# How do you solve x^2-2x+9=0?

Jul 18, 2016

$x = 1 + i 2 \sqrt{2}$ or $x = 1 - i 2 \sqrt{2}$

#### Explanation:

In the equation ${x}^{2} - 2 x + 9 = 0$, the coefficients are rational but as the discriminant is (-2)^2-4×1×9=4-36=-32 is negative, the roots of the equation are pair of complex conjugate numbers. Hence, we can use quadratic formula to get roots.

Quadratic formula gives the roots of $a {x}^{2} + b x + c = 0$ as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. Note that ${b}^{2} - 4 a c$ is the discriminant.

Hence solution of ${x}^{2} - 2 x + 9 = 0$ is given by
x=(-(-2)+-sqrt((-2)^2-4×1×9))/(2×1) or

$x = \frac{2 \pm \sqrt{- 32}}{2}$ i.e.
$x = \frac{2 + i 4 \sqrt{2}}{2}$ or $x = \frac{2 - i 4 \sqrt{2}}{2}$ i.e.

$x = 1 + i 2 \sqrt{2}$ or $x = 1 - i 2 \sqrt{2}$