Find All Complex Solutions square root of x^2-6x+9+2x=9

$\sqrt{x^{2} - 6 x + 9} + 2 x = 9$

Subtract $9$ from both sides of the equation.

$\sqrt{x^{2} - 6 x + 9} + 2 x - 9 = 0$

Simplify $\sqrt{x^{2} - 6 x + 9} + 2 x - 9$ .

Simplify each term.

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$\sqrt{x^{2} - 6 x + 3^{2}} + 2 x - 9 = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

$\sqrt{x^{2} - 2 \cdot x \cdot 3 + 3^{2}} + 2 x - 9 = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

$\sqrt{{(x - 3)}^{2}} + 2 x - 9 = 0$

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

$x - 3 + 2 x - 9 = 0$

Add $x$ and $2 x$ .

$3 x - 3 - 9 = 0$

Subtract $9$ from $- 3$ .

$3 x - 12 = 0$

Factor $3$ out of $3 x - 12$ .

Factor $3$ out of $3 x$ .

$3 (x) - 12 = 0$

Factor $3$ out of $- 12$ .

$3 x + 3 \cdot - 4 = 0$

Factor $3$ out of $3 x + 3 \cdot - 4$ .

$3 (x - 4) = 0$

Divide each term in $3 (x - 4) = 0$ by $3$ and simplify.

Divide each term in $3 (x - 4) = 0$ by $3$ .

$\frac{3 (x - 4)}{3} = \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (x - 4)}{3} = \frac{0}{3}$

Divide $x - 4$ by $1$ .

$x - 4 = \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$