Solve for x 4 square root of 2x-3<12

$4 \sqrt{2 x - 3} < 12$

Step 1

To remove the radical on the left side of the inequality, square both sides of the inequality.

${(4 \sqrt{2 x - 3})}^{2} < 12^{2}$

Step 2

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x - 3}$ as ${(2 x - 3)}^{\frac{1}{2}}$ .

${(4 {(2 x - 3)}^{\frac{1}{2}})}^{2} < 12^{2}$

Simplify the left side.

Simplify ${(4 {(2 x - 3)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $4 {(2 x - 3)}^{\frac{1}{2}}$ .

$4^{2} {({(2 x - 3)}^{\frac{1}{2}})}^{2} < 12^{2}$

Raise $4$ to the power of $2$ .

$16 {({(2 x - 3)}^{\frac{1}{2}})}^{2} < 12^{2}$

Multiply the exponents in ${({(2 x - 3)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$16 {(2 x - 3)}^{\frac{1}{2} \cdot 2} < 12^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$16 {(2 x - 3)}^{\frac{1}{2} \cdot 2} < 12^{2}$

Rewrite the expression.

$16 {(2 x - 3)}^{1} < 12^{2}$

Simplify.

$16 (2 x - 3) < 12^{2}$

Apply the distributive property.

$16 (2 x) + 16 \cdot - 3 < 12^{2}$

Multiply.

Multiply $2$ by $16$ .

$32 x + 16 \cdot - 3 < 12^{2}$

Multiply $16$ by $- 3$ .

$32 x - 48 < 12^{2}$

Simplify the right side.

Raise $12$ to the power of $2$ .

$32 x - 48 < 144$

Step 3

Solve for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Add $48$ to both sides of the inequality.

$32 x < 144 + 48$

Add $144$ and $48$ .

$32 x < 192$

Divide each term in $32 x < 192$ by $32$ and simplify.

Divide each term in $32 x < 192$ by $32$ .

$\frac{32 x}{32} < \frac{192}{32}$

Simplify the left side.

Cancel the common factor of $32$ .

Cancel the common factor.

$\frac{32 x}{32} < \frac{192}{32}$

Divide $x$ by $1$ .

$x < \frac{192}{32}$

Simplify the right side.

Divide $192$ by $32$ .

$x < 6$

Step 4

Find the domain of $4 \sqrt{2 x - 3} - 12$ .

Set the radicand in $\sqrt{2 x - 3}$ greater than or equal to $0$ to find where the expression is defined.

$2 x - 3 \geq 0$

Solve for $x$ .

Add $3$ to both sides of the inequality.

$2 x \geq 3$

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

Divide each term in $2 x \geq 3$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

The domain is all values of $x$ that make the expression defined.

$[\frac{3}{2}, \infty)$

Step 5

Use each root to create test intervals.

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

$\frac{3}{2} < x < 6$

$x > 6$

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.

$4 \sqrt{2 (0) - 3} < 12$

The left side is not equal to the right side, which means that the given statement is false.

False

Test a value on the interval $\frac{3}{2} < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $\frac{3}{2} < x < 6$ and see if this value makes the original inequality true.

$x = 4$

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

$4 \sqrt{2 (4) - 3} < 12$

The left side $8.9442719$ is less than the right side $12$ , which means that the given statement is always true.

True

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

$4 \sqrt{2 (8) - 3} < 12$

The left side $14.4222051$ is not less than the right side $12$ , which means that the given statement is false.

False

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

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

$\frac{3}{2} < x < 6$ True

$x > 6$ False

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

$\frac{3}{2} < x < 6$ True

$x > 6$ False

Step 7

The solution consists of all of the true intervals.

$\frac{3}{2} < x < 6$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$\frac{3}{2} < x < 6$

Interval Notation:

$(\frac{3}{2}, 6)$

Step 9

Solve for x 4 square root of 2x-3<12

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