Solve for x 2 square root of x+3<=8

$2 \sqrt{x} + 3 \leq 8$

Step 1

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

Subtract $3$ from both sides of the inequality.

$2 \sqrt{x} \leq 8 - 3$

Subtract $3$ from $8$ .

$2 \sqrt{x} \leq 5$

Step 2

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

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

Step 3

Simplify each side of the inequality.

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

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

Simplify the left side.

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$4 x \leq 5^{2}$

Simplify the right side.

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

$4 x \leq 25$

Step 4

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

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

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 5

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.

$[0, \infty)$

Step 6

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{25}{4}$

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

Step 7

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < 0$ to see if it makes the inequality true.

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$2 \sqrt{- 2} + 3 \leq 8$

The left side is not equal to the right side, which means that the given statement is false.

False

Test a value on the interval $0 < x < \frac{25}{4}$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < \frac{25}{4}$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

$2 \sqrt{3} + 3 \leq 8$

The left side $6.46410161$ is less than the right side $8$ , which means that the given statement is always true.

True

Test a value on the interval $x > \frac{25}{4}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{25}{4}$ and see if this value makes the original inequality true.

$x = 9$

Replace $x$ with $9$ in the original inequality.

$2 \sqrt{9} + 3 \leq 8$

The left side $9$ is greater than the right side $8$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ False

$0 < x < \frac{25}{4}$ True

$x > \frac{25}{4}$ False

$x < 0$ False

$0 < x < \frac{25}{4}$ True

$x > \frac{25}{4}$ False

Step 8

The solution consists of all of the true intervals.

$0 \leq x \leq \frac{25}{4}$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$0 \leq x \leq \frac{25}{4}$

Interval Notation:

$[0, \frac{25}{4}]$

Step 10

Solve for x 2 square root of x+3<=8

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