Solve for x 2 square root of x-9>=-4

$2 \sqrt{x} - 9 \geq - 4$

Move all terms not containing $2 \sqrt{x}$ to the right side of the inequality.

Add $9$ to both sides of the inequality.

$2 \sqrt{x} \geq - 4 + 9$

Add $- 4$ and $9$ .

$2 \sqrt{x} \geq 5$

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

${(2 \sqrt{x})}^{2} \geq 5^{2}$

Simplify each side of the inequality.

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

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

Simplify the left side.

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

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

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

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

$4 {(x^{\frac{1}{2}})}^{2} \geq 5^{2}$

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

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

$4 x^{\frac{1}{2} \cdot 2} \geq 5^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 x^{\frac{1}{2} \cdot 2} \geq 5^{2}$

Rewrite the expression.

$4 x^{1} \geq 5^{2}$

Simplify.

$4 x \geq 5^{2}$

Simplify the right side.

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

$4 x \geq 25$

Divide each term in $4 x \geq 25$ by $4$ and simplify.

Divide each term in $4 x \geq 25$ by $4$ .

$\frac{4 x}{4} \geq \frac{25}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} \geq \frac{25}{4}$

Divide $x$ by $1$ .

$x \geq \frac{25}{4}$

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

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.

$x \geq \frac{25}{4}$

The result can be shown in multiple forms.

Inequality Form:

$x \geq \frac{25}{4}$

Interval Notation:

$[\frac{25}{4}, \infty)$

Solve for x 2 square root of x-9>=-4

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