# How do you solve x^2-6x+25=0?

Jul 4, 2015

This has no real solution. It does have imaginary solutions however.

#### Explanation:

You can see this in two ways.

1) Without the discriminant
${x}^{2} - 6 x + 9 + 16 = 0 \setminus \textcolor{w h i t e}{\ldots \ldots \ldots \ldots .}$(Completing the square)
${\left(x - 3\right)}^{2} + 16 = 0$
${\left(x - 3\right)}^{2} = - 16$
$\left(x - 3\right) = \pm \sqrt{- 16}$

Since $\sqrt{- 16}$ doesn't exist (at least not with real numbers), this equation has no real solutions.

If you want the imaginary solutions:
$\left(x - 3\right) = \pm \sqrt{- 16}$
$x - 3 = \pm \sqrt{16} \cdot \sqrt{- 1}$
$x - 3 = \pm 4 \cdot i$
$x = \pm 4 i + 3$

2) With the discriminant
The discriminant is: $D = {b}^{2} - 4 a c$ if you have a quadratic equation of the form $a {x}^{2} + b x + c = 0$.
It can be proven that if this discriminant is less than 0, the equation has no solutions:
$D = {\left(- 6\right)}^{2} - 4 \cdot 1 \cdot 25 = 36 - 100 = - 64 < 0$