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

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

Step 1

Subtract $8$ from both sides of the equation.

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

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 $b$ .

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

Step 3

Add the term to each side of the equation.

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

Step 4

Simplify the equation.

Simplify the left side.

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

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

Simplify the right side.

Simplify $- 8 + {(- 2)}^{2}$ .

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

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

Add $- 8$ and $4$ .

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

Step 5

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

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

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.

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

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x - 2 = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x - 2 = \pm \sqrt{- 1} \cdot \sqrt{4}$

Rewrite $\sqrt{- 1}$ as $i$ .

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

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

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

Pull terms out from under the radical, assuming positive real numbers.

$x - 2 = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x - 2 = \pm 2 i$

Move all terms not containing $x$ to the right side of the equation.

Add $2$ to both sides of the equation.

$x = \pm 2 i + 2$

Reorder $\pm 2 i$ and $2$ .

$x = 2 \pm 2 i$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.