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

**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) 2xÂ² â€“ 8x + 5 = 0**

**(ii) 3xÂ² â€“ 2âˆš6 x + 2 = 0**

**(iii) 5xÂ² â€“ 4x + 1 = 0**

**(iv) 5x (x â€“ 2) + 6 = 0**

**(v) 12xÂ² â€“ 4âˆš15 x + 5 = 0**

**(vi) xÂ² â€“ x + 2 = 0**

**Solution:**

**(i)2xÂ² â€“ 8x + 5 = 0**

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

a = 2, b = â€“ 8, c = 5

Using Discriminant Formula:

D = b^2 â€“ 4ac

= (â€“ 8)^2 â€“ 4.2.5

= 64 â€“ 40

= 24 > 0

Hence the roots of equation are real and unequal.

**(ii) 3xÂ² â€“ 2âˆš6 x + 2 = 0**

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

a = 3, b = â€“ 2âˆš6, c = 2

Using Discriminant Formula:

D = b^2 â€“ 4ac

= (â€“ 2âˆš6)^2 â€“ 4.3.2

=24 – 24

= 0

Roots of equation are real and equal.

**(iii) 5xÂ² â€“ 4x + 1 = 0**

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

a = 5, b = â€“ 4, c = 1

Discriminant:

D = b^2 â€“ 4ac

= (â€“ 4)^2 â€“ 4.5.1

= 16 â€“ 20

= â€“ 4 < 0

Equation has no real roots.

**(iv) 5x (x â€“ 2) + 6 = 0**

5x^2 â€“ 10x + 6 = 0

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

a = 5, b = â€“ 10, c = 6

Discriminant:

D = b^2 â€“ 4ac

= (â€“ 10)2 â€“ 4.5.6

= 100 â€“ 120

= â€“ 20 < 0

Equation has no real roots.

**(v) 12xÂ² â€“ 4âˆš15 x + 5 = 0**

12xÂ² â€“ 4âˆš15 x + 5 = 0

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

a = 12, b = â€“ 4âˆš15, c = 5

Discriminant:

D = b^2 â€“ 4ac

= (â€“ 4âˆš15)^2 â€“ 4.12.5

= 240 – 240

= 0

Equation has real and equal roots.

**(vi) xÂ² â€“ x + 2 = 0**

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

a = 1, b = â€“ 1, c = 2

Discriminant:

D = b^2 â€“ 4ac

= (â€“ 1)^2 â€“ 4.1.2

= 1 â€“ 8

= â€“ 7 < 0

Equation has no real roots.

**Question 2:**

**If a and b are distinct real numbers, show that the quadratic equation 2(aÂ² + bÂ²) xÂ² + 2 (a + b) x + 1 = 0 has no real roots:**

**Solution:**

2(aÂ² + bÂ²) xÂ² + 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 â€“ 4ac

=[2(a + b)]^2 â€“ 4. 2 (a^2 + b^2).1

= 4a^2 + 4b^2 + 8ab â€“ 8a^2 â€“ 8b^2

= â€“ 4a^2 â€“ 4b^2 + 8ab

= â€“ 4(a^2 + b^2 â€“ 2ab)

= â€“ 4(a â€“ b)^2

< 0

Hence the equation has no real roots.

**Question 3:**

**Show that the roots of the equation xÂ² + px â€“ qÂ² = 0 are real for all real values of p and q**

**Solution:**

xÂ² + px â€“ qÂ² = 0

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

a = 1, b = p, c = -qÂ²

Using discriminant formula:

D = bÂ² â€“ 4ac

= (p)Â² â€“ 4 x 1 x (-qÂ²)

= pÂ² + 4 qÂ² > 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 3xÂ² + 2kx + 27 = 0 real and equal? **

**Solution:**

3xÂ² + 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 = bÂ² â€“ 4ac

= (2k)Â² â€“ 4 x 3 x 27

= (2k)Â² â€“ 324 >0

Roots are real and equal

Find the value of k:

(2k)Â² â€“ 324 = 0

(2k)Â² â€“ (18)Â² = 0

(k)Â² â€“ (9)Â² = 0

(k + 9) (k â€“ 9) = 0

Either k + 9 = 0 or k â€“ 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 â€“ 2âˆš5) + 10 = 0 real and equal?**

