Solve by Completing the Square x^2+6x+41=0

$x^{2} + 6 x + 41 = 0$

Step 1

Subtract $41$ from both sides of the equation.

$x^{2} + 6 x = - 41$

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 $b$ .

${(\frac{b}{2})}^{2} = {(3)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + 6 x + {(3)}^{2} = - 41 + {(3)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $3$ to the power of $2$ .

$x^{2} + 6 x + 9 = - 41 + {(3)}^{2}$

Simplify the right side.

Simplify $- 41 + {(3)}^{2}$ .

Raise $3$ to the power of $2$ .

$x^{2} + 6 x + 9 = - 41 + 9$

Add $- 41$ and $9$ .

$x^{2} + 6 x + 9 = - 32$

Step 5

Factor the perfect trinomial square into ${(x + 3)}^{2}$ .

${(x + 3)}^{2} = - 32$

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.

$x + 3 = \pm \sqrt{- 32}$

Simplify $\pm \sqrt{- 32}$ .

Rewrite $- 32$ as $- 1 (32)$ .

$x + 3 = \pm \sqrt{- 1 (32)}$

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$x + 3 = \pm \sqrt{- 1} \cdot \sqrt{32}$

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

$x + 3 = \pm i \cdot \sqrt{32}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$x + 3 = \pm i \cdot \sqrt{16 (2)}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

$x + 3 = \pm i \cdot (4 \sqrt{2})$

Move $4$ to the left of $i$ .

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

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

Subtract $3$ from both sides of the equation.

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

Reorder $\pm 4 i \sqrt{2}$ and $- 3$ .

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

Solve by Completing the Square x^2+6x+41=0

Related Questions

Support LCMGCF.com by disabling your adblocker.