# How do you find all the real and complex roots of x^2 + 6x + 13 = 0?

Jul 9, 2016

The Zeroes of given quadr. eqn. are $- 3 \pm 2 i .$

#### Explanation:

There are two Methods to solve this problem :-

Method I : Method of Complete Square :

We rewrite the given eqn. as $: {x}^{2} + 6 x = - 13.$

To make the LHS a complete square, we add

last term $= {\left(m i \mathrm{dl} . t e r m\right)}^{2} / \left(4 \cdot f i r s t t e r m\right) = {\left(6 x\right)}^{2} / \left(4 \cdot {x}^{2}\right) = 9$ on both sides.

$\therefore {x}^{2} + 6 x + 9 = - 13 + 9$

$\therefore {\left(x + 3\right)}^{2} = - 4 = {\left(2 i\right)}^{2}$

$\therefore x + 3 = \pm 2 i$

$\therefore x = - 3 \pm 2 i$

The Zeroes of given quadr. eqn. are $- 3 \pm 2 i .$

Method I : based on Formula :

The Zeroes, $\alpha , \beta$ of a quadr. eqn. $: a {x}^{2} + b x + c = 0$ are $\frac{- b \pm \sqrt{\Delta}}{2 a} , w h e r e , \Delta = {b}^{2} - 4 a c .$

In our case, $a = 1 , b = 6 , c = 13$, so that, $\Delta = {6}^{2} - 4 \cdot 1 \cdot 13 = 36 - 52 = - 16 < 0. .$

Hence, the roots are complex.

$\alpha = \frac{- 6 + \sqrt{- 16}}{2 \cdot 1} = \frac{- 6 + 4 i}{2} = - 3 + 2 i .$

$\beta = \frac{- 6 - 4 i}{2} = - 3 - 2 i .$

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