Solve by Completing the Square x^2-8x+4=0

$x^{2} - 8 x + 4 = 0$

Step 1

Subtract $4$ from both sides of the equation.

$x^{2} - 8 x = - 4$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- 4)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} - 8 x + {(- 4)}^{2} = - 4 + {(- 4)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $- 4$ to the power of $2$ .

$x^{2} - 8 x + 16 = - 4 + {(- 4)}^{2}$

Simplify the right side.

Simplify $- 4 + {(- 4)}^{2}$ .

Raise $- 4$ to the power of $2$ .

$x^{2} - 8 x + 16 = - 4 + 16$

Add $- 4$ and $16$ .

$x^{2} - 8 x + 16 = 12$

Step 5

Factor the perfect trinomial square into ${(x - 4)}^{2}$ .

${(x - 4)}^{2} = 12$

Step 6

Solve the equation for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x - 4 = \pm \sqrt{12}$

Simplify $\pm \sqrt{12}$ .

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x - 4 = \pm \sqrt{4 (3)}$

Rewrite $4$ as $2^{2}$ .

$x - 4 = \pm \sqrt{2^{2} \cdot 3}$

Pull terms out from under the radical.

$x - 4 = \pm 2 \sqrt{3}$

Add $4$ to both sides of the equation.

$x = \pm 2 \sqrt{3} + 4$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm 2 \sqrt{3} + 4$

Decimal Form:

$x = 7.46410161 \dots, 0.53589838 \dots$