a$x^{2}$ + bx + c = 0 , a,b,c $\epsilon$ R and a $\neq$ 0.

Multiplying both sides by a , we get

$a^{2}x^{2}$ +abx + ac = 0

$a^{2}x^{2}$ +abx = - ac ( subtract ac on both sides )

$a^{2}x^{2} +abx + \frac{b^{2}}{4} = \frac{b^{2}}{4}$ - ac ( add $\frac{b^{2}}{4}$ on both sides)

$\left ( ax + \frac{b}{2} \right )^{2} = \frac{b^{2}-4ac}{4}$

$\left ( ax + \frac{b}{2} \right) = \pm \frac{\sqrt{b^{2}-4ac}}{2}$

∴ ax = $\frac{- b}{2}\pm \frac{\sqrt{b^{2}-4ac}}{2}$

ax = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2}$

x = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

So the roots of quadratic equation $\alpha$ and $\beta$, given by

$\alpha = \frac{-b +\sqrt{b^{2}-4ac}}{2a}$ and $\beta = \frac{-b - \sqrt{b^{2}-4ac}}{2a}$

The nature of roots of the quadratic equation is dependent on $b^{2}$-4ac. This is called discriminant and is denoted by 'D'

If $b^{2}$-4ac < 0 then the roots are not real.

And if $b^{2}$-4ac > 0 then the roots are real.

$\alpha = \frac{-b +i\sqrt{4ac - b^{2}}}{2a}$ and $\beta = \frac{-b - i\sqrt{4ac - b^{2}}}{2a}$

4$x^{2}$ + 9 = 0

Here, a = 4 , b = 0 and c = 9

$b^{2}$-4ac = $0^{2}-4\times 4 \times$ 9 = -64 < 0 so the roots are imaginary.

$\alpha = \frac{-b +i\sqrt{4ac - b^{2}}}{2a}$ and $\beta = \frac{-b - i\sqrt{4ac - b^{2}}}{2a}$

$\alpha =\frac{0 + i\sqrt{144}}{2\times 4}$ = $\frac{12i}{8} = \frac{3}{2}i$

and $\beta =\frac{0 - i\sqrt{144}}{2\times 4}$ = $\frac{-12i}{8} = \frac{-3}{2}i$

2) Solve the equation $x^{2}$ - 4x + 13 = 0 by factorization method.

$x^{2}$ - 4x + 4 + 9 = 0

$(x - 2)^{2}$ + 9 = 0

$(x - 2)^{2} - 9i^{2}$ = 0

$(x - 2)^{2} - (3i)^{2}$ = 0

[(x - 2) -3i][(x -2) + 3i] = 0

(x - 2 - 3i)(x - 2 + 3i) = 0

∴ x = 2 + 3i or x = 2 - 3i

From Quadratic equations with real coefficients to Home

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