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:

Consider the quadratic equation x^{2} – 2x + 1 = 0
This is in the form of ax^{2} + bx + c = 0; a = 1 , b = 2 and c = 1

Consider the equation x^{2} – 2x – 3 = 0
This is in the form ax^{2} + bx + c = 0, where a = 1 , b = 2 , c = 3

Consider the quadratic equation x^{2} – 2x + 3 = 0
This is in the form ax^{2} + bx + c = 0, where a = 1 , b = 2 , c = 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 b^{2} – 4ac determines the nature of the roots. We say the nature of roots depends on the values of b^{2} – 4ac.
The value of the expression b^{2} – 4ac discriminates the nature of the roots of ax^{2} + 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 ax^{2} + bx + c = 0 are
The above results are presented in the table given below:
Discriminant  Nature of roots 
∆ = 0  real and equal 
∆ > 0  real and distinct 
∆ < 0  no real roots(imaginary roots) 
Example 1: Determine the nature of the roots of the equation 2x^{2} – 5x – 1 = 0
Solution:
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 2, b = 5, c = 1
∆ = b^{2} – 4ac = (5)^{2} – 4(2)(1) = 25 + 8 = 33
Example 2: Determine the nature of the roots of the eqation 4x^{2} – 4x + 1 = 0
Solution:
Consider the equation 4x^{2} – 4x + 1 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 4, b = 4, c = 1
∆ = b^{2} – 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 x^{2} + mx + 4 = 0 are (i) equal (ii) distinct
Solution:
Consider the equation x^{2} + mx + 4 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 1, b = m, c = 4
∆ = b^{2} – 4ac
∆ = m^{2} – 4(1)(4)
∆ = m^{2} – 16
(i) If roots are equal then ∆ = 0
m^{2} – 16 = 0
m^{2} = 16
m = ±4
(ii) If roots are distinct, then ∆ > 0
m^{2} – 16 >0
m^{2} > 16
m > √16
m > 4
Quadratic equation – Exercise 9.6 – Class X
 Discuss the nature of roots of the following equations:
(i) y^{2} – 7y + 2 = 0
(ii) x^{2} – 2x + 3 = 0
(iii) 2n^{2} + 5n – 1 = 0
(iv) a^{2} + 4a + 4 = 0
(v) x^{2} + 3x – 4 = 0
(vi) 3d^{2} – 2d + 1 = 0
2. For what positive values of m roots of the following equation are
a) Equal b) Distinct c) Imaginary
(i) a^{2} – ma + 1 = 0
(ii) x^{2} – mx + 9 = 0
(iii) r^{2} – (m + 1)r + 4 = 0
(iv) mk^{2} – 3k + 1 = 0
3. Find the value of ‘p’ for which the quadratic equations have equal roots.
(i) x^{2} – px + 9 = 0
(ii) 2a^{2} + 3a + p = 0
(iii) pk^{2} – 12k + 9 = 0
(iv) 2y^{2} – py + 1 = 0
(v) (p + 1)n^{2} + 2(p + 3)n + (p + 8) = 0
(vi) (3p + 1)c^{2} + 2(p + 1)c + p = 0
Quadratic equation – Exercise 9.6 – Class X
 Discuss the nature of roots of the following equations:
(i) y^{2} – 7y + 2 = 0
Solution:
Consider the equation y^{2} – 7y + 2 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 1, b = 7, c = 2
∆ = b^{2} – 4ac
∆ = (7)^{2} – 4(1)(2)
∆ = 49 – 8
∆ = 41
Hence, ∆ = 41 , therefore, ∆ > 0
Therefore, the roots of the equation y^{2} – 7y + 2 = 0 are real and distinct.
(ii) x^{2} – 2x + 3 = 0
Solution:
Consider the equation x^{2} – 2x + 3 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 1, b = 2, c = 3
∆ = b^{2} – 4ac
∆ = (2)^{2} – 4(1)(3)
∆ = 4 – 12
∆ = 8
Hence, ∆ = 8 , therefore, ∆ < 0
Thus, the roots of the equation x^{2} – 2x + 3 = 0 are imaginary.
(iii) 2n^{2} + 5n – 1 = 0
Solution:
Consider the equation 2n^{2} + 5n – 1 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 2, b = 5, c = 1
∆ = b^{2} – 4ac
∆ = (5)^{2} – 4(2)(1)
∆ = 25 + 8
∆ = 33
Hence, ∆ = 33 , therefore, ∆ > 0
Therefore, the roots of the equation 2n^{2} + 5n – 1 = 0 are real and distinct.
(iv) a^{2} + 4a + 4 = 0
Solution:
Consider the equation a^{2} + 4a + 4 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 1, b = 4, c = 4
∆ = b^{2} – 4ac
∆ = (4)^{2} – 4(1)(4)
∆ = 16 – 16
∆ = 0
Hence, ∆ = 0
Therefore, the roots of the equation a^{2} + 4a + 4 = 0 are real and equal
(v) x^{2} + 3x – 4 = 0
Solution:
Consider the equation x^{2} + 3x – 4 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 1, b = 3, c = 4
∆ = b^{2} – 4ac
∆ = (3)^{2} – 4(1)(4)
∆ = 9 + 16
∆ = 25
Hence, ∆ = 25 , therefore, ∆ > 0
Therefore, the roots of the equation x^{2} + 3x – 4 = 0 are real and distinct.
(vi) 3d^{2} – 2d + 1 = 0
Solution:
