Solve for x 5- square root of 20x+4>=-3

$5 - \sqrt{20 x + 4} \geq - 3$

Step 1

Move all terms not containing $- \sqrt{20 x + 4}$ to the right side of the inequality.

Subtract $5$ from both sides of the inequality.

$- \sqrt{20 x + 4} \geq - 3 - 5$

Subtract $5$ from $- 3$ .

$- \sqrt{20 x + 4} \geq - 8$

Step 2

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

${(- \sqrt{20 x + 4})}^{2} \geq {(- 8)}^{2}$

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{20 x + 4}$ as ${(20 x + 4)}^{\frac{1}{2}}$ .

${(- {(20 x + 4)}^{\frac{1}{2}})}^{2} \geq {(- 8)}^{2}$

Simplify the left side.

Simplify ${(- {(20 x + 4)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- {(20 x + 4)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(20 x + 4)}^{\frac{1}{2}})}^{2} \geq {(- 8)}^{2}$

Raise $- 1$ to the power of $2$ .

$1 {({(20 x + 4)}^{\frac{1}{2}})}^{2} \geq {(- 8)}^{2}$

Multiply ${({(20 x + 4)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(20 x + 4)}^{\frac{1}{2}})}^{2} \geq {(- 8)}^{2}$

Multiply the exponents in ${({(20 x + 4)}^{\frac{1}{2}})}^{2}$ .

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

${(20 x + 4)}^{\frac{1}{2} \cdot 2} \geq {(- 8)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(20 x + 4)}^{\frac{1}{2} \cdot 2} \geq {(- 8)}^{2}$

Rewrite the expression.

${(20 x + 4)}^{1} \geq {(- 8)}^{2}$

Simplify.

$20 x + 4 \geq {(- 8)}^{2}$

Simplify the right side.

Raise $- 8$ to the power of $2$ .

$20 x + 4 \geq 64$

Step 4

Solve for $x$ .

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

Subtract $4$ from both sides of the inequality.

$20 x \geq 64 - 4$

Subtract $4$ from $64$ .

$20 x \geq 60$

Divide each term in $20 x \geq 60$ by $20$ and simplify.

Divide each term in $20 x \geq 60$ by $20$ .

$\frac{20 x}{20} \geq \frac{60}{20}$

Simplify the left side.

Cancel the common factor of $20$ .

Cancel the common factor.

$\frac{20 x}{20} \geq \frac{60}{20}$

Divide $x$ by $1$ .

$x \geq \frac{60}{20}$

Simplify the right side.

Divide $60$ by $20$ .

$x \geq 3$

Step 5

Find the domain of $8 - 2 \sqrt{5 x + 1}$ .

Set the radicand in $\sqrt{5 x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$5 x + 1 \geq 0$

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$5 x \geq - 1$

Divide each term in $5 x \geq - 1$ by $5$ and simplify.

Divide each term in $5 x \geq - 1$ by $5$ .

$\frac{5 x}{5} \geq \frac{- 1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} \geq \frac{- 1}{5}$

Divide $x$ by $1$ .

$x \geq \frac{- 1}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{1}{5}$

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

$[- \frac{1}{5}, \infty)$

Step 6

Use each root to create test intervals.

$x < - \frac{1}{5}$

$- \frac{1}{5} < x < 3$

$x > 3$

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 $x < - \frac{1}{5}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{1}{5}$ and see if this value makes the original inequality true.

$x = - 3$

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

$5 - \sqrt{20 (- 3) + 4} \geq - 3$

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 $- \frac{1}{5} < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{1}{5} < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$5 - \sqrt{20 (0) + 4} \geq - 3$

The left side $3$ is greater than the right side $- 3$ , which means that the given statement is always true.

True

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$5 - \sqrt{20 (6) + 4} \geq - 3$

The left side $- 6.13552872$ is less than the right side $- 3$ , which means that the given statement is false.

False

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

$x < - \frac{1}{5}$ False

$- \frac{1}{5} < x < 3$ True

$x > 3$ False

$x < - \frac{1}{5}$ False

$- \frac{1}{5} < x < 3$ True

$x > 3$ False

Step 8

The solution consists of all of the true intervals.

$- \frac{1}{5} \leq x \leq 3$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$- \frac{1}{5} \leq x \leq 3$

Interval Notation:

$[- \frac{1}{5}, 3]$

Step 10

Solve for x 5- square root of 20x+4>=-3

Related Questions

Support LCMGCF.com by disabling your adblocker.