Quadratic Equation – Exercise 9.6 – Class X

So far we have learnt to find the roots of given quadratic equations by different methods. We see that the roots are all real numbers.

Is it possible to determine the nature of roots of a given quadratic equation, without actually finding them? Now, let us learn about this.

Nature of the roots of quadratic equations -Quadratic Equation:

Study the following examples:

  1. Consider the quadratic equation x2 – 2x + 1 = 0

This is in the form of ax2 + bx + c = 0; a = 1 , b = -2 and c = 1

Quadratic Equations - Exercise 9.4 - Class x

  1. Consider the equation x2 – 2x – 3 = 0

This is in the form ax2 + bx + c = 0, where a = 1 , b = -2 , c = -3

2

  1. Consider the quadratic equation x2 – 2x + 3 = 0

This is in the form ax2 + bx + c = 0, where a = 1 , b = -2 , c = 3

3

From the above examples, it is evident that the roots of a quadratic equation can be real and equal, real and distinct or imaginary.

Also, observe that the value of b2 – 4ac determines the nature of the roots. We say the nature of roots depends on the values of b2 – 4ac.

The value of the expression b2 – 4ac discriminates the nature of the roots of ax2 + bx + c = 0 and so it is called the discriminant  of the quadratic equation. It is denoted by symbol ∆ and real as delta.

In general, the roots of the quadratic equation ax2 + bx + c = 0 areQuadratic Equation - Exercise 9.6 - Class X

The above results are presented in the table given below:

DiscriminantNature of roots
∆ = 0real and equal
∆ > 0real and distinct
∆ < 0no real roots(imaginary roots)

Example 1: Determine the nature of the roots of the equation 2x2 – 5x – 1 = 0

Solution:

This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = -5, c = -1

∆ = b2 – 4ac = (-5)2 – 4(2)(-1) = 25 + 8 = 33


Example 2: Determine the nature of the roots of the eqation 4x2 – 4x + 1 = 0

Solution:

Consider the equation 4x2 – 4x + 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 4, b = -4, c = 1

∆ = b2 – 4ac = (-4)2 – 4(4)(1) = 16 – 16 = 0

Therefore, the roots of are real and equal.


Example 3: For what positive values of m, roots of the equation x2 + mx + 4 = 0 are (i) equal (ii) distinct

Solution:

Consider the equation x2 + mx + 4 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = m, c = 4

∆ = b2 – 4ac

∆ = m2 – 4(1)(4)

∆ = m2 – 16

(i) If roots are equal then ∆ = 0

m2 – 16 = 0

m2 = 16

m = ±4

(ii) If roots are distinct, then ∆ > 0

m2 – 16 >0

m2 > 16

m > √16

m > 4


Quadratic equation – Exercise 9.6 – Class X

  1. Discuss the nature of roots of the following equations:

(i) y2 – 7y + 2 = 0

(ii) x2 – 2x + 3 = 0

(iii) 2n2 + 5n – 1 = 0

(iv)  a2 + 4a + 4 = 0

(v) x2 + 3x – 4 = 0

(vi) 3d2 – 2d + 1 = 0

2. For what positive values of m roots of the following equation are

a) Equal b) Distinct c) Imaginary

(i) a2 – ma + 1 = 0

(ii) x2  – mx + 9 = 0

(iii) r2 – (m + 1)r + 4 = 0

(iv) mk2 – 3k + 1 = 0

3. Find the value of ‘p’ for which the quadratic equations have equal roots.

(i) x2 – px + 9 = 0

(ii) 2a2 + 3a + p = 0

(iii) pk2 – 12k + 9 = 0

(iv) 2y2 – py + 1 = 0

(v) (p + 1)n2 + 2(p + 3)n + (p + 8) = 0

(vi) (3p + 1)c2 + 2(p + 1)c + p = 0


Quadratic equation – Exercise 9.6 – Class X

  1. Discuss the nature of roots of the following equations:

(i) y2 – 7y + 2 = 0

Solution:

Consider the equation y2 – 7y + 2 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -7, c = 2

∆ = b2 – 4ac

∆ = (-7)2 – 4(1)(2)

∆ = 49 – 8

∆ = 41

Hence, ∆ = 41 , therefore, ∆ > 0

Therefore, the roots of the equation y2 – 7y + 2 = 0 are real and distinct.


