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Nature of roots in quadratic equation

The nature of roots in quadratic equation is dependent on discriminant($b^{2}$ - 4ac).
The standard form of quadratic equation is a$x^{2}$ + bx + c = 0 where a,b,c $\epsilon$ R and a $\ne$ 0. The roots are given by
x= $\frac{-b\pm \sqrt{b^{2} -4ac}}{2a}$
OR
x = x= $\frac{-b \pm \sqrt{D}}{2a}$, where, D = $b^{2}$ -4ac
(i) Roots are real and equal: If $b^{2}$ -4ac = 0 or D = 0 then roots are real and equal.
Example : Find the roots of $x^{2}$ -4x + 4 = 0
Here, a = 1, b = -4 and c = 4
x= $\frac{-b\pm \sqrt{b^{2} -4ac}}{2a}$

x= $\frac{-(-4)\pm \sqrt{(-4)^{2} -4.1.4}}{2.1}$

x = $\frac{-(-4)\pm \sqrt{16 -16}}{2.1}$ Here D = 0 so $\sqrt{D}$ = 0

x = $\frac{4}{2}$
x = 2
So the roots are equal which is 2.

(ii) Roots are rational and unequal:
If a,b,c are rational numbers and $b^{2}$ -4ac is positive and perfect square then $\sqrt{b^{2} -4ac}$ is a rational number then the roots are rational and unequal.
Example : 5$x^{2}$ + 6x + 1 = 0
Here, a = 5, b = 6 and c = 1
x= $\frac{-b\pm \sqrt{b^{2} -4ac}}{2a}$

x= $\frac{-6\pm \sqrt{(6)^{2} -4.5.1}}{2.5}$

x = $\frac{-6\pm \sqrt{36 -20}}{10}$

x = $\frac{-6\pm \sqrt{16}}{10}$

x = $\frac{-6 + 4 }{10}$        or x = $\frac{-6 - 4 }{10}$

x = $\frac{-2}{10} = \frac{-1}{5}$        or x = -1
So the roots are rational and unequal which is $\frac{-1}{5}$ and -1.

(iii) Roots are irrational and unequal: If a,b,c are rational numbers and $b^{2}$ -4ac is not a perfect square,then the roots are irrational and unequal.Irrational roots are always in conjugate pair.
Example : $x^{2}$ + 4x + 1 = 0
Here, a = 1, b = 4 and c = 1
x= $\frac{-b\pm \sqrt{b^{2} -4ac}}{2a}$

x= $\frac{-4\pm \sqrt{(4)^{2} -4.1.1}}{2.1}$

x = $\frac{-4\pm \sqrt{16 -4}}{2}$

x = $\frac{-4\pm \sqrt{12}}{2}$

x = $\frac{-4 + 2\sqrt{3} }{2}$        or x = $\frac{-4 - 2\sqrt{3} }{2}$

x = -2 + $\sqrt{3}$        or x = -2 - $\sqrt{3}$
So the roots are irrational and unequal which is -2 + $\sqrt{3}$ and -2 - $\sqrt{3}$ .

(iv) Roots are imaginary and unequal: If a,b,c are rational numbers and $b^{2}$ -4ac < 0 ,then the roots are imaginary and unequal.Imaginary roots are always in complex conjugate of each other.
Example : $x^{2}$ + 2x + 5 = 0
Here, a = 1, b = 2 and c = 5
x= $\frac{-b\pm \sqrt{b^{2} -4ac}}{2a}$

x= $\frac{-2\pm \sqrt{(2)^{2} -4.1.5}}{2.1}$

x = $\frac{-2\pm \sqrt{4 -20}}{2}$

x = $\frac{-2\pm \sqrt{-20}}{2}$

x = $\frac{-2 + 2\sqrt{-5} }{2}$        or x = $\frac{-2 - 2\sqrt{-5} }{2}$

x = -1 + $\sqrt{5}i$        or x = -1 - $\sqrt{5}i$
So the roots are imaginary and unequal which is -1 + $\sqrt{5}i$ and
-1 - $\sqrt{5}i$ .

Practice on nature of roots in quadratic equation

Q.1 Without finding the factors of the given quadratic equation, state the nature of the roots.
1) $x^{2}$ + 6x + 8 = 0
2) $3x^{2}$ - 7x + 1 = 0
3) $x^{2}$ + 2ix - 9 = 0
4) $7x^{2}$ + - 8 = 0
5) $12x^{2}$ + 10x + 25 = 0

11th grade math

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