Solve for x 5 square root of x+2>17

$5 \sqrt{x} + 2 > 17$

Move all terms not containing $5 \sqrt{x}$ to the right side of the inequality.

Subtract $2$ from both sides of the inequality.

$5 \sqrt{x} > 17 - 2$

Subtract $2$ from $17$ .

$5 \sqrt{x} > 15$

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

${(5 \sqrt{x})}^{2} > 15^{2}$

Simplify each side of the inequality.

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

${(5 x^{\frac{1}{2}})}^{2} > 15^{2}$

Simplify the left side.

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

Apply the product rule to $5 x^{\frac{1}{2}}$ .

$5^{2} {(x^{\frac{1}{2}})}^{2} > 15^{2}$

Raise $5$ to the power of $2$ .

$25 {(x^{\frac{1}{2}})}^{2} > 15^{2}$

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

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

$25 x^{\frac{1}{2} \cdot 2} > 15^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$25 x^{\frac{1}{2} \cdot 2} > 15^{2}$

Rewrite the expression.

$25 x^{1} > 15^{2}$

Simplify.

$25 x > 15^{2}$

Simplify the right side.

Raise $15$ to the power of $2$ .

$25 x > 225$

Divide each term in $25 x > 225$ by $25$ and simplify.

Divide each term in $25 x > 225$ by $25$ .

$\frac{25 x}{25} > \frac{225}{25}$

Simplify the left side.

Cancel the common factor of $25$ .

Cancel the common factor.

$\frac{25 x}{25} > \frac{225}{25}$

Divide $x$ by $1$ .

$x > \frac{225}{25}$

Simplify the right side.

Divide $225$ by $25$ .

$x > 9$

Find the domain of $5 \sqrt{x} - 15$ .

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

$x \geq 0$

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

$[0, \infty)$

The solution consists of all of the true intervals.

$x > 9$

The result can be shown in multiple forms.

Inequality Form:

$x > 9$

Interval Notation:

$(9, \infty)$

Solve for x 5 square root of x+2>17

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