**Solution:**

kx (x â€“ 2âˆš5) x + 10 = 0

kxÂ² â€“ 2âˆš5 kx + 10 = 0

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

a = k, b = -2âˆš5 k, c = 10

Find discriminant:

D = bÂ² â€“ 4ac

= (-2 k)Â² â€“ 4 x k x 10 = 20kÂ² â€“ 40k

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

20kÂ² â€“ 40k = 0

kÂ² â€“ 2k = 0

k (k â€“ 2) = 0

Either, k = 0 or k â€“ 2 = 0

Hence k = 0 or k = 2

**Question 6:**

**For what values of p are the roots of the equation 4xÂ²+ px + 3 = 0 real-and equal?**

**Solution:**

4xÂ² + 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 = bÂ² â€“ 4ac

= pÂ² â€“ 4 x 4 x 3

= pÂ²- 48

Since roots are real and equal (given)

Put D = 0

pÂ² â€“ 48 = 0

pÂ² = 48 = (Â±4âˆš3)Â²

p = Â± 4âˆš3

Hence p= 4âˆš3 or p = -4âˆš3

**Question 7:**

**Find the nonzero value of k for which the roots of the quadratic equation 9xÂ² â€“ 3kx + k â€“ 0 are real and equal.**

**Solution:**

9xÂ² â€“ 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 = bÂ² â€“ 4ac

= (-3k)Â² â€“ 4 x 9 x k

= 9kÂ² â€“ 36k

Since roots are real and equal (given)

Put D = 0

9kÂ² â€“ 36k = 0

9k (k â€“ 4) = 0

Either, k = 0 or k â€“ 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) xÂ² + 2 (k + 1) x + 1 = 0 has real and equal roots**

**(ii) Find the value of k for which the equation xÂ² + k (2x + k â€“ 1) + 2 = 0 has real and equal roots **

**Solution:**

(i) (3k + 1) xÂ² + 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 = bÂ² â€“ 4 ac

= (2(k + 1))Â² â€“ 4(3k + 1) x 1

= 4kÂ² + 4 + 8k â€“ 12k â€“ 4

= 4k (k â€“ 1)

Since roots are real and equal (given)

Put D = 0

4k (k â€“ 1) = 0

Either, k = 0 or k â€“ 1 = 0

k = 0, k = 1

(ii) xÂ² + k(2x + k â€“ 1) + 2 = 0

Simplify above equation:

xÂ² + 2kx + (kÂ² â€“ 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 = (kÂ² â€“ k + 2)

Find Discriminant:

D = bÂ² â€“ 4ac

= (2k)Â² â€“ 4 x 1 x (kÂ² â€“ k + 2)

= 4kÂ² â€“ 4kÂ² + 4k â€“ 8

= 4k â€“ 8

Since roots are real and equal (given)

Put D = 0

4k â€“ 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) xÂ² â€“ (7p + 2) x + (7p â€“ 3) = 0 has real and equal roots.**

**Solution:**

(2p + 1) xÂ² â€“ (7p + 2) x + (7p â€“ 3) = 0

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

a = (2p + 1), b = â€“ (7p + 2) and c = (7p â€“ 3)

Discriminant:

D = b^2 â€“ 4ac

= (â€“ (7p + 2))^2 â€“ 4.(2p + 1).(7p â€“ 3)

= (49p^2 + 28p + 4) â€“ 4(14p^2 + p â€“ 3)

= 49p^2 + 28p + 4 â€“ 56p^2 â€“ 4p + 12

= – 7p2 + 24p + 16

Since roots are real and equal (given)

Put D = 0

7p^2 â€“ 24p â€“ 16 = 0

7p^2 â€“ 28p + 4p â€“ 16 = 0

7p(p â€“ 4) + 4(p â€“ 4) = 0

(7p + 4)(p â€“ 4) = 0

Either (7p + 4) = 0 or (p â€“ 4) = 0

p = -4/7 or p = 4

**Question 10:**

**Find the values of p for which the quadratic equation (p + 1) xÂ² â€“ 6(p + 1) x + 3(p + 9) = 0, p â‰ -1 has equal roots: Hence, find the roots of the equation.**

**Solution:**

The given quadratic equation is

(p + 1) xÂ² â€“ 6(p + 1) x + 3(p + 9) = 0, p â‰ -1

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

a = (p + 1), b = â€“ 6(p + 1) and c = 3(p + 9)

