Solve by Completing the Square y^2+2y+17=0

$y^{2} + 2 y + 17 = 0$

Step 1

Subtract $17$ from both sides of the equation.

$y^{2} + 2 y = - 17$

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 $b$ .

${(\frac{b}{2})}^{2} = {(1)}^{2}$

Step 3

Add the term to each side of the equation.

$y^{2} + 2 y + {(1)}^{2} = - 17 + {(1)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

One to any power is one.

$y^{2} + 2 y + 1 = - 17 + {(1)}^{2}$

Simplify the right side.

Simplify $- 17 + {(1)}^{2}$ .

One to any power is one.

$y^{2} + 2 y + 1 = - 17 + 1$

Add $- 17$ and $1$ .

$y^{2} + 2 y + 1 = - 16$

Step 5

Factor the perfect trinomial square into ${(y + 1)}^{2}$ .

${(y + 1)}^{2} = - 16$

Step 6

Solve the equation for $y$ .

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

$y + 1 = \pm \sqrt{- 16}$

Simplify $\pm \sqrt{- 16}$ .

Rewrite $- 16$ as $- 1 (16)$ .

$y + 1 = \pm \sqrt{- 1 (16)}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$y + 1 = \pm \sqrt{- 1} \cdot \sqrt{16}$

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

$y + 1 = \pm i \cdot \sqrt{16}$

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

$y + 1 = \pm i \cdot \sqrt{4^{2}}$

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

$y + 1 = \pm i \cdot 4$

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

$y + 1 = \pm 4 i$

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

Subtract $1$ from both sides of the equation.

$y = \pm 4 i - 1$

Reorder $\pm 4 i$ and $- 1$ .

$y = - 1 \pm 4 i$

Solve by Completing the Square y^2+2y+17=0

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