Solve for x square root of 4x+2 = square root of 3x+4

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

Step 1

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

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

Step 2

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$4 x + 2 = 3 x + 4$

Step 3

Solve for $x$ .

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

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

$4 x + 2 - 3 x = 4$

Subtract $3 x$ from $4 x$ .

$x + 2 = 4$

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

Subtract $2$ from both sides of the equation.

$x = 4 - 2$

Subtract $2$ from $4$ .

$x = 2$

Solve for x square root of 4x+2 = square root of 3x+4

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