Solve y+1/16=sqrt{y/4-(y-2)^2}

$$y + \frac { 1 } { 16 } = \sqrt { \frac { y } { 4 } - ( y - 2 ) ^ { 2 } }$$

Solve for y

$$y=\frac{33}{32}+\frac{31}{32}i=1.03125+0.96875i$$

$$y=\frac{33}{32}-\frac{31}{32}i=1.03125-0.96875i$$

Solution Steps

Square both sides of the equation.

$$\left(y+\frac{1}{16}\right)^{2}=\left(\sqrt{\frac{y}{4}-\left(y-2\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+\frac{1}{16}\right)^{2}.

$$y^{2}+\frac{1}{8}y+\frac{1}{256}=\left(\sqrt{\frac{y}{4}-\left(y-2\right)^{2}}\right)^{2}$$

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

$$y^{2}+\frac{1}{8}y+\frac{1}{256}=\left(\sqrt{\frac{y}{4}-\left(y^{2}-4y+4\right)}\right)^{2}$$

To find the opposite of y^{2}-4y+4, find the opposite of each term.

$$y^{2}+\frac{1}{8}y+\frac{1}{256}=\left(\sqrt{\frac{y}{4}-y^{2}+4y-4}\right)^{2}$$

Combine \frac{y}{4} and 4y to get \frac{17}{4}y.

$$y^{2}+\frac{1}{8}y+\frac{1}{256}=\left(\sqrt{\frac{17}{4}y-y^{2}-4}\right)^{2}$$

Calculate \sqrt{\frac{17}{4}y-y^{2}-4} to the power of 2 and get \frac{17}{4}y-y^{2}-4.

$$y^{2}+\frac{1}{8}y+\frac{1}{256}=\frac{17}{4}y-y^{2}-4$$

Subtract \frac{17}{4}y from both sides.

$$y^{2}+\frac{1}{8}y+\frac{1}{256}-\frac{17}{4}y=-y^{2}-4$$

Combine \frac{1}{8}y and -\frac{17}{4}y to get -\frac{33}{8}y.

$$y^{2}-\frac{33}{8}y+\frac{1}{256}=-y^{2}-4$$

Add y^{2} to both sides.

$$y^{2}-\frac{33}{8}y+\frac{1}{256}+y^{2}=-4$$

Combine y^{2} and y^{2} to get 2y^{2}.

$$2y^{2}-\frac{33}{8}y+\frac{1}{256}=-4$$

Add 4 to both sides.

$$2y^{2}-\frac{33}{8}y+\frac{1}{256}+4=0$$

Add \frac{1}{256} and 4 to get \frac{1025}{256}.

$$2y^{2}-\frac{33}{8}y+\frac{1025}{256}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -\frac{33}{8} for b, and \frac{1025}{256} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{33}{8}\right)±\sqrt{\left(-\frac{33}{8}\right)^{2}-4\times 2\times \frac{1025}{256}}}{2\times 2}$$

Square -\frac{33}{8} by squaring both the numerator and the denominator of the fraction.

$$y=\frac{-\left(-\frac{33}{8}\right)±\sqrt{\frac{1089}{64}-4\times 2\times \frac{1025}{256}}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-\left(-\frac{33}{8}\right)±\sqrt{\frac{1089}{64}-8\times \frac{1025}{256}}}{2\times 2}$$

Multiply -8 times \frac{1025}{256}.

$$y=\frac{-\left(-\frac{33}{8}\right)±\sqrt{\frac{1089}{64}-\frac{1025}{32}}}{2\times 2}$$

Add \frac{1089}{64} to -\frac{1025}{32} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{33}{8}\right)±\sqrt{-\frac{961}{64}}}{2\times 2}$$

Take the square root of -\frac{961}{64}.

$$y=\frac{-\left(-\frac{33}{8}\right)±\frac{31}{8}i}{2\times 2}$$

The opposite of -\frac{33}{8} is \frac{33}{8}.

$$y=\frac{\frac{33}{8}±\frac{31}{8}i}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{\frac{33}{8}±\frac{31}{8}i}{4}$$

Now solve the equation y=\frac{\frac{33}{8}±\frac{31}{8}i}{4} when ± is plus. Add \frac{33}{8} to \frac{31}{8}i.

$$y=\frac{\frac{33}{8}+\frac{31}{8}i}{4}$$

Divide \frac{33}{8}+\frac{31}{8}i by 4.

$$y=\frac{33}{32}+\frac{31}{32}i$$

Now solve the equation y=\frac{\frac{33}{8}±\frac{31}{8}i}{4} when ± is minus. Subtract \frac{31}{8}i from \frac{33}{8}.

$$y=\frac{\frac{33}{8}-\frac{31}{8}i}{4}$$

Divide \frac{33}{8}-\frac{31}{8}i by 4.

$$y=\frac{33}{32}-\frac{31}{32}i$$

The equation is now solved.

$$y=\frac{33}{32}+\frac{31}{32}i y=\frac{33}{32}-\frac{31}{32}i$$

Substitute \frac{33}{32}+\frac{31}{32}i for y in the equation y+\frac{1}{16}=\sqrt{\frac{y}{4}-\left(y-2\right)^{2}}.

$$\frac{33}{32}+\frac{31}{32}i+\frac{1}{16}=\sqrt{\frac{\frac{33}{32}+\frac{31}{32}i}{4}-\left(\frac{33}{32}+\frac{31}{32}i-2\right)^{2}}$$

Simplify. The value y=\frac{33}{32}+\frac{31}{32}i satisfies the equation.

$$\frac{35}{32}+\frac{31}{32}i=\frac{35}{32}+\frac{31}{32}i$$

Substitute \frac{33}{32}-\frac{31}{32}i for y in the equation y+\frac{1}{16}=\sqrt{\frac{y}{4}-\left(y-2\right)^{2}}.

$$\frac{33}{32}-\frac{31}{32}i+\frac{1}{16}=\sqrt{\frac{\frac{33}{32}-\frac{31}{32}i}{4}-\left(\frac{33}{32}-\frac{31}{32}i-2\right)^{2}}$$

Simplify. The value y=\frac{33}{32}-\frac{31}{32}i satisfies the equation.

$$\frac{35}{32}-\frac{31}{32}i=\frac{35}{32}-\frac{31}{32}i$$

List all solutions of y+\frac{1}{16}=\sqrt{\frac{y}{4}-\left(y-2\right)^{2}}.

$$y=\frac{33}{32}+\frac{31}{32}i y=\frac{33}{32}-\frac{31}{32}i$$