Solve by Completing the Square x^2+8x-48=0

$x^{2} + 8 x - 48 = 0$

Step 1

Add $48$ to both sides of the equation.

$x^{2} + 8 x = 48$

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.

$x^{2} + 8 x + {(4)}^{2} = 48 + {(4)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$x^{2} + 8 x + 16 = 48 + {(4)}^{2}$

Simplify the right side.

Simplify $48 + {(4)}^{2}$ .

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

$x^{2} + 8 x + 16 = 48 + 16$

Add $48$ and $16$ .

$x^{2} + 8 x + 16 = 64$

Step 5

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

${(x + 4)}^{2} = 64$

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.

$x + 4 = \pm \sqrt{64}$

Simplify $\pm \sqrt{64}$ .

Rewrite $64$ as $8^{2}$ .

$x + 4 = \pm \sqrt{8^{2}}$

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

$x + 4 = \pm 8$

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.

$x + 4 = 8$

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

Subtract $4$ from both sides of the equation.

$x = 8 - 4$

Subtract $4$ from $8$ .

$x = 4$

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

$x + 4 = - 8$

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

Subtract $4$ from both sides of the equation.

$x = - 8 - 4$

Subtract $4$ from $- 8$ .

$x = - 12$

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

$x = 4, - 12$

Solve by Completing the Square x^2+8x-48=0

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