Solve by Completing the Square x^2-8x+25=0

$x^{2} - 8 x + 25 = 0$

Step 1

Subtract $25$ from both sides of the equation.

$x^{2} - 8 x = - 25$

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.

$x^{2} - 8 x + {(- 4)}^{2} = - 25 + {(- 4)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$x^{2} - 8 x + 16 = - 25 + {(- 4)}^{2}$

Simplify the right side.

Simplify $- 25 + {(- 4)}^{2}$ .

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

$x^{2} - 8 x + 16 = - 25 + 16$

Add $- 25$ and $16$ .

$x^{2} - 8 x + 16 = - 9$

Step 5

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

${(x - 4)}^{2} = - 9$

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.

$x - 4 = \pm \sqrt{- 9}$

Simplify $\pm \sqrt{- 9}$ .

Rewrite $- 9$ as $- 1 (9)$ .

$x - 4 = \pm \sqrt{- 1 (9)}$

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x - 4 = \pm \sqrt{- 1} \cdot \sqrt{9}$

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

$x - 4 = \pm i \cdot \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

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

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

$x - 4 = \pm i \cdot 3$

Move $3$ to the left of $i$ .

$x - 4 = \pm 3 i$

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

Add $4$ to both sides of the equation.

$x = \pm 3 i + 4$

Reorder $\pm 3 i$ and $4$ .

$x = 4 \pm 3 i$

Solve by Completing the Square x^2-8x+25=0

Related Questions

Support LCMGCF.com by disabling your adblocker.