Solve by Completing the Square x^2-4x+40=0

$x^{2} - 4 x + 40 = 0$

Step 1

Subtract $40$ from both sides of the equation.

$x^{2} - 4 x = - 40$

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.

$x^{2} - 4 x + {(- 2)}^{2} = - 40 + {(- 2)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$x^{2} - 4 x + 4 = - 40 + {(- 2)}^{2}$

Simplify the right side.

Simplify $- 40 + {(- 2)}^{2}$ .

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

$x^{2} - 4 x + 4 = - 40 + 4$

Add $- 40$ and $4$ .

$x^{2} - 4 x + 4 = - 36$

Step 5

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

${(x - 2)}^{2} = - 36$

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.

$x - 2 = \pm \sqrt{- 36}$

Simplify $\pm \sqrt{- 36}$ .

Rewrite $- 36$ as $- 1 (36)$ .

$x - 2 = \pm \sqrt{- 1 (36)}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$x - 2 = \pm \sqrt{- 1} \cdot \sqrt{36}$

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

$x - 2 = \pm i \cdot \sqrt{36}$

Rewrite $36$ as $6^{2}$ .

$x - 2 = \pm i \cdot \sqrt{6^{2}}$

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

$x - 2 = \pm i \cdot 6$

Move $6$ to the left of $i$ .

$x - 2 = \pm 6 i$

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

Add $2$ to both sides of the equation.

$x = \pm 6 i + 2$

Reorder $\pm 6 i$ and $2$ .

$x = 2 \pm 6 i$

Solve by Completing the Square x^2-4x+40=0

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