Evaluate square root of x+3 = square root of 2x-3

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

Step 1

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

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

Step 2

Simplify each side of the equation.

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

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

Simplify the left side.

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

Multiply the exponents in ${({(x + 3)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$x + 3 = 2 x - 3$

Step 3

Solve for $x$ .

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

Subtract $2 x$ from both sides of the equation.

$x + 3 - 2 x = - 3$

Subtract $2 x$ from $x$ .

$- x + 3 = - 3$

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

Subtract $3$ from both sides of the equation.

$- x = - 3 - 3$

Subtract $3$ from $- 3$ .

$- x = - 6$

Divide each term in $- x = - 6$ by $- 1$ and simplify.

Divide each term in $- x = - 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 6}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 6}{- 1}$

Simplify the right side.

Divide $- 6$ by $- 1$ .

$x = 6$

Evaluate square root of x+3 = square root of 2x-3

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