 (ii) x2 – 2x + 3 = 0

Solution:

Consider the equation x2 – 2x + 3 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -2, c = 3

∆ = b2 – 4ac

∆ = (-2)2 – 4(1)(3)

∆ = 4 – 12

∆ = -8

Hence, ∆ = -8 , therefore, ∆ < 0

Thus, the roots of the equation x2 – 2x + 3 = 0 are imaginary.


 (iii) 2n2 + 5n – 1 = 0

Solution:

Consider the equation 2n2 + 5n – 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = 5, c = -1

∆ = b2 – 4ac

∆ = (5)2 – 4(2)(-1)

∆ = 25 + 8

∆ = 33

Hence, ∆ = 33 , therefore, ∆ > 0

Therefore, the roots of the equation 2n2 + 5n – 1 = 0 are real and distinct.


 (iv)  a2 + 4a + 4 = 0

Solution:

Consider the equation a2 + 4a + 4 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = 4, c = 4

∆ = b2 – 4ac

∆ = (4)2 – 4(1)(4)

∆ = 16 – 16

∆ = 0

Hence, ∆ = 0

Therefore, the roots of the equation a2 + 4a + 4 = 0 are real and equal


 (v) x2 + 3x – 4 = 0

Solution:

Consider the equation x2 + 3x – 4 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = 3, c = -4

∆ = b2 – 4ac

∆ = (3)2 – 4(1)(-4)

∆ = 9 + 16

∆ = 25

Hence, ∆ = 25 , therefore, ∆ > 0

Therefore, the roots of the equation x2 + 3x – 4 = 0 are real and distinct.


 (vi) 3d2 – 2d + 1 = 0

Solution:

Consider the equation 3d2 – 2d + 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 3, b = -2, c = 1

∆ = b2 – 4ac

∆ = (-2)2 – 4(3)(1)

∆ = 4 – 12

∆ = -8

Hence, ∆ = -8 , therefore, ∆ < 0

Therefore, the roots of the equation 3d2 – 2d + 1 = 0 are imaginary.


2. For what positive values of m roots of the following equation are

a) Equal b)Distinct c)Imaginary

(i) a2 – ma + 1 = 0

Solution:

Consider the equation a2 – ma + 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -m, c = 1

∆ = b2 – 4ac

∆ = (-m)2 – 4(1)(1)

∆ = m2 – 4

(i) If roots are equal then ∆ = 0

m2 – 4 = 0

m2 = 4

Therefore, at m = 2 the roots of the quadratic equation a2 – ma + 1 = 0 are equal.

(ii) If roots are distinct, then ∆ > 0

m2 – 4 >0

m2 > 4

m > √4

⸫ m > +2

(iii) If roots are imaginary, then ∆ < 0

m2 – 4 < 0

m2 < 4

m < √4

⸫ m < -2

 

 (ii) x2  – mx + 9 = 0

Solution:

Consider the equation x2 – mx + 9 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -m, c = 9

∆ = b2 – 4ac

∆ = (-m)2 – 4(1)(9)

∆ = m2 – 36

(i) If roots are equal then ∆ = 0

m2 – 36 = 0

m2 = 36

m = 6

(ii) If roots are distinct, then ∆ > 0

m2 – 36 >0

m2 > 36

m > √36

⸫ m > +6

(iii) If roots are imaginary, then ∆ < 0

m2 – 36 < 0

m2 < 36

m < √36

⸫ m < -6

 

 (iii) r2 – (m + 1)r + 4 = 0

Solution:

Consider the equation r2 – (m + 1)r + 4 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -(m+1), c = 4

∆ = b2 – 4ac

∆ = [-(m+1)]2 – 4(1)(4)

∆ = (m + 1)2 – 16

∆ = m2 + 2m + 1 – 16

∆ = m2 + 2m – 15

(i) If roots are equal then ∆ = 0

m2 + 2m – 15 =0

m2 + 5m – 3m – 15 = 0

m(m + 5)-3(m + 5) = 0

(m – 3)(m + 5) = 0

m – 3 = 0 or m + 5 = 0

m = 3  or m = -5

Therefore, at no value m can be the quadratic equation mk2 – 3k + 1 = 0 which has equal roots.

