Algebra Examples

\sqrt[4]{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

$4$ .

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

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

Step 2

Simplify each side of the inequality.

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Step 2.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt[4]{x + 2}

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

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

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

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

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify

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

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

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Step 2.2.1.1

Multiply the exponents in

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

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

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Step 2.2.1.1.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

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

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

Step 2.2.1.1.2

Cancel the common factor of

4

$4$ .

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Step 2.2.1.1.2.1

Cancel the common factor.

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

Step 2.2.1.1.2.2

Rewrite the expression.

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

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

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

Step 2.2.1.2

Simplify.

$x + 2 > 2^{4}$

$x + 2 > 2^{4}$

$x + 2 > 2^{4}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Raise

$2$ to the power of

$4$ .

$x + 2 > 16$

$x + 2 > 16$

$x + 2 > 16$

Step 3

Move all terms not containing

$x$ to the right side of the inequality.

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Step 3.1

Subtract

$2$ from both sides of the inequality.

$x > 16 - 2$

Step 3.2

Subtract

$2$ from

$16$ .

$x > 14$

$x > 14$

Step 4

Find the domain of

$\sqrt[4]{x + 2} - 2$ .

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Step 4.1

Set the radicand in

$\sqrt[4]{x + 2}$ greater than or equal to

$0$ to find where the expression is defined.

$x + 2 \geq 0$

Step 4.2

Subtract

$2$ from both sides of the inequality.

$x \geq - 2$

Step 4.3

The domain is all values of

$x$ that make the expression defined.

$[- 2, \infty)$

$[- 2, \infty)$

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

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$