Solve by Completing the Square y^2+10y=-9

$y^{2} + 10 y = - 9$

Step 1

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(5)}^{2}$

Step 2

Add the term to each side of the equation.

$y^{2} + 10 y + {(5)}^{2} = - 9 + {(5)}^{2}$

Step 3

Simplify the equation.

Simplify the left side.

Raise $5$ to the power of $2$ .

$y^{2} + 10 y + 25 = - 9 + {(5)}^{2}$

Simplify the right side.

Simplify $- 9 + {(5)}^{2}$ .

Raise $5$ to the power of $2$ .

$y^{2} + 10 y + 25 = - 9 + 25$

Add $- 9$ and $25$ .

$y^{2} + 10 y + 25 = 16$

Step 4

Factor the perfect trinomial square into ${(y + 5)}^{2}$ .

${(y + 5)}^{2} = 16$

Step 5

Solve the equation for $y$ .

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

$y + 5 = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

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

$y + 5 = \pm \sqrt{4^{2}}$

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

$y + 5 = \pm 4$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y + 5 = 4$

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

Subtract $5$ from both sides of the equation.

$y = 4 - 5$

Subtract $5$ from $4$ .

$y = - 1$

Next, use the negative value of the $\pm$ to find the second solution.

$y + 5 = - 4$

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

Subtract $5$ from both sides of the equation.

$y = - 4 - 5$

Subtract $5$ from $- 4$ .

$y = - 9$

The complete solution is the result of both the positive and negative portions of the solution.

$y = - 1, - 9$

Solve by Completing the Square y^2+10y=-9

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