Solve the Inequality for x 4x^2-12x+9>0

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

Step 1

Convert the inequality to an equation.

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

Step 2

Factor using the perfect square rule.

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

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

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

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

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

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

Rewrite the polynomial.

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

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

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

Step 3

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

$2 x - 3 = 0$

Step 4

Solve for $x$ .

Add $3$ to both sides of the equation.

$2 x = 3$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 5

Use each root to create test intervals.

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

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

Step 6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < \frac{3}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{3}{2}$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$4 {(0)}^{2} - 12 \cdot 0 + 9 > 0$

The left side $9$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > \frac{3}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{3}{2}$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$4 {(4)}^{2} - 12 \cdot 4 + 9 > 0$

The left side $25$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < \frac{3}{2}$ True

$x > \frac{3}{2}$ True

$x < \frac{3}{2}$ True

$x > \frac{3}{2}$ True

Step 7

The solution consists of all of the true intervals.

$x < \frac{3}{2}$ or $x > \frac{3}{2}$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{3}{2} or x > \frac{3}{2}$

Interval Notation:

$(- \infty, \frac{3}{2}) \cup (\frac{3}{2}, \infty)$

Step 9

Solve the Inequality for x 4x^2-12x+9>0

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