Solve y=(x/4)^2+(64-x/4)^2

$$y = ( \frac { x } { 4 } ) ^ { 2 } + ( \frac { 64 - x } { 4 } ) ^ { 2 }$$

Solve for y

$$y=\frac{x^{2}-64x+2048}{8}$$

Short Solution Steps

To raise \frac{x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$y=\frac{x^{2}}{4^{2}}+\left(\frac{64-x}{4}\right)^{2}$$

To raise \frac{64-x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$y=\frac{x^{2}}{4^{2}}+\frac{\left(64-x\right)^{2}}{4^{2}}$$

Since \frac{x^{2}}{4^{2}} and \frac{\left(64-x\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$y=\frac{x^{2}+\left(64-x\right)^{2}}{4^{2}}$$

Do the multiplications in x^{2}+\left(64-x\right)^{2}.

$$y=\frac{x^{2}+4096-128x+x^{2}}{4^{2}}$$

Combine like terms in x^{2}+4096-128x+x^{2}.

$$y=\frac{2x^{2}+4096-128x}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$y=\frac{2x^{2}+4096-128x}{16}$$

Divide each term of 2x^{2}+4096-128x by 16 to get \frac{1}{8}x^{2}+256-8x.

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

Solve for x (complex solution)

$$x=-2\sqrt{2y-256}+32$$

$$x=2\sqrt{2y-256}+32$$

Solve for x

$$x=-2\sqrt{2y-256}+32$$

$$x=2\sqrt{2y-256}+32\text{, }y\geq 128$$