Solve for X square root of 2a-6<4

$\sqrt{2 a - 6} < 4$

Step 1

Factor $2$ out of $2 a - 6$ .

Factor $2$ out of $2 a$ .

$\sqrt{2 (a) - 6} < 4$

Factor $2$ out of $- 6$ .

$\sqrt{2 a + 2 \cdot - 3} < 4$

Factor $2$ out of $2 a + 2 \cdot - 3$ .

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

Step 2

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

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

Step 3

Simplify each side of the inequality.

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

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

Simplify the left side.

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

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

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

${(2 (a - 3))}^{\frac{1}{2} \cdot 2} < 4^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 (a - 3))}^{\frac{1}{2} \cdot 2} < 4^{2}$

Rewrite the expression.

${(2 (a - 3))}^{1} < 4^{2}$

Apply the distributive property.

${(2 a + 2 \cdot - 3)}^{1} < 4^{2}$

Multiply.

Multiply $2$ by $- 3$ .

${(2 a - 6)}^{1} < 4^{2}$

Simplify.

$2 a - 6 < 4^{2}$

Simplify the right side.

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

$2 a - 6 < 16$

Step 4

Solve for $a$ .

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

Add $6$ to both sides of the inequality.

$2 a < 16 + 6$

Add $16$ and $6$ .

$2 a < 22$

Divide each term in $2 a < 22$ by $2$ and simplify.

Divide each term in $2 a < 22$ by $2$ .

$\frac{2 a}{2} < \frac{22}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} < \frac{22}{2}$

Divide $a$ by $1$ .

$a < \frac{22}{2}$

Simplify the right side.

Divide $22$ by $2$ .

$a < 11$

Step 5

Find the domain of $\sqrt{2 (a - 3)} - 4$ .

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

$2 (a - 3) \geq 0$

Solve for $a$ .

Divide each term in $2 (a - 3) \geq 0$ by $2$ and simplify.

Divide each term in $2 (a - 3) \geq 0$ by $2$ .

$\frac{2 (a - 3)}{2} \geq \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (a - 3)}{2} \geq \frac{0}{2}$

Divide $a - 3$ by $1$ .

$a - 3 \geq \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$a - 3 \geq 0$

Add $3$ to both sides of the inequality.

$a \geq 3$

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

$[3, \infty)$

Step 6

Use each root to create test intervals.

$a < 3$

$3 < a < 11$

$a > 11$

Step 7

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 $a < 3$ to see if it makes the inequality true.

Choose a value on the interval $a < 3$ and see if this value makes the original inequality true.

$a = 0$

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

$\sqrt{2 (0) - 6} < 4$

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 $3 < a < 11$ to see if it makes the inequality true.

Choose a value on the interval $3 < a < 11$ and see if this value makes the original inequality true.

$a = 7$

Replace $a$ with $7$ in the original inequality.

$\sqrt{2 (7) - 6} < 4$

The left side $2.82842712$ is less than the right side $4$ , which means that the given statement is always true.

True

Test a value on the interval $a > 11$ to see if it makes the inequality true.

Choose a value on the interval $a > 11$ and see if this value makes the original inequality true.

$a = 14$

Replace $a$ with $14$ in the original inequality.

$\sqrt{2 (14) - 6} < 4$

The left side $4.69041575$ is not less than the right side $4$ , which means that the given statement is false.

False

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

$a < 3$ False

$3 < a < 11$ True

$a > 11$ False

$a < 3$ False

$3 < a < 11$ True

$a > 11$ False

Step 8

The solution consists of all of the true intervals.

$3 < a < 11$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$3 < a < 11$

Interval Notation:

$(3, 11)$

Step 10

Solve for X square root of 2a-6<4

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