Solve for x square root of 3x+6-1>=5

$\sqrt{3 x + 6} - 1 \geq 5$

Step 1

Move all terms not containing $\sqrt{3 x + 6}$ to the right side of the inequality.

Add $1$ to both sides of the inequality.

$\sqrt{3 x + 6} \geq 5 + 1$

Add $5$ and $1$ .

$\sqrt{3 x + 6} \geq 6$

Step 2

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

${\sqrt{3 x + 6}}^{2} \geq 6^{2}$

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x + 6}$ as ${(3 x + 6)}^{\frac{1}{2}}$ .

${({(3 x + 6)}^{\frac{1}{2}})}^{2} \geq 6^{2}$

Simplify the left side.

Simplify ${({(3 x + 6)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 x + 6)}^{\frac{1}{2}})}^{2}$ .

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

${(3 x + 6)}^{\frac{1}{2} \cdot 2} \geq 6^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 x + 6)}^{\frac{1}{2} \cdot 2} \geq 6^{2}$

Rewrite the expression.

${(3 x + 6)}^{1} \geq 6^{2}$

Simplify.

$3 x + 6 \geq 6^{2}$

Simplify the right side.

Raise $6$ to the power of $2$ .

$3 x + 6 \geq 36$

Step 4

Solve for $x$ .

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

Subtract $6$ from both sides of the inequality.

$3 x \geq 36 - 6$

Subtract $6$ from $36$ .

$3 x \geq 30$

Divide each term in $3 x \geq 30$ by $3$ and simplify.

Divide each term in $3 x \geq 30$ by $3$ .

$\frac{3 x}{3} \geq \frac{30}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{30}{3}$

Divide $x$ by $1$ .

$x \geq \frac{30}{3}$

Simplify the right side.

Divide $30$ by $3$ .

$x \geq 10$

Step 5

Find the domain of $\sqrt{3 (x + 2)} - 6$ .

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

$3 (x + 2) \geq 0$

Solve for $x$ .

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

Divide each term in $3 (x + 2) \geq 0$ by $3$ .

$\frac{3 (x + 2)}{3} \geq \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (x + 2)}{3} \geq \frac{0}{3}$

Divide $x + 2$ by $1$ .

$x + 2 \geq \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$x + 2 \geq 0$

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

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

$[- 2, \infty)$

Step 6

The solution consists of all of the true intervals.

$x \geq 10$

Step 7

The result can be shown in multiple forms.

Inequality Form:

$x \geq 10$

Interval Notation:

$[10, \infty)$

Step 8

Solve for x square root of 3x+6-1>=5

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