 = 0 has no real roots:

Solution:

2(a² + b²) x² + 2 (a + b) x + 1 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 2 (a^2 + b^2), b = 2(a + b), c = 1

Discriminant:

D = b^2 – 4ac

=[2(a + b)]^2 – 4. 2 (a^2 + b^2).1

= 4a^2 + 4b^2 + 8ab – 8a^2 – 8b^2

= – 4a^2 – 4b^2 + 8ab

= – 4(a^2 + b^2 – 2ab)

= – 4(a – b)^2

< 0

Hence the equation has no real roots.

Question 3:

Show that the roots of the equation x² + px – q² = 0 are real for all real values of p and q

Solution:

x² + px – q² = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 1, b = p, c = -q²

Using discriminant formula:

D = b² – 4ac

= (p)² – 4 x 1 x (-q²)

= p² + 4 q² > 0

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

Question 4:

For what values of k are the roots of the quadratic equation 3x² + 2kx + 27 = 0 real and equal?

Solution:

3x² + 2kx + 27 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

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

Find discriminant:

D = b² – 4ac

= (2k)² – 4 x 3 x 27

= (2k)² – 324 >0

Roots are real and equal

Find the value of k:

(2k)² – 324 = 0

(2k)² – (18)² = 0

(k)² – (9)² = 0

(k + 9) (k – 9) = 0

Either k + 9 = 0 or k – 9 = 0

k = -9 or k = 9

Hence, the value of k is k = 9 or -9

Question 5:

For what values of k are the roots of the quadratic equation kx (x – 2√5) + 10 = 0 real and equal?

Solution:

kx (x – 2√5) x + 10 = 0

kx² – 2√5 kx + 10 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = k, b = -2√5 k, c = 10

Find discriminant:

D = b² – 4ac

= (-2 k)² – 4 x k x 10 = 20k² – 40k

Since roots are real and equal (given), put D = 0

20k² – 40k = 0

k² – 2k = 0

k (k – 2) = 0

Either, k = 0 or k – 2 = 0

Hence k = 0 or k = 2

Question 6:

For what values of p are the roots of the equation 4x²+ px + 3 = 0 real-and equal?

Solution:

4x² + px + 3 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 4, b = p, c = 3

Find discriminant:

D = b² – 4ac

= p² – 4 x 4 x 3

= p²- 48

Since roots are real and equal (given)

Put D = 0

p² – 48 = 0

p² = 48 = (±4√3)²

p = ± 4√3

Hence p= 4√3 or p = -4√3

Question 7:

Find the nonzero value of k for which the roots of the quadratic equation 9x² – 3kx + k – 0 are real and equal.

Solution:

9x² – 3kx + k = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

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

Find discriminant:

D = b² – 4ac

= (-3k)² – 4 x 9 x k

= 9k² – 36k

Since roots are real and equal (given)

Put D = 0

9k² – 36k = 0

9k (k – 4) = 0

Either, k = 0 or k – 4 = 0

As, value of k is non-zero:

So, k = 4

Question 8:

(i) Find the values of k for which the quadratic equation (3k + 1) x² + 2 (k + 1) x + 1 = 0 has real and equal roots

(ii) Find the value of k for which the equation x² + k (2x + k – 1) + 2 = 0 has real and equal roots

Solution:

(i) (3k + 1) x² + 2(k + 1) x + 1 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

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

Find discriminant:

D = b² – 4 ac

= (2(k + 1))² – 4(3k + 1) x 1

= 4k² + 4 + 8k – 12k – 4

= 4k (k – 1)

Since roots are real and equal (given)

Put D = 0

4k (k – 1) = 0

Either, k = 0 or k – 1 = 0

k = 0, k = 1

(ii) x² + k(2x + k – 1) + 2 = 0

Simplify above equation:

x² + 2kx + (k² – k + 2) = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

Here, a = 1, b = 2k, c = (k² – k + 2)

Find Discriminant:

D = b² – 4ac

= (2k)² – 4 x 1 x (k² – k + 2)

= 4k² – 4k² + 4k – 8

= 4k – 8

Since roots are real and equal (given)

Put D = 0

4k – 8 = 0

k = 2

Hence, the value of k is 2.

Question 9:

Find the values of p for which the quadratic equation (2p + 1) x² – (7p + 2) x + (7p – 3) = 0 has real and equal roots.

