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

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

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

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

Simplify each side of the equation.

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

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $1$ by $1$ .

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

Simplify.

$4 - x = 3 x + 4$

Solve for $x$ .

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

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

$4 - x - 3 x = 4$

Subtract $3 x$ from $- x$ .

$4 - 4 x = 4$

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

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Subtract $4$ from both sides of the equation.

$- 4 x = 4 - 4$

Subtract $4$ from $4$ .

$- 4 x = 0$

Divide each term by $- 4$ and simplify.

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Divide each term in $- 4 x = 0$ by $- 4$ .

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

Cancel the common factor of $- 4$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

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

Divide $0$ by $- 4$ .

$x = 0$

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