# How do you solve  x^2-14x+5=0 by completing the square?

Mar 21, 2018

$x = 7 - 2 \sqrt{11}$ and $x = 7 + 2 \sqrt{11}$

#### Explanation:

Half the second term of $x$ put it in brackets and square it.

${x}^{2} - 14 x + 5 \to {\left(x - 7\right)}^{2} + 5$

Take away the squared number of the number inside the bracket.

${7}^{2} = 49$ $\therefore$ $- 49$

$\to {\left(x - 7\right)}^{2} - 49 + 5$

Simplify:

${\left(x - 7\right)}^{2} - 44$

Solving:

Set equal to $0$:

${\left(x - 7\right)}^{2} - 44 = 0$

Add $44$:

${\left(x - 7\right)}^{2} = 44$

Square root:

$x - 7 = \pm \sqrt{44}$

Add $7$:

$x = 7 \pm \sqrt{44}$

Simplifying:

$\sqrt{44} = \sqrt{4} \times \sqrt{11}$

$\sqrt{4} = 2$

$\sqrt{11} = \sqrt{11}$, It cannot be simplified as it is a prime number.

$\therefore$ $x = 7 \pm \sqrt{44} \to x = 7 \pm 2 \sqrt{11}$

$x = 7 - 2 \sqrt{11}$
$x = 7 + 2 \sqrt{11}$