Solve for y y-4 = square root of 2y

$y - 4 = \sqrt{2 y}$

Since the radical is on the right side of the equation, switch the sides so it is on the left side of the equation.

$\sqrt{2 y} = y - 4$

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

${\sqrt{2 y}}^{2} = {(y - 4)}^{2}$

Simplify each side of the equation.

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

${({(2 y)}^{\frac{1}{2}})}^{2} = {(y - 4)}^{2}$

Simplify the left side.

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

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

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

${(2 y)}^{\frac{1}{2} \cdot 2} = {(y - 4)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 y)}^{\frac{1}{2} \cdot 2} = {(y - 4)}^{2}$

Rewrite the expression.

${(2 y)}^{1} = {(y - 4)}^{2}$

Simplify.

$2 y = {(y - 4)}^{2}$

Simplify the right side.

Simplify ${(y - 4)}^{2}$ .

Rewrite ${(y - 4)}^{2}$ as $(y - 4) (y - 4)$ .

$2 y = (y - 4) (y - 4)$

Expand $(y - 4) (y - 4)$ using the FOIL Method.

Apply the distributive property.

$2 y = y (y - 4) - 4 (y - 4)$

Apply the distributive property.

$2 y = y \cdot y + y \cdot - 4 - 4 (y - 4)$

Apply the distributive property.

$2 y = y \cdot y + y \cdot - 4 - 4 y - 4 \cdot - 4$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$2 y = y^{2} + y \cdot - 4 - 4 y - 4 \cdot - 4$

Move $- 4$ to the left of $y$ .

$2 y = y^{2} - 4 \cdot y - 4 y - 4 \cdot - 4$

Multiply $- 4$ by $- 4$ .

$2 y = y^{2} - 4 y - 4 y + 16$

Subtract $4 y$ from $- 4 y$ .

$2 y = y^{2} - 8 y + 16$

Solve for $y$ .

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$y^{2} - 8 y + 16 = 2 y$

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

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

$y^{2} - 8 y + 16 - 2 y = 0$

Subtract $2 y$ from $- 8 y$ .

$y^{2} - 10 y + 16 = 0$

Factor $y^{2} - 10 y + 16$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $16$ and whose sum is $- 10$ .

$- 8, - 2$

Write the factored form using these integers.

$(y - 8) (y - 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y - 8 = 0$

$y - 2 = 0$

Set $y - 8$ equal to $0$ and solve for $y$ .

Set $y - 8$ equal to $0$ .

$y - 8 = 0$

Add $8$ to both sides of the equation.

$y = 8$

Set $y - 2$ equal to $0$ and solve for $y$ .

Set $y - 2$ equal to $0$ .

$y - 2 = 0$

Add $2$ to both sides of the equation.

$y = 2$

The final solution is all the values that make $(y - 8) (y - 2) = 0$ true.

$y = 8, 2$

Exclude the solutions that do not make $y - 4 = \sqrt{2 y}$ true.

$y = 8$

Solve for y y-4 = square root of 2y

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