Discriminant:

D = b^2 â€“ 4ac

= (-6(p + 1))^2 â€“ 4.(p + 1).3(p + 9)

= 36(p + 1)(p + 1) â€“ 12(p + 1)(p + 9)

= 12(p + 1) (3p + 3 â€“ p â€“ 9)

= 12(p + 1)(2p â€“ 6)

Since roots are real and equal (given)

Put D = 0

12(p + 1)(2p â€“ 6) = 0

either (p + 1) = 0 or (2p â€“ 6) = 0

p = -1 or p = 3

**Question 11:**

**If -5 is a root of the quadratic equation 2xÂ² + px â€“ 15 = 0 and the quadratic equation p(xÂ² + x) + k = 0 has equal roots, find the value of k.**

**Solution:**

Given: -5 is a root of the quadratic equation 2xÂ² + px â€“ 15 = 0

Substitute the value of x = -5

2(-5)Â² + p(-5) â€“ 15 = 0

50 â€“ 5p â€“ 15 = 0

35 â€“ 5p = 0

p = 7

Again,

In quadratic equation p(xÂ² + x) + k = 0

7 (xÂ² + x) + k = 0 (put value of p = 7)

7xÂ² + 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 = bÂ² â€“ 4ac

= (7)Â² â€“ 4 x 7 x k

= 49 â€“ 28k

Since roots are real and equal, put D = 0

49 â€“ 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 xÂ² â€“ x + k â€“ 0, find the value of p so that the roots of the equation xÂ² + k (2x + k + 2) + p = 0 are equal.**

**Solution:**

Given: 3 is a root of equation xÂ² â€“ x + k = 0

Substitute the value of x = 3

(3)Â² â€“ (3) + k = 0

9 â€“ 3 + k = 0

k = -6

Now, xÂ² + k (2x + k + 2) + p = 0

xÂ² + (-6)(2x â€“ 6 + 2) + p = 0

xÂ² â€“ 12x + 36 â€“ 12 + p = 0

xÂ² â€“ 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 = bÂ² â€“ 4ac

= (-12)Â² â€“ 4 x 1 x (24 + p)

= 144 â€“ 96 â€“ 4p = 48 â€“ 4p

Since roots are real and equal, put D = 0

48 â€“ 4p = 0

4p = 48

p = 12

The value of p is 12.

**Question 13:**

**If -4 is a root of the equation xÂ² + 2x + 4p = 0, find the value of k for which the quadratic equation xÂ² + px (1 + 3k) + 7 (3 + 2k) = 0 has equal roots.**

**Solution:**

Given: -4 is a root of the equation xÂ² + 2x + 4p = 0

Substitute the value of x = -4

(-4)Â² + 2(-4) + 4p = 0

16 â€“ 8 + 4p = 0

8 + 4p = 0

4p = -8

or p = -2

In the quadratic equation xÂ² + px (1 + 3k) + 7(3 + 2k) = 0

xÂ² â€“ 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 = bÂ² â€“ 4ac

= (-2(1 + 3k))Â² â€“ 4 x 1 x 7(3 + 2k)

= 4(1 + 9kÂ² + 6k) â€“ 28(3 + 2k)

= 36kÂ² â€“ 32k â€“ 80

Since roots are real and equal, put D = 0

36kÂ² â€“ 32k â€“ 80 = 0

9kÂ² â€“ 8k â€“ 20 = 0

9kÂ² â€“ 18k + 10k â€“ 20 = 0

9k (k â€“ 2) + 10(k â€“ 2) = 0

(k â€“ 2) (9k + 10) = 0

Either, k â€“ 2 = 0 or 9k + 10 = 0

k = 2 or k = -10/9

**Question 14:**

**If the quadratic equation (1 + mÂ²) xÂ² + 2mcx + cÂ² â€“ aÂ² = 0 has equal roots, prove that cÂ² = aÂ² (1 + mÂ²).**

**Solution:**

(1 + mÂ²) xÂ² + 2mcx + cÂ² â€“ aÂ² = 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 â€“ a^2

Since roots are equal, so D = 0

(2mc)^{2} â€“ 4.(1 + m^{2})(c^{2 }â€“ a^{2}) = 0

4 m^{2}c^{2} â€“ 4c^{2} + 4a^{2} â€“ 4 m^{2}c^{2} + 4 m^{2}a^{2} = 0

a^{2} + m^{2}a^{2} = c^{2}

or c^{2} = a^{2} (1 + m^{2})

