# How do you solve by completing the square: x^2- 4x-11=0?

Apr 5, 2015
• First, we Transpose the Constant to one side of the equation.
Transposing $- 11$ to the other side we get:
${x}^{2} - 4 x = 11$

• Application of ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$
We look at the Co-efficient of $x$. It's $- 4$
We take half of this number (including the sign), giving us –2
We square this value to get ${\left(- 2\right)}^{2} = 4$. We add this number to BOTH sides of the Equation.
${x}^{2} - 4 x + 4 = 11 + 4$
${x}^{2} - 4 x + 4 = 15$
The Left Hand side ${x}^{2} - 4 x + 4$ is in the form ${a}^{2} - 2 a b + {b}^{2}$
where $a$ is $x$, and $b$ is $2$

• The equation can be written as
${\left(x - 2\right)}^{2} = 15$

So $\left(x - 2\right)$ can take either $\sqrt{15}$ or $- \sqrt{15}$ as a value. That's because squaring either will give us 15.

$x - 2 = \sqrt{15}$ (or) $x - 2 = - \sqrt{15}$
$x = 2 + \sqrt{15}$ (or) $x = 2 - \sqrt{15}$

• Solution : $x = 2 + \sqrt{15} , 2 - \sqrt{15}$

• Verify your answer by substituting these values in the Original Equation ${x}^{2} - 4 x - 11 = 0$
You will see that the Solution is correct.