Solve by Completing the Square x^2-10x+9=0

$x^{2} - 10 x + 9 = 0$

Step 1

Subtract $9$ from both sides of the equation.

$x^{2} - 10 x = - 9$

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.

$x^{2} - 10 x + {(- 5)}^{2} = - 9 + {(- 5)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$x^{2} - 10 x + 25 = - 9 + {(- 5)}^{2}$

Simplify the right side.

Simplify $- 9 + {(- 5)}^{2}$ .

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

$x^{2} - 10 x + 25 = - 9 + 25$

Add $- 9$ and $25$ .

$x^{2} - 10 x + 25 = 16$

Step 5

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

${(x - 5)}^{2} = 16$

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.

$x - 5 = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

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

$x - 5 = \pm \sqrt{4^{2}}$

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

$x - 5 = \pm 4$

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.

$x - 5 = 4$

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

Add $5$ to both sides of the equation.

$x = 4 + 5$

Add $4$ and $5$ .

$x = 9$

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

$x - 5 = - 4$

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

Add $5$ to both sides of the equation.

$x = - 4 + 5$

Add $- 4$ and $5$ .

$x = 1$

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

$x = 9, 1$

Solve by Completing the Square x^2-10x+9=0

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