Solve (x/4)^2+(40-x/4)^2=58

$$( \frac { x } { 4 } ) ^ { 2 } + ( \frac { 40 - x } { 4 } ) ^ { 2 } = 58$$

Solve for x

$$x=12$$

$$x=28$$

Steps Using the Quadratic Formula

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

$$\frac{x^{2}}{4^{2}}+\left(\frac{40-x}{4}\right)^{2}=58$$

To raise \frac{40-x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{4^{2}}+\frac{\left(40-x\right)^{2}}{4^{2}}=58$$

Since \frac{x^{2}}{4^{2}} and \frac{\left(40-x\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}+\left(40-x\right)^{2}}{4^{2}}=58$$

Do the multiplications in x^{2}+\left(40-x\right)^{2}.

$$\frac{x^{2}+1600-80x+x^{2}}{4^{2}}=58$$

Combine like terms in x^{2}+1600-80x+x^{2}.

$$\frac{2x^{2}+1600-80x}{4^{2}}=58$$

Calculate 4 to the power of 2 and get 16.

$$\frac{2x^{2}+1600-80x}{16}=58$$

Divide each term of 2x^{2}+1600-80x by 16 to get \frac{1}{8}x^{2}+100-5x.

$$\frac{1}{8}x^{2}+100-5x=58$$

Subtract 58 from both sides.

$$\frac{1}{8}x^{2}+100-5x-58=0$$

Subtract 58 from 100 to get 42.

$$\frac{1}{8}x^{2}+42-5x=0$$

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.

$$\frac{1}{8}x^{2}-5x+42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{8} for a, -5 for b, and 42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times \frac{1}{8}\times 42}}{2\times \frac{1}{8}}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times \frac{1}{8}\times 42}}{2\times \frac{1}{8}}$$

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

$$x=\frac{-\left(-5\right)±\sqrt{25-\frac{1}{2}\times 42}}{2\times \frac{1}{8}}$$

Multiply -\frac{1}{2} times 42.

$$x=\frac{-\left(-5\right)±\sqrt{25-21}}{2\times \frac{1}{8}}$$

Add 25 to -21.

$$x=\frac{-\left(-5\right)±\sqrt{4}}{2\times \frac{1}{8}}$$

Take the square root of 4.

$$x=\frac{-\left(-5\right)±2}{2\times \frac{1}{8}}$$

The opposite of -5 is 5.

$$x=\frac{5±2}{2\times \frac{1}{8}}$$

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

$$x=\frac{5±2}{\frac{1}{4}}$$

Now solve the equation x=\frac{5±2}{\frac{1}{4}} when ± is plus. Add 5 to 2.

$$x=\frac{7}{\frac{1}{4}}$$

Divide 7 by \frac{1}{4} by multiplying 7 by the reciprocal of \frac{1}{4}.

$$x=28$$

Now solve the equation x=\frac{5±2}{\frac{1}{4}} when ± is minus. Subtract 2 from 5.

$$x=\frac{3}{\frac{1}{4}}$$

Divide 3 by \frac{1}{4} by multiplying 3 by the reciprocal of \frac{1}{4}.

$$x=12$$

The equation is now solved.

$$x=28 x=12$$

Steps for Completing the Square

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

$$\frac{x^{2}}{4^{2}}+\left(\frac{40-x}{4}\right)^{2}=58$$

To raise \frac{40-x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{4^{2}}+\frac{\left(40-x\right)^{2}}{4^{2}}=58$$

Since \frac{x^{2}}{4^{2}} and \frac{\left(40-x\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}+\left(40-x\right)^{2}}{4^{2}}=58$$

Do the multiplications in x^{2}+\left(40-x\right)^{2}.

$$\frac{x^{2}+1600-80x+x^{2}}{4^{2}}=58$$

Combine like terms in x^{2}+1600-80x+x^{2}.

$$\frac{2x^{2}+1600-80x}{4^{2}}=58$$

Calculate 4 to the power of 2 and get 16.

$$\frac{2x^{2}+1600-80x}{16}=58$$

Divide each term of 2x^{2}+1600-80x by 16 to get \frac{1}{8}x^{2}+100-5x.

$$\frac{1}{8}x^{2}+100-5x=58$$

Subtract 100 from both sides.

$$\frac{1}{8}x^{2}-5x=58-100$$

Subtract 100 from 58 to get -42.

$$\frac{1}{8}x^{2}-5x=-42$$

Multiply both sides by 8.

$$\frac{\frac{1}{8}x^{2}-5x}{\frac{1}{8}}=-\frac{42}{\frac{1}{8}}$$

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

$$x^{2}+\left(-\frac{5}{\frac{1}{8}}\right)x=-\frac{42}{\frac{1}{8}}$$

Divide -5 by \frac{1}{8} by multiplying -5 by the reciprocal of \frac{1}{8}.

$$x^{2}-40x=-\frac{42}{\frac{1}{8}}$$

Divide -42 by \frac{1}{8} by multiplying -42 by the reciprocal of \frac{1}{8}.

$$x^{2}-40x=-336$$

Divide -40, the coefficient of the x term, by 2 to get -20. Then add the square of -20 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-40x+\left(-20\right)^{2}=-336+\left(-20\right)^{2}$$

Square -20.

$$x^{2}-40x+400=-336+400$$

Add -336 to 400.

$$x^{2}-40x+400=64$$

Factor x^{2}-40x+400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-20\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-20\right)^{2}}=\sqrt{64}$$

Simplify.

$$x-20=8 x-20=-8$$

Add 20 to both sides of the equation.

$$x=28 x=12$$