Solution:

(2p + 1) x² – (7p + 2) x + (7p – 3) = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = (2p + 1), b = – (7p + 2) and c = (7p – 3)

Discriminant:

D = b^2 – 4ac

= (– (7p + 2))^2 – 4.(2p + 1).(7p – 3)

= (49p^2 + 28p + 4) – 4(14p^2 + p – 3)

= 49p^2 + 28p + 4 – 56p^2 – 4p + 12

= – 7p2 + 24p + 16

Since roots are real and equal (given)

Put D = 0

7p^2 – 24p – 16 = 0

7p^2 – 28p + 4p – 16 = 0

7p(p – 4) + 4(p – 4) = 0

(7p + 4)(p – 4) = 0

Either (7p + 4) = 0 or (p – 4) = 0

p = -4/7 or p = 4

Question 10:

Find the values of p for which the quadratic equation (p + 1) x² – 6(p + 1) x + 3(p + 9) = 0, p ≠ -1 has equal roots: Hence, find the roots of the equation.

Solution:

The given quadratic equation is

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

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = (p + 1), b = – 6(p + 1) and c = 3(p + 9)

Discriminant:

D = b^2 – 4ac

= (-6(p + 1))^2 – 4.(p + 1).3(p + 9)

= 36(p + 1)(p + 1) – 12(p + 1)(p + 9)

= 12(p + 1) (3p + 3 – p – 9)

= 12(p + 1)(2p – 6)

Since roots are real and equal (given)

Put D = 0

12(p + 1)(2p – 6) = 0

either (p + 1) = 0 or (2p – 6) = 0

p = -1 or p = 3

Question 11:

If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.

Solution:

Given: -5 is a root of the quadratic equation 2x² + px – 15 = 0

Substitute the value of x = -5

2(-5)² + p(-5) – 15 = 0

50 – 5p – 15 = 0

35 – 5p = 0

p = 7

Again,

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

7 (x² + x) + k = 0 (put value of p = 7)

7x² + 7x + k = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 7, b = 7, c = k

Find Discriminant:

D = b² – 4ac

= (7)² – 4 x 7 x k

= 49 – 28k

Since roots are real and equal, put D = 0

49 – 28k = 0

28k = 49

k = 7 / 4

The value of k is 7/4

Question 12:

If 3 is a root of the quadratic equation x² – x + k – 0, find the value of p so that the roots of the equation x² + k (2x + k + 2) + p = 0 are equal.

Solution:

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

Substitute the value of x = 3

(3)² – (3) + k = 0

9 – 3 + k = 0

k = -6

Now, 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

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

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

Find Discriminant:

D = b² – 4ac

= (-12)² – 4 x 1 x (24 + p)

= 144 – 96 – 4p = 48 – 4p

Since roots are real and equal, put D = 0

48 – 4p = 0

4p = 48

p = 12

The value of p is 12.

Question 13:

If -4 is a root of the equation x² + 2x + 4p = 0, find the value of k for which the quadratic equation x² + px (1 + 3k) + 7 (3 + 2k) = 0 has equal roots.

Solution:

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

Substitute the value of x = -4

(-4)² + 2(-4) + 4p = 0

16 – 8 + 4p = 0

8 + 4p = 0

4p = -8

or p = -2

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

x² – 2x (1 + 3k) + 7(3 + 2k) = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

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

Find Discriminant:

D = b² – 4ac

= (-2(1 + 3k))² – 4 x 1 x 7(3 + 2k)

= 4(1 + 9k² + 6k) – 28(3 + 2k)

= 36k² – 32k – 80

Since roots are real and equal, put 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 or 9k + 10 = 0

k = 2 or k = -10/9

Question 14:

If the quadratic equation (1 + m²) x² + 2mcx + c² – a² = 0 has equal roots, prove that c² = a² (1 + m²).

