# How do you solve x^2 - 2x + 10 = 0?

May 14, 2018

$x = 1 \pm 3 i$

#### Explanation:

$\text{the equation is in standard form } \textcolor{w h i t e}{x} a {x}^{2} + b x + c = 0$

$\text{with "a=1,b=-2" and } c = 10$

$\text{there are no whole number values which allow us to}$
$\text{factor the quadratic}$

$\text{check the value of the "color(blue)"discriminant}$

•color(white)(x)Delta=b^2-4ac

${b}^{2} - 4 a c = {\left(- 2\right)}^{2} - \left(4 \times 1 \times 10\right) = 4 - 40 = - 36$

$\text{since "Delta<0" then roots are complex}$

$\text{to calculate use the "color(blue)"quadratic formula}$

•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)

$\Rightarrow x = \frac{2 \pm \sqrt{- 36}}{2} = \frac{2 \pm 6 i}{2}$

$\Rightarrow x = 1 \pm 3 i$

May 14, 2018

Solution: $x = 1 + 3 i \mathmr{and} x = 1 - 3 i$

#### Explanation:

${x}^{2} - 2 x + 10 = 0 \mathmr{and} {x}^{2} - 2 x = - 10$ or

${x}^{2} - 2 x + 1 = 1 - 10$ or

${\left(x - 1\right)}^{2} = - 9 \mathmr{and} \left(x - 1\right) = \pm \sqrt{- 9}$

It has complex roots.

$\therefore \left(x - 1\right) = \pm \sqrt{9 {i}^{2}} \left[{i}^{2} = - 1\right]$ or

$\left(x - 1\right) = \pm 3 i \mathmr{and} x = 1 \pm 3 i$

Solution: $x = 1 + 3 i \mathmr{and} x = 1 - 3 i$ [Ans]