Solve by Completing the Square x^2-10x=-21

$x^{2} - 10 x = - 21$

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 $b$ .

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

Step 2

Add the term to each side of the equation.

$x^{2} - 10 x + {(- 5)}^{2} = - 21 + {(- 5)}^{2}$

Step 3

Simplify the equation.

Simplify the left side.

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

$x^{2} - 10 x + 25 = - 21 + {(- 5)}^{2}$

Simplify the right side.

Simplify $- 21 + {(- 5)}^{2}$ .

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

$x^{2} - 10 x + 25 = - 21 + 25$

Add $- 21$ and $25$ .

$x^{2} - 10 x + 25 = 4$

Step 4

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

${(x - 5)}^{2} = 4$

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.

$x - 5 = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

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

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

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

$x - 5 = \pm 2$

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.

$x - 5 = 2$

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

Add $5$ to both sides of the equation.

$x = 2 + 5$

Add $2$ and $5$ .

$x = 7$

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

$x - 5 = - 2$

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

Add $5$ to both sides of the equation.

$x = - 2 + 5$

Add $- 2$ and $5$ .

$x = 3$

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

$x = 7, 3$

Solve by Completing the Square x^2-10x=-21

Related Questions

Support LCMGCF.com by disabling your adblocker.