Solve for g square root of g+2 = square root of 2g-5

$\sqrt{g + 2} = \sqrt{2 g - 5}$

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

${\sqrt{g + 2}}^{2} = {\sqrt{2 g - 5}}^{2}$

Simplify each side of the equation.

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

${({(g + 2)}^{\frac{1}{2}})}^{2} = {\sqrt{2 g - 5}}^{2}$

Simplify the left side.

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

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

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

${(g + 2)}^{\frac{1}{2} \cdot 2} = {\sqrt{2 g - 5}}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(g + 2)}^{\frac{1}{2} \cdot 2} = {\sqrt{2 g - 5}}^{2}$

Rewrite the expression.

${(g + 2)}^{1} = {\sqrt{2 g - 5}}^{2}$

Simplify.

$g + 2 = {\sqrt{2 g - 5}}^{2}$

Simplify the right side.

Rewrite ${\sqrt{2 g - 5}}^{2}$ as $2 g - 5$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 g - 5}$ as ${(2 g - 5)}^{\frac{1}{2}}$ .

$g + 2 = {({(2 g - 5)}^{\frac{1}{2}})}^{2}$

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

$g + 2 = {(2 g - 5)}^{\frac{1}{2} \cdot 2}$

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

$g + 2 = {(2 g - 5)}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$g + 2 = {(2 g - 5)}^{\frac{2}{2}}$

Rewrite the expression.

$g + 2 = {(2 g - 5)}^{1}$

Simplify.

$g + 2 = 2 g - 5$

Solve for $g$ .

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

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

$g + 2 - 2 g = - 5$

Subtract $2 g$ from $g$ .

$- g + 2 = - 5$

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

Subtract $2$ from both sides of the equation.

$- g = - 5 - 2$

Subtract $2$ from $- 5$ .

$- g = - 7$

Divide each term in $- g = - 7$ by $- 1$ and simplify.

Divide each term in $- g = - 7$ by $- 1$ .

$\frac{- g}{- 1} = \frac{- 7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{g}{1} = \frac{- 7}{- 1}$

Divide $g$ by $1$ .

$g = \frac{- 7}{- 1}$

Simplify the right side.

Divide $- 7$ by $- 1$ .

$g = 7$

Solve for g square root of g+2 = square root of 2g-5

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