Solve by Completing the Square x^2-6x=7

$x^{2} - 6 x = 7$

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 $b$ .

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

Step 2

Add the term to each side of the equation.

$x^{2} - 6 x + {(- 3)}^{2} = 7 + {(- 3)}^{2}$

Step 3

Simplify the equation.

Simplify the left side.

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

$x^{2} - 6 x + 9 = 7 + {(- 3)}^{2}$

Simplify the right side.

Simplify $7 + {(- 3)}^{2}$ .

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

$x^{2} - 6 x + 9 = 7 + 9$

Add $7$ and $9$ .

$x^{2} - 6 x + 9 = 16$

Step 4

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

${(x - 3)}^{2} = 16$

Step 5

Solve the equation for $x$ .

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

$x - 3 = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

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

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

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

$x - 3 = \pm 4$

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.

$x - 3 = 4$

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

Add $3$ to both sides of the equation.

$x = 4 + 3$

Add $4$ and $3$ .

$x = 7$

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

$x - 3 = - 4$

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

Add $3$ to both sides of the equation.

$x = - 4 + 3$

Add $- 4$ and $3$ .

$x = - 1$

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

$x = 7, - 1$

Solve by Completing the Square x^2-6x=7

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