Solve (x/4)^2+(80-x/4)^2=200

$$( \frac { x } { 4 } ) ^ { 2 } + ( \frac { 80 - x } { 4 } ) ^ { 2 } = 200$$

Solve for x

$$x=40$$

Steps Using the Quadratic Formula

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

$$\frac{x^{2}}{4^{2}}+\left(\frac{80-x}{4}\right)^{2}=200$$

To raise \frac{80-x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{4^{2}}+\frac{\left(80-x\right)^{2}}{4^{2}}=200$$

Since \frac{x^{2}}{4^{2}} and \frac{\left(80-x\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}+\left(80-x\right)^{2}}{4^{2}}=200$$

Do the multiplications in x^{2}+\left(80-x\right)^{2}.

$$\frac{x^{2}+6400-160x+x^{2}}{4^{2}}=200$$

Combine like terms in x^{2}+6400-160x+x^{2}.

$$\frac{2x^{2}+6400-160x}{4^{2}}=200$$

Calculate 4 to the power of 2 and get 16.

$$\frac{2x^{2}+6400-160x}{16}=200$$

Divide each term of 2x^{2}+6400-160x by 16 to get \frac{1}{8}x^{2}+400-10x.

$$\frac{1}{8}x^{2}+400-10x=200$$

Subtract 200 from both sides.

$$\frac{1}{8}x^{2}+400-10x-200=0$$

Subtract 200 from 400 to get 200.

$$\frac{1}{8}x^{2}+200-10x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\frac{1}{8}x^{2}-10x+200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{8} for a, -10 for b, and 200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times \frac{1}{8}\times 200}}{2\times \frac{1}{8}}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times \frac{1}{8}\times 200}}{2\times \frac{1}{8}}$$

Multiply -4 times \frac{1}{8}.

$$x=\frac{-\left(-10\right)±\sqrt{100-\frac{1}{2}\times 200}}{2\times \frac{1}{8}}$$

Multiply -\frac{1}{2} times 200.

$$x=\frac{-\left(-10\right)±\sqrt{100-100}}{2\times \frac{1}{8}}$$

Add 100 to -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2\times \frac{1}{8}}$$

Take the square root of 0.

$$x=-\frac{-10}{2\times \frac{1}{8}}$$

The opposite of -10 is 10.

$$x=\frac{10}{2\times \frac{1}{8}}$$

Multiply 2 times \frac{1}{8}.

$$x=\frac{10}{\frac{1}{4}}$$

Divide 10 by \frac{1}{4} by multiplying 10 by the reciprocal of \frac{1}{4}.

$$x=40$$

Steps for Completing the Square

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

$$\frac{x^{2}}{4^{2}}+\left(\frac{80-x}{4}\right)^{2}=200$$

To raise \frac{80-x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{4^{2}}+\frac{\left(80-x\right)^{2}}{4^{2}}=200$$

Since \frac{x^{2}}{4^{2}} and \frac{\left(80-x\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}+\left(80-x\right)^{2}}{4^{2}}=200$$

Do the multiplications in x^{2}+\left(80-x\right)^{2}.

$$\frac{x^{2}+6400-160x+x^{2}}{4^{2}}=200$$

Combine like terms in x^{2}+6400-160x+x^{2}.

$$\frac{2x^{2}+6400-160x}{4^{2}}=200$$

Calculate 4 to the power of 2 and get 16.

$$\frac{2x^{2}+6400-160x}{16}=200$$

Divide each term of 2x^{2}+6400-160x by 16 to get \frac{1}{8}x^{2}+400-10x.

$$\frac{1}{8}x^{2}+400-10x=200$$

Subtract 400 from both sides.

$$\frac{1}{8}x^{2}-10x=200-400$$

Subtract 400 from 200 to get -200.

$$\frac{1}{8}x^{2}-10x=-200$$

Multiply both sides by 8.

$$\frac{\frac{1}{8}x^{2}-10x}{\frac{1}{8}}=-\frac{200}{\frac{1}{8}}$$

Dividing by \frac{1}{8} undoes the multiplication by \frac{1}{8}.

$$x^{2}+\left(-\frac{10}{\frac{1}{8}}\right)x=-\frac{200}{\frac{1}{8}}$$

Divide -10 by \frac{1}{8} by multiplying -10 by the reciprocal of \frac{1}{8}.

$$x^{2}-80x=-\frac{200}{\frac{1}{8}}$$

Divide -200 by \frac{1}{8} by multiplying -200 by the reciprocal of \frac{1}{8}.

$$x^{2}-80x=-1600$$

Divide -80, the coefficient of the x term, by 2 to get -40. Then add the square of -40 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-80x+\left(-40\right)^{2}=-1600+\left(-40\right)^{2}$$

Square -40.

$$x^{2}-80x+1600=-1600+1600$$

Add -1600 to 1600.

$$x^{2}-80x+1600=0$$

Factor x^{2}-80x+1600. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-40\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-40\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-40=0 x-40=0$$

Add 40 to both sides of the equation.

$$x=40 x=40$$

The equation is now solved. Solutions are the same.

$$x=40$$