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

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.

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.

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.

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 - $\sqrt{3}$ .

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 - $\sqrt{5}i$ .

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

From Nature of roots in quadratic equation to Home

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