Solve the Inequality for x square root of 3x+1>=4

$\sqrt{3 x + 1} \geq 4$

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

${\sqrt{3 x + 1}}^{2} \geq 4^{2}$

Simplify each side of the inequality.

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

${({(3 x + 1)}^{\frac{1}{2}})}^{2} \geq 4^{2}$

Simplify the left side.

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

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

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

${(3 x + 1)}^{\frac{1}{2} \cdot 2} \geq 4^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 x + 1)}^{\frac{1}{2} \cdot 2} \geq 4^{2}$

Rewrite the expression.

${(3 x + 1)}^{1} \geq 4^{2}$

Simplify.

$3 x + 1 \geq 4^{2}$

Simplify the right side.

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

$3 x + 1 \geq 16$

Solve for $x$ .

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

Subtract $1$ from both sides of the inequality.

$3 x \geq 16 - 1$

Subtract $1$ from $16$ .

$3 x \geq 15$

Divide each term in $3 x \geq 15$ by $3$ and simplify.

Divide each term in $3 x \geq 15$ by $3$ .

$\frac{3 x}{3} \geq \frac{15}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{15}{3}$

Divide $x$ by $1$ .

$x \geq \frac{15}{3}$

Simplify the right side.

Divide $15$ by $3$ .

$x \geq 5$

Find the domain of $\sqrt{3 x + 1} - 4$ .

Set the radicand in $\sqrt{3 x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$3 x + 1 \geq 0$

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$3 x \geq - 1$

Divide each term in $3 x \geq - 1$ by $3$ and simplify.

Divide each term in $3 x \geq - 1$ by $3$ .

$\frac{3 x}{3} \geq \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{- 1}{3}$

Divide $x$ by $1$ .

$x \geq \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{1}{3}$

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

$[- \frac{1}{3}, \infty)$

The solution consists of all of the true intervals.

$x \geq 5$

The result can be shown in multiple forms.

Inequality Form:

$x \geq 5$

Interval Notation:

$[5, \infty)$

Solve the Inequality for x square root of 3x+1>=4

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