Solve Using the Quadratic Formula 9x^2-18x+9=0

$9 x^{2} - 18 x + 9 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 9$ , $b = - 18$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{18 \pm \sqrt{{(- 18)}^{2} - 4 \cdot (9 \cdot 9)}}{2 \cdot 9}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 18$ to the power of $2$ .

$x = \frac{18 \pm \sqrt{324 - 4 \cdot 9 \cdot 9}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot 9$ .

Multiply $- 4$ by $9$ .

$x = \frac{18 \pm \sqrt{324 - 36 \cdot 9}}{2 \cdot 9}$

Multiply $- 36$ by $9$ .

$x = \frac{18 \pm \sqrt{324 - 324}}{2 \cdot 9}$

Subtract $324$ from $324$ .

$x = \frac{18 \pm \sqrt{0}}{2 \cdot 9}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{18 \pm \sqrt{0^{2}}}{2 \cdot 9}$

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

$x = \frac{18 \pm 0}{2 \cdot 9}$

$18$ plus or minus $0$ is $18$ .

$x = \frac{18}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{18}{18}$

Divide $18$ by $18$ .

$x = 1$

Step 4

The final answer is the combination of both solutions.

$x = 1$ Double roots