How do you find the roots of x^ { 2} + 3x + 8= 0?

Jul 25, 2018

The roots are $x = \frac{- 3 \pm i \sqrt{23}}{2}$.

Explanation:

${x}^{2} + 3 x + 8 = 0$

To find the roots (solutions of $x$), use the quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

We know that $a = 1$, $b = 3$, and $c = 8$, so let's plug them into the formula:

$x = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \left(1\right) \left(8\right)}}{2 \left(1\right)}$

$x = \frac{- 3 \pm \sqrt{9 - 32}}{2}$

$x = \frac{- 3 \pm \sqrt{- 23}}{2}$

Since a square root of a negative number has no real solution, the roots are imaginary.

We know that $i$ (imaginary number) is equal to $\sqrt{- 1}$, so we can take that out and have an imaginary solution:
$x = \frac{- 3 \pm i \sqrt{23}}{2}$

Hope this helps!