Solve the Inequality for x fourth root of x+2>2

$\sqrt[4]{x + 2} > 2$

Step 1

To remove the radical on the left side of the inequality, raise both sides of the inequality to the power of $4$ .

${\sqrt[4]{x + 2}}^{4} > 2^{4}$

Step 2

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{x + 2}$ as ${(x + 2)}^{\frac{1}{4}}$ .

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

Simplify the left side.

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

Multiply the exponents in ${({(x + 2)}^{\frac{1}{4}})}^{4}$ .

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

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

${(x + 2)}^{1} > 2^{4}$

Simplify.

$x + 2 > 2^{4}$

Simplify the right side.

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

$x + 2 > 16$

Step 3

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

Subtract $2$ from both sides of the inequality.

$x > 16 - 2$

Subtract $2$ from $16$ .

$x > 14$

Step 4

Find the domain of $\sqrt[4]{x + 2} - 2$ .

Set the radicand in $\sqrt[4]{x + 2}$ greater than or equal to $0$ to find where the expression is defined.

$x + 2 \geq 0$

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 5

The solution consists of all of the true intervals.

$x > 14$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$x > 14$

Interval Notation:

$(14, \infty)$

Step 7

Solve the Inequality for x fourth root of x+2>2

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