Algebra Examples

\sqrt{2 x + 5} - \sqrt{9 + x} > 0

$\sqrt{2 x + 5} - \sqrt{9 + x} > 0$

Step 1

Add

\sqrt{9 + x}

$\sqrt{9 + x}$ to both sides of the inequality.

\sqrt{2 x + 5} > \sqrt{9 + x}

$\sqrt{2 x + 5} > \sqrt{9 + x}$

Step 2

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

{\sqrt{2 x + 5}}^{2} > {\sqrt{9 + x}}^{2}

${\sqrt{2 x + 5}}^{2} > {\sqrt{9 + x}}^{2}$

Step 3

Simplify each side of the inequality.

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Step 3.1

Use

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

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

\sqrt{2 x + 5}

$\sqrt{2 x + 5}$ as

{(2 x + 5)}^{\frac{1}{2}}

${(2 x + 5)}^{\frac{1}{2}}$ .

{({(2 x + 5)}^{\frac{1}{2}})}^{2} > {\sqrt{9 + x}}^{2}

${({(2 x + 5)}^{\frac{1}{2}})}^{2} > {\sqrt{9 + x}}^{2}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify

{({(2 x + 5)}^{\frac{1}{2}})}^{2}

${({(2 x + 5)}^{\frac{1}{2}})}^{2}$ .

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Step 3.2.1.1

Multiply the exponents in

{({(2 x + 5)}^{\frac{1}{2}})}^{2}

${({(2 x + 5)}^{\frac{1}{2}})}^{2}$ .

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Step 3.2.1.1.1

Apply the power rule and multiply exponents,

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

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

{(2 x + 5)}^{\frac{1}{2} \cdot 2} > {\sqrt{9 + x}}^{2}

${(2 x + 5)}^{\frac{1}{2} \cdot 2} > {\sqrt{9 + x}}^{2}$

Step 3.2.1.1.2

Cancel the common factor of

2

$2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

${(2 x + 5)}^{\frac{1}{2} \cdot 2} > {\sqrt{9 + x}}^{2}$

Step 3.2.1.1.2.2

Rewrite the expression.

${(2 x + 5)}^{1} > {\sqrt{9 + x}}^{2}$

Step 3.2.1.2

Simplify.

$2 x + 5 > {\sqrt{9 + x}}^{2}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Rewrite

${\sqrt{9 + x}}^{2}$ as

$9 + x$ .

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Step 3.3.1.1

Use

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

$\sqrt{9 + x}$ as

${(9 + x)}^{\frac{1}{2}}$ .

$2 x + 5 > {({(9 + x)}^{\frac{1}{2}})}^{2}$

Step 3.3.1.2

Apply the power rule and multiply exponents,

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

$2 x + 5 > {(9 + x)}^{\frac{1}{2} \cdot 2}$

Step 3.3.1.3

Combine

$\frac{1}{2}$ and

$2$ .

$2 x + 5 > {(9 + x)}^{\frac{2}{2}}$

Step 3.3.1.4

Cancel the common factor of

$2$ .

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Step 3.3.1.4.1

Cancel the common factor.

$2 x + 5 > {(9 + x)}^{\frac{2}{2}}$

Step 3.3.1.4.2

Rewrite the expression.

$2 x + 5 > {(9 + x)}^{1}$

Step 3.3.1.5

Simplify.

$2 x + 5 > 9 + x$

Step 4

Solve for

$x$ .

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Step 4.1

Move all terms containing

$x$ to the left side of the inequality.

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Step 4.1.1

Subtract

$x$ from both sides of the inequality.

$2 x + 5 - x > 9$

Step 4.1.2

Subtract

$x$ from

$2 x$ .

$x + 5 > 9$

Step 4.2

Move all terms not containing

$x$ to the right side of the inequality.

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Step 4.2.1

Subtract

$5$ from both sides of the inequality.

$x > 9 - 5$

Step 4.2.2

Subtract

$5$ from

$9$ .

$x > 4$

Step 5

Find the domain of

$\sqrt{2 x + 5} - \sqrt{9 + x}$ .

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Step 5.1

Set the radicand in

$\sqrt{2 x + 5}$ greater than or equal to

$0$ to find where the expression is defined.

$2 x + 5 \geq 0$

Step 5.2

Solve for

$x$ .

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Step 5.2.1

Subtract

$5$ from both sides of the inequality.

$2 x \geq - 5$

Step 5.2.2

Divide each term in

$2 x \geq - 5$ by

$2$ and simplify.

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Step 5.2.2.1

Divide each term in

$2 x \geq - 5$ by

$2$ .

$\frac{2 x}{2} \geq \frac{- 5}{2}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of

$2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{- 5}{2}$

Step 5.2.2.2.1.2

Divide

$x$ by

$1$ .

$x \geq \frac{- 5}{2}$

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

Move the negative in front of the fraction.

$x \geq - \frac{5}{2}$

Step 5.3

Set the radicand in

$\sqrt{9 + x}$ greater than or equal to

$0$ to find where the expression is defined.

$9 + x \geq 0$

Step 5.4

Subtract

$9$ from both sides of the inequality.

$x \geq - 9$

Step 5.5

The domain is all values of

$x$ that make the expression defined.

$[- \frac{5}{2}, \infty)$

Step 6

The solution consists of all of the true intervals.

$x > 4$

Step 7

The result can be shown in multiple forms.

Inequality Form:

$x > 4$

Interval Notation:

$(4, \infty)$

Step 8