Solve the Inequality for x square root of 2x+5- square root of 9+x>0

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

Step 1

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

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

Step 2

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

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

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x + 5}$ as ${(2 x + 5)}^{\frac{1}{2}}$ .

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

Simplify the left side.

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

Multiply the exponents in ${({(2 x + 5)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{9 + x}$ as ${(9 + x)}^{\frac{1}{2}}$ .

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

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

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$2 x + 5 > 9 + x$

Step 4

Solve for $x$ .

Move all terms containing $x$ to the left side of the inequality.

Subtract $x$ from both sides of the inequality.

$2 x + 5 - x > 9$

Subtract $x$ from $2 x$ .

$x + 5 > 9$

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

Subtract $5$ from both sides of the inequality.

$x > 9 - 5$

Subtract $5$ from $9$ .

$x > 4$

Step 5

Find the domain of $\sqrt{2 x + 5} - \sqrt{9 + x}$ .

Set the radicand in $\sqrt{2 x + 5}$ greater than or equal to $0$ to find where the expression is defined.

$2 x + 5 \geq 0$

Solve for $x$ .

Subtract $5$ from both sides of the inequality.

$2 x \geq - 5$

Divide each term in $2 x \geq - 5$ by $2$ and simplify.

Divide each term in $2 x \geq - 5$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Set the radicand in $\sqrt{9 + x}$ greater than or equal to $0$ to find where the expression is defined.

$9 + x \geq 0$

Subtract $9$ from both sides of the inequality.

$x \geq - 9$

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

Solve the Inequality for x square root of 2x+5- square root of 9+x>0

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