Solve by Factoring square root of 4x^2-12x+9<=9

$\sqrt{4 x^{2} - 12 x + 9} \leq 9$

Step 1

Subtract $9$ from both sides of the inequality.

$\sqrt{4 x^{2} - 12 x + 9} - 9 \leq 0$

Step 2

Simplify $\sqrt{4 x^{2} - 12 x + 9} - 9$ .

Simplify each term.

Factor using the perfect square rule.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\sqrt{{(2 x)}^{2} - 12 x + 9} - 9 \leq 0$

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

$\sqrt{{(2 x)}^{2} - 12 x + 3^{2}} - 9 \leq 0$

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

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

Rewrite the polynomial.

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

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

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

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

$2 x - 3 - 9 \leq 0$

Subtract $9$ from $- 3$ .

$2 x - 12 \leq 0$

Step 3

Factor $2$ out of $2 x - 12$ .

Factor $2$ out of $2 x$ .

$2 (x) - 12 \leq 0$

Factor $2$ out of $- 12$ .

$2 x + 2 \cdot - 6 \leq 0$

Factor $2$ out of $2 x + 2 \cdot - 6$ .

$2 (x - 6) \leq 0$

Step 4

Divide each term in $2 (x - 6) \leq 0$ by $2$ and simplify.

Divide each term in $2 (x - 6) \leq 0$ by $2$ .

$\frac{2 (x - 6)}{2} \leq \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (x - 6)}{2} \leq \frac{0}{2}$

Divide $x - 6$ by $1$ .

$x - 6 \leq \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x - 6 \leq 0$

Step 5

Add $6$ to both sides of the inequality.

$x \leq 6$

Solve by Factoring square root of 4x^2-12x+9<=9

Related Questions

Support LCMGCF.com by disabling your adblocker.