Solve by Completing the Square x^2-6x+1=17

$x^{2} - 6 x + 1 = 17$

Step 1

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

Subtract $1$ from both sides of the equation.

$x^{2} - 6 x = 17 - 1$

Subtract $1$ from $17$ .

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

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 $b$ .

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

Step 3

Add the term to each side of the equation.

$x^{2} - 6 x + {(- 3)}^{2} = 16 + {(- 3)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$x^{2} - 6 x + 9 = 16 + {(- 3)}^{2}$

Simplify the right side.

Simplify $16 + {(- 3)}^{2}$ .

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

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

Add $16$ and $9$ .

$x^{2} - 6 x + 9 = 25$

Step 5

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

${(x - 3)}^{2} = 25$

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.

$x - 3 = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$x - 3 = \pm \sqrt{5^{2}}$

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

$x - 3 = \pm 5$

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.

$x - 3 = 5$

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

Add $3$ to both sides of the equation.

$x = 5 + 3$

Add $5$ and $3$ .

$x = 8$

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

$x - 3 = - 5$

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

Add $3$ to both sides of the equation.

$x = - 5 + 3$

Add $- 5$ and $3$ .

$x = - 2$

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

$x = 8, - 2$

Solve by Completing the Square x^2-6x+1=17

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