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

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

Step 1

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

${\sqrt{4 - x}}^{2} \geq {\sqrt{3 x + 4}}^{2}$

Step 2

Simplify each side of the inequality.

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

${({(4 - x)}^{\frac{1}{2}})}^{2} \geq {\sqrt{3 x + 4}}^{2}$

Simplify the left side.

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

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

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

${(4 - x)}^{\frac{1}{2} \cdot 2} \geq {\sqrt{3 x + 4}}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(4 - x)}^{\frac{1}{2} \cdot 2} \geq {\sqrt{3 x + 4}}^{2}$

Rewrite the expression.

${(4 - x)}^{1} \geq {\sqrt{3 x + 4}}^{2}$

Simplify.

$4 - x \geq {\sqrt{3 x + 4}}^{2}$

Simplify the right side.

Rewrite ${\sqrt{3 x + 4}}^{2}$ as $3 x + 4$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x + 4}$ as ${(3 x + 4)}^{\frac{1}{2}}$ .

$4 - x \geq {({(3 x + 4)}^{\frac{1}{2}})}^{2}$

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

$4 - x \geq {(3 x + 4)}^{\frac{1}{2} \cdot 2}$

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

$4 - x \geq {(3 x + 4)}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 - x \geq {(3 x + 4)}^{\frac{2}{2}}$

Rewrite the expression.

$4 - x \geq {(3 x + 4)}^{1}$

Simplify.

$4 - x \geq 3 x + 4$

Step 3

Solve for $x$ .

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

Subtract $3 x$ from both sides of the inequality.

$4 - x - 3 x \geq 4$

Subtract $3 x$ from $- x$ .

$4 - 4 x \geq 4$

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

Subtract $4$ from both sides of the inequality.

$- 4 x \geq 4 - 4$

Subtract $4$ from $4$ .

$- 4 x \geq 0$

Divide each term in $- 4 x \geq 0$ by $- 4$ and simplify.

Divide each term in $- 4 x \geq 0$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $0$ by $- 4$ .

$x \leq 0$

Step 4

Find the domain of $\sqrt{4 - x} - \sqrt{3 x + 4}$ .

Set the radicand in $\sqrt{4 - x}$ greater than or equal to $0$ to find where the expression is defined.

$4 - x \geq 0$

Solve for $x$ .

Subtract $4$ from both sides of the inequality.

$- x \geq - 4$

Divide each term in $- x \geq - 4$ by $- 1$ and simplify.

Divide each term in $- x \geq - 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- 4}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x \leq 4$

Set the radicand in $\sqrt{3 x + 4}$ greater than or equal to $0$ to find where the expression is defined.

$3 x + 4 \geq 0$

Solve for $x$ .

Subtract $4$ from both sides of the inequality.

$3 x \geq - 4$

Divide each term in $3 x \geq - 4$ by $3$ and simplify.

Divide each term in $3 x \geq - 4$ by $3$ .

$\frac{3 x}{3} \geq \frac{- 4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{- 4}{3}$

Divide $x$ by $1$ .

$x \geq \frac{- 4}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{4}{3}$

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

$[- \frac{4}{3}, 4]$

Step 5

Use each root to create test intervals.

$x < - \frac{4}{3}$

$- \frac{4}{3} < x < 0$

$0 < x < 4$

$x > 4$

Step 6

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 $x < - \frac{4}{3}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{4}{3}$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

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

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 $- \frac{4}{3} < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{4}{3} < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

$\sqrt{4 - (- 1)} \geq \sqrt{3 (- 1) + 4}$

The left side $2.23606797$ is greater than the right side $1$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 4$ to see if it makes the inequality true.

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

$x = 2$

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

$\sqrt{4 - (2)} \geq \sqrt{3 (2) + 4}$

The left side $1.41421356$ is less than the right side $3.16227766$ , which means that the given statement is false.

False

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

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

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

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

False

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

$x < - \frac{4}{3}$ False

$- \frac{4}{3} < x < 0$ True

$0 < x < 4$ False

$x > 4$ False

$x < - \frac{4}{3}$ False

$- \frac{4}{3} < x < 0$ True

$0 < x < 4$ False

$x > 4$ False

Step 7

The solution consists of all of the true intervals.

$- \frac{4}{3} \leq x \leq 0$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$- \frac{4}{3} \leq x \leq 0$

Interval Notation:

$[- \frac{4}{3}, 0]$

Step 9

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

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