Solve by Completing the Square x^2-10x-2=17

$x^{2} - 10 x - 2 = 17$

Step 1

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

Add $2$ to both sides of the equation.

$x^{2} - 10 x = 17 + 2$

Add $17$ and $2$ .

$x^{2} - 10 x = 19$

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 $b$ .

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

Step 3

Add the term to each side of the equation.

$x^{2} - 10 x + {(- 5)}^{2} = 19 + {(- 5)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$x^{2} - 10 x + 25 = 19 + {(- 5)}^{2}$

Simplify the right side.

Simplify $19 + {(- 5)}^{2}$ .

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

$x^{2} - 10 x + 25 = 19 + 25$

Add $19$ and $25$ .

$x^{2} - 10 x + 25 = 44$

Step 5

Factor the perfect trinomial square into ${(x - 5)}^{2}$ .

${(x - 5)}^{2} = 44$

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.

$x - 5 = \pm \sqrt{44}$

Simplify $\pm \sqrt{44}$ .

Rewrite $44$ as $2^{2} \cdot 11$ .

Factor $4$ out of $44$ .

$x - 5 = \pm \sqrt{4 (11)}$

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

$x - 5 = \pm \sqrt{2^{2} \cdot 11}$

Pull terms out from under the radical.

$x - 5 = \pm 2 \sqrt{11}$

Add $5$ to both sides of the equation.

$x = \pm 2 \sqrt{11} + 5$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm 2 \sqrt{11} + 5$

Decimal Form:

$x = 11.63324958 \dots, - 1.63324958 \dots$

Solve by Completing the Square x^2-10x-2=17

Related Questions

Support LCMGCF.com by disabling your adblocker.