Solve sqrt{4-y}-y=2

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

Solve for y

$$y=0$$

Solution Steps

Subtract -y from both sides of the equation.

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

Square both sides of the equation.

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

Calculate \sqrt{4-y} to the power of 2 and get 4-y.

$$4-y=\left(2+y\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+y\right)^{2}.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Subtract 4y from both sides.

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

Combine -y and -4y to get -5y.

$$-5y=y^{2}$$

Subtract y^{2} from both sides.

$$-5y-y^{2}=0$$

Factor out y.

$$y\left(-5-y\right)=0$$

To find equation solutions, solve y=0 and -5-y=0.

$$y=0 y=-5$$

Substitute 0 for y in the equation \sqrt{4-y}-y=2.

$$\sqrt{4-0}-0=2$$

Simplify. The value y=0 satisfies the equation.

$$2=2$$

Substitute -5 for y in the equation \sqrt{4-y}-y=2.

$$\sqrt{4-\left(-5\right)}-\left(-5\right)=2$$

Simplify. The value y=-5 does not satisfy the equation.

$$8=2$$

Equation \sqrt{4-y}=y+2 has a unique solution.

$$y=0$$