Solution:

(1 + m²) x² + 2mcx + c² – a² = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = (1 + m^2), b = 2mc and c = c^2 – a^2

Since roots are equal, so D = 0

(2mc)2 – 4.(1 + m2)(c2 – a2) = 0

4 m2c2 – 4c2 + 4a2 – 4 m2c2 + 4 m2a2 = 0

a2 + m2a2 = c2

or c2 = a2 (1 + m2)

Hence Proved

Question 15:

If the roots of the equation (c² – ab) x² – 2(a² – bc)x + (b² – ac) = 0 are real and equal, show that either a = 0 or (a3+ b3 + c3) = 3abc

Solution:

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = (c^2 – ab) b = – 2(a^2 – bc) c = (b^2 – ac)

Since roots are equal, so D = 0

(– 2(a2 – bc)) 2 – 4(c2 – ab) (b2 – ac) = 0

4(a4 – 2a2bc + b2c2) – 4(b2c2 – ac3 – ab3 + a2bc) = 0

a4 – 3a2bc + ac3 + ab3 = 0

a (a3 – 3abc + c3 + b3) = 0

either a = 0 or (a3 – 3abc + c3 + b3) = 0

a = 0 or a3 + c3 + b3 = 3abc

Hence Proved.

Question 16:

Find the values of p for which the quadratic equation 2x² + px + 8 = 0 has real roots.

Solution:

2x² + px + 8 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 2, b = p, c = 8

Find D:

D = b2 – 4ac

= p² – 4 x 2 x 8

= p² – 64

Since roots are real, so D ≥ 0

p² – 64 ≥ 0

p² ≥ 64

≥ (±8)²

Either p ≥ 8 or p ≤ -8

Question 17:

Find the value of a for which the equation (α – 12) x² + 2(α – 12) x + 2 = 0 has equal roots.

Solution:

(α – 12) x² + 2(α – 12) x + 2 = 0

Roots of given equation are equal ( given)

So, D = 0

4(α – 12) (α – 14) = 0

α – 14 = 0 {(α – 12) ≠ 0}

α = 14

Hence the value of α is 14

Question 18:

Find the value of k for which the roots of 9x² + 8kx + 16 = 0 are real and equal.

Solution:

9x² + 8kx + 16 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

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

Find D:

D = b² – 4ac

= (8k)² – 4 x 9 x 16

= 64k² – 576

Roots of given equation are equal ( given)

So, D = 0

64k² – 576 = 0

64k² = 576

k² = 9

k = ±3

Answer: k = 3, k = -3

Question 19:

Find the values of k for which the given quadratic equation has real and distinct roots.

(i) kx² + 6x + 1 = 0

(ii) x² – kx + 9 = 0

(iii) 9x² + 3kx + 4 = 0

(iv) 5x² – kx + 1 = 0

Solution:

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

(i) a = k, b = 6, c = 1

For real and distinct roots, then D > 0

6^2 – 4k > 0

36 – 4k > 0

k < 9

(ii)

a = 1, b = – k, c = 9

For real and distinct roots, then D > 0

(-k)^2 – 36 > 0

k > 6 or k < -6

(iii)

a = 9 ,b = 3k ,c = 4

For real and distinct roots, then D > 0

(3k)^2 – 144 > 0

9k^2 > 144

k^2 > 16

k > 4 or k < – 4

(iv)

a = 5, b = – k, c = 1

For real and distinct roots, then D > 0

k^2 – 20 > 0

k2 > 20

k > 2√5 or k < –2√5

Question 20:

If a and b are real and a ≠ b then show that the roots of the equation (a – b) x² + 5(a + b) x – 2 (a – b) = 0 are real and unequal.

Solution:

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = (a – b), b = 5(a + b), c = – 2(a – b)

Find Discriminant:

D = b^2 – 4ac

= (5(a + b))^2 – 4(a – b)(– 2(a – b))

= 25(a + b)^2 + 8(a – b)^2

= 17(a + b)^2 + {8(a + b)^2 + 8(a – b)^2 }

= 17(a + b)^2 + 16(a^2 + b^2)

Which is always greater than zero.

Equation has real and unequal roots.

R S Aggarwal Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4D

Class 10 Maths Chapter 4 Quadratic Equations Exercise 4D is based on the nature of roots of a quadratic equation.

Discriminant is represented as D. Formula to find the D is D = b^2-4ac.

  • Case 1: When D>0

Roots are real and distinct.

  • Case 2: When D = 0

Roots are real and equal.

  • Case 3: D < 0

Roots are imaginary. Also we can say that the equation has no real roots.

You can also practice numerous problems on R S Aggarwal Class 10 Maths and master quadratic equations.