Hence Proved

**Question 15:**

**If the roots of the equation (cÂ² â€“ ab) xÂ² â€“ 2(aÂ² â€“ bc)x + (bÂ² â€“ ac) = 0 are real and equal, show that either a = 0 or (a ^{3}+ b^{3} + c^{3}) = 3abc**

**Solution:**

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

a = (c^2 â€“ ab) b = â€“ 2(a^2 â€“ bc) c = (b^2 â€“ ac)

Since roots are equal, so D = 0

(â€“ 2(a^{2} â€“ bc))^{ 2} â€“ 4(c^{2} â€“ ab) (b^{2} â€“ ac) = 0

4(a^{4} â€“ 2a^{2}bc + b^{2}c^{2}) â€“ 4(b^{2}c^{2} â€“ ac^{3} â€“ ab3 + a^{2}bc) = 0

a^{4} â€“ 3a^{2}bc + ac^{3} + ab^{3} = 0

a (a^{3} â€“ 3abc + c^{3} + b^{3}) = 0

either a = 0 or (a^{3} â€“ 3abc + c^{3} + b^{3}) = 0

a = 0 or a^{3} + c^{3} + b^{3} = 3abc

Hence Proved.

**Question 16:**

**Find the values of p for which the quadratic equation 2xÂ² + px + 8 = 0 has real roots.**

**Solution:**

2xÂ² + 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 = b^{2} â€“ 4ac

= pÂ² â€“ 4 x 2 x 8

= pÂ² â€“ 64

Since roots are real, so D â‰¥ 0

pÂ² â€“ 64 â‰¥ 0

pÂ² â‰¥ 64

â‰¥ (Â±8)Â²

Either p â‰¥ 8 or p â‰¤ -8

**Question 17:**

**Find the value of a for which the equation (Î± â€“ 12) xÂ² + 2(Î± â€“ 12) x + 2 = 0 has equal roots.**

**Solution:**

(Î± â€“ 12) xÂ² + 2(Î± â€“ 12) x + 2 = 0

Roots of given equation are equal ( given)

So, D = 0

4(Î± â€“ 12) (Î± â€“ 14) = 0

Î± â€“ 14 = 0 {(Î± â€“ 12) â‰ 0}

Î± = 14

Hence the value of Î± is 14

**Question 18:**

**Find the value of k for which the roots of 9xÂ² + 8kx + 16 = 0 are real and equal.**

**Solution:**

9xÂ² + 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 = bÂ² â€“ 4ac

= (8k)Â² â€“ 4 x 9 x 16

= 64kÂ² â€“ 576

Roots of given equation are equal ( given)

So, D = 0

64kÂ² â€“ 576 = 0

64kÂ² = 576

kÂ² = 9

k = Â±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) kxÂ² + 6x + 1 = 0**

**(ii) xÂ² â€“ kx + 9 = 0**

**(iii) 9xÂ² + 3kx + 4 = 0**

**(iv) 5xÂ² â€“ 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 â€“ 4k > 0

36 â€“ 4k > 0

k < 9

(ii)

a = 1, b = â€“ k, c = 9

For real and distinct roots, then D > 0

(-k)^2 â€“ 36 > 0

k > 6 or k < -6

(iii)

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

For real and distinct roots, then D > 0

(3k)^2 â€“ 144 > 0

9k^2 > 144

k^2 > 16

k > 4 or k < â€“ 4

(iv)

a = 5, b = â€“ k, c = 1

For real and distinct roots, then D > 0

k^2 â€“ 20 > 0

k2 > 20

k > 2âˆš5 or k < â€“2âˆš5

**Question 20:**

**If a and b are real and a â‰ b then show that the roots of the equation (a â€“ b) xÂ² + 5(a + b) x â€“ 2 (a â€“ 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 â€“ b), b = 5(a + b), c = â€“ 2(a â€“ b)

Find Discriminant:

D = b^2 â€“ 4ac

= (5(a + b))^2 â€“ 4(a â€“ b)(â€“ 2(a â€“ b))

= 25(a + b)^2 + 8(a â€“ b)^2

= 17(a + b)^2 + {8(a + b)^2 + 8(a â€“ 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.