Solve by Factoring 9x^2-12x+4=0

$9 x^{2} - 12 x + 4 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

${(3 x)}^{2} - 12 x + 4 = 0$

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

${(3 x)}^{2} - 12 x + 2^{2} = 0$

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

$12 x = 2 \cdot (3 x) \cdot 2$

Rewrite the polynomial.

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

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

${(3 x - 2)}^{2} = 0$

Step 2

Set the $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Step 3

Solve for $x$ .

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{2}{3}$