(ii) If roots are distinct, then ∆ > 0

m2 + 2m – 15 >0

m2 – 3m + 5m – 15 > 0

m(m – 3)-1(m +3) > 0

(m + 3)(m – 1) > 0

m + 3 > 0 or m – 1 > 0

⸫ m > 1

(iii) If roots are imaginary, then ∆ < 0

m2 + 2m – 3 < 0

m2 +3m – m – 3 < 0

m(m + 3)-1(m +3) < 0

(m + 3)(m – 1) < 0

m + 3 < 0 or m – 1 < 0

⸫ m < -3

 

 (iv) mk2 – 3k + 1 = 0

Solution:

Consider the equation mk2 – 3k + 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = m, b = -3, c = 1

∆ = b2 – 4ac

∆ = [-3]2 – 4(m)(1)

∆ = 9 – 4m

(i) If roots are equal then ∆ = 0

9 – 4m =0

-4m = – 9

m = 9/4

Therefore, at m = 9/4 the quadratic equation mk2 – 3k + 1 = 0 has equal roots

(ii) If roots are distinct, then ∆ > 0

9 – 4m > 0

-4m > – 9

⸫ m > 9/4

(iii) If roots are imaginary, then ∆ < 0

9 – 4m < 0

-4m < – 9

⸫ m < 9/4


3.. Find the value of ‘p’ for which the quadratic equations have equal roots.

(i) x2 – px + 9 = 0

Solution:

x2 – px + 9 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -p, c = 9

∆ = (-p)2 – 4x1x9

∆ = [p]2 – 36

∆ = p2 – 36

If roots of quadratic equation x2 – px + 9 = 0 then ∆ = 0

p2 – 36 = 0

p2 = 36

p = √36 = 6


 (ii) 2a2 + 3a + p = 0

Solution:

2a2 + 3a + p = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = 3, c = p

∆ = (3)2 – 4x2xp

∆ = [3]2 – 8p

∆ = 9 – 8p

If roots of quadratic equation 2a2 + 3a + p = 0 then ∆ = 0

9 – 8p = 0

9 = 8p

p = 9/8


 (iii) pk2 – 12k + 9 = 0

Solution:

pk2 – 12k + 9 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = p, b = -12, c = 9

∆ = b2 – 4ac

∆ = [-12]2 – 4(p)(9)

∆ = 144 – 36p

If roots of quadratic equation pk2 -12k + 9 = 0 then ∆ = 0

144 – 36p = 0

36p = 144

p = 144/36 = 24/6 = 4

Therefore, p = 4 .


 (iv) 2y2 – py + 1 = 0

Solution:

2y2 – py + 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = -p, c = 1

∆ = b2 – 4ac

∆ = [-p]2 – 4(2)(1)

∆ = p2 – 8

If roots of quadratic equation 2y2 – py + 1 = 0 then ∆ = 0

p2 – 8 = 0

p2 = 8

p = √8 = 2√2

Therefore, p = 2√2


 (v) (p + 1)n2 + 2(p + 3)n + (p + 8) = 0

Solution:

Consider the quadratic equation (p + 1)n2 + 2(p + 3)n + (p + 8) = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = (p + 1), b = 2(p + 3), c = (p + 8)

∆ = b2 – 4ac

∆ = [2(p+3)]2 – 4(p+1)(p+8)

∆ = 4(p + 3)2 – 4[p2 + 8p + p + 8]

∆ = p2 + 6p + 9 – p2 – 9p – 8

∆ =-3p + 1

If roots of quadratic equation (p + 1)n2 + 2(p + 3)n + (p + 8) = 0 then ∆ = 0

-3p + 1 = 0

-3p = – 1

p = 1/3


 (vi) (3p + 1)c2 + 2(p + 1)c + p = 0

Solution:

Consider the quadratic equation (3p + 1)c2 + 2(p + 1)c + p = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = (3p + 1), b = 2(p + 1), c = p

∆ = b2 – 4ac

∆ = [2(p + 1)]2 – 4(3p+1)p

∆ = 4[p2 + 2p + 1] – 12p2 – 4p

∆ = 4p2 + 8p + 4 – 12p2 – 4p

∆ = -8p2 + 4p + 4

∆ = 8p2 – 4p – 4

If roots of quadratic equation (3p + 1)c2 + 2(p + 1)c + p = 0 then ∆ = 0

8p2 – 4p – 4 = 0

8p2 – 8p + 4p – 4 = 0

8p(p – 1)+4(p – 1) = 0

(8p + 4)(p – 1) = 0

8p + 4 = 0 or p – 1 = 0

8p = -4 or p = 1

p = –4/8 or p = 1

Therefore, p = –1/­2 or p = 1


Quadratic Equations – Exercise 9.1 – Class X

Quadratic Equations – Exercise 9.2 – Class X

Quadratic Equations – Exercise 9.3 – Class X

Quadratic Equations – Exercise 9.4 – Class X

Quadratic Equations – Exercise 9.5 – Class X