Consider the equation 3d^{2} – 2d + 1 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 3, b = 2, c = 1
∆ = b^{2} – 4ac
∆ = (2)^{2} – 4(3)(1)
∆ = 4 – 12
∆ = 8
Hence, ∆ = 8 , therefore, ∆ < 0
Therefore, the roots of the equation 3d^{2} – 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) a^{2} – ma + 1 = 0
Solution:
Consider the equation a^{2} – ma + 1 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 1, b = m, c = 1
∆ = b^{2} – 4ac
∆ = (m)^{2} – 4(1)(1)
∆ = m^{2} – 4
(i) If roots are equal then ∆ = 0
m^{2} – 4 = 0
m^{2} = 4
Therefore, at m = 2 the roots of the quadratic equation a^{2} – ma + 1 = 0 are equal.
(ii) If roots are distinct, then ∆ > 0
m^{2} – 4 >0
m^{2} > 4
m > √4
⸫ m > +2
(iii) If roots are imaginary, then ∆ < 0
m^{2} – 4 < 0
m^{2} < 4
m < √4
⸫ m < 2
(ii) x^{2} – mx + 9 = 0
Solution:
Consider the equation x^{2} – mx + 9 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 1, b = m, c = 9
∆ = b^{2} – 4ac
∆ = (m)^{2} – 4(1)(9)
∆ = m^{2} – 36
(i) If roots are equal then ∆ = 0
m^{2} – 36 = 0
m^{2} = 36
m = 6
(ii) If roots are distinct, then ∆ > 0
m^{2} – 36 >0
m^{2} > 36
m > √36
⸫ m > +6
(iii) If roots are imaginary, then ∆ < 0
m^{2} – 36 < 0
m^{2} < 36
m < √36
⸫ m < 6
(iii) r^{2} – (m + 1)r + 4 = 0
Solution:
Consider the equation r^{2} – (m + 1)r + 4 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 1, b = (m+1), c = 4
∆ = b^{2} – 4ac
∆ = [(m+1)]^{2} – 4(1)(4)
∆ = (m + 1)^{2} – 16
∆ = m^{2} + 2m + 1 – 16
∆ = m^{2} + 2m – 15
(i) If roots are equal then ∆ = 0
m^{2} + 2m – 15 =0
m^{2} + 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 mk^{2} – 3k + 1 = 0 which has equal roots.
(ii) If roots are distinct, then ∆ > 0
m^{2} + 2m – 15 >0
m^{2} – 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
m^{2} + 2m – 3 < 0
m^{2} +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) mk^{2} – 3k + 1 = 0
Solution:
Consider the equation mk^{2} – 3k + 1 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = m, b = 3, c = 1
∆ = b^{2} – 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 mk^{2} – 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) x^{2} – px + 9 = 0
Solution:
x^{2} – px + 9 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 1, b = p, c = 9
∆ = (p)^{2} – 4x1x9
∆ = [p]^{2} – 36
∆ = p^{2} – 36
If roots of quadratic equation x^{2} – px + 9 = 0 then ∆ = 0
p^{2} – 36 = 0
p^{2} = 36
p = √36 = 6
(ii) 2a^{2} + 3a + p = 0
Solution:
2a^{2} + 3a + p = 0
This is in the form ax^{2} + 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 2a^{2} + 3a + p = 0 then ∆ = 0
9 – 8p = 0
9 = 8p
p = ^{9}/_{8}
(iii) pk^{2} – 12k + 9 = 0
Solution:
pk^{2} – 12k + 9 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = p, b = 12, c = 9
∆ = b^{2} – 4ac
∆ = [12]^{2} – 4(p)(9)
∆ = 144 – 36p
If roots of quadratic equation pk^{2} 12k + 9 = 0 then ∆ = 0
144 – 36p = 0
36p = 144
p = ^{144}/_{36} = ^{24}/_{6} = 4
Therefore, p = 4 .
(iv) 2y^{2} – py + 1 = 0
Solution:
2y^{2} – py + 1 = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = 2, b = p, c = 1
∆ = b^{2} – 4ac
∆ = [p]^{2} – 4(2)(1)
∆ = p^{2} – 8
If roots of quadratic equation 2y^{2} – py + 1 = 0 then ∆ = 0
p^{2} – 8 = 0
p^{2} = 8
p = √8 = 2√2
Therefore, p = 2√2
(v) (p + 1)n^{2} + 2(p + 3)n + (p + 8) = 0
Solution:
Consider the quadratic equation (p + 1)n^{2} + 2(p + 3)n + (p + 8) = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = (p + 1), b = 2(p + 3), c = (p + 8)
∆ = b^{2} – 4ac
∆ = [2(p+3)]^{2} – 4(p+1)(p+8)
∆ = 4(p + 3)^{2} – 4[p^{2} + 8p + p + 8]
∆ = p^{2} + 6p + 9 – p^{2} – 9p – 8
∆ =3p + 1
If roots of quadratic equation (p + 1)n^{2} + 2(p + 3)n + (p + 8) = 0 then ∆ = 0
3p + 1 = 0
3p = – 1
p = ^{1}/_{3}
(vi) (3p + 1)c^{2} + 2(p + 1)c + p = 0
Solution:
Consider the quadratic equation (3p + 1)c^{2} + 2(p + 1)c + p = 0
This is in the form ax^{2} + bx + c = 0. The coefficient are a = (3p + 1), b = 2(p + 1), c = p
∆ = b^{2} – 4ac
∆ = [2(p + 1)]^{2} – 4(3p+1)p
∆ = 4[p^{2} + 2p + 1] – 12p^{2} – 4p
∆ = 4p^{2} + 8p + 4 – 12p^{2} – 4p
∆ = 8p^{2} + 4p + 4
∆ = 8p^{2} – 4p – 4
If roots of quadratic equation (3p + 1)c^{2} + 2(p + 1)c + p = 0 then ∆ = 0
8p^{2} – 4p – 4 = 0
8p^{2} – 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