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

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

Step 1

Add $48$ to both sides of the equation.

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

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 $b$ .

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

Step 3

Add the term to each side of the equation.

$x^{2} + 2 x + {(1)}^{2} = 48 + {(1)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

One to any power is one.

$x^{2} + 2 x + 1 = 48 + {(1)}^{2}$

Simplify the right side.

Simplify $48 + {(1)}^{2}$ .

One to any power is one.

$x^{2} + 2 x + 1 = 48 + 1$

Add $48$ and $1$ .

$x^{2} + 2 x + 1 = 49$

Step 5

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

${(x + 1)}^{2} = 49$

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.

$x + 1 = \pm \sqrt{49}$

Simplify $\pm \sqrt{49}$ .

Rewrite $49$ as $7^{2}$ .

$x + 1 = \pm \sqrt{7^{2}}$

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

$x + 1 = \pm 7$

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.

$x + 1 = 7$

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

Subtract $1$ from both sides of the equation.

$x = 7 - 1$

Subtract $1$ from $7$ .

$x = 6$

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

$x + 1 = - 7$

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

Subtract $1$ from both sides of the equation.

$x = - 7 - 1$

Subtract $1$ from $- 7$ .

$x = - 8$

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

$x = 6, - 8$

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

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