Solve {left(1/4xright)}^2+left(40-xright)^2=58

$${ \left( \frac{ 1 }{ 4 } x \right) }^{ 2 } + { \left(40-x \right) }^{ 2 } = 58$$

Solve for x (complex solution)

$$x=\frac{640+4\sqrt{614}i}{17}\approx 37.647058824+5.830358444i$$

$$x=\frac{-4\sqrt{614}i+640}{17}\approx 37.647058824-5.830358444i$$

Steps Using the Quadratic Formula

Expand \left(\frac{1}{4}x\right)^{2}.

$$\left(\frac{1}{4}\right)^{2}x^{2}+\left(40-x\right)^{2}=58$$

Calculate \frac{1}{4} to the power of 2 and get \frac{1}{16}.

$$\frac{1}{16}x^{2}+\left(40-x\right)^{2}=58$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(40-x\right)^{2}.

$$\frac{1}{16}x^{2}+1600-80x+x^{2}=58$$

Combine \frac{1}{16}x^{2} and x^{2} to get \frac{17}{16}x^{2}.

$$\frac{17}{16}x^{2}+1600-80x=58$$

Subtract 58 from both sides.

$$\frac{17}{16}x^{2}+1600-80x-58=0$$

Subtract 58 from 1600 to get 1542.

$$\frac{17}{16}x^{2}+1542-80x=0$$

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.

$$\frac{17}{16}x^{2}-80x+1542=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{17}{16} for a, -80 for b, and 1542 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times \frac{17}{16}\times 1542}}{2\times \frac{17}{16}}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times \frac{17}{16}\times 1542}}{2\times \frac{17}{16}}$$

Multiply -4 times \frac{17}{16}.

$$x=\frac{-\left(-80\right)±\sqrt{6400-\frac{17}{4}\times 1542}}{2\times \frac{17}{16}}$$

Multiply -\frac{17}{4} times 1542.

$$x=\frac{-\left(-80\right)±\sqrt{6400-\frac{13107}{2}}}{2\times \frac{17}{16}}$$

Add 6400 to -\frac{13107}{2}.

$$x=\frac{-\left(-80\right)±\sqrt{-\frac{307}{2}}}{2\times \frac{17}{16}}$$

Take the square root of -\frac{307}{2}.

$$x=\frac{-\left(-80\right)±\frac{\sqrt{614}i}{2}}{2\times \frac{17}{16}}$$

The opposite of -80 is 80.

$$x=\frac{80±\frac{\sqrt{614}i}{2}}{2\times \frac{17}{16}}$$

Multiply 2 times \frac{17}{16}.

$$x=\frac{80±\frac{\sqrt{614}i}{2}}{\frac{17}{8}}$$

Now solve the equation x=\frac{80±\frac{\sqrt{614}i}{2}}{\frac{17}{8}} when ± is plus. Add 80 to \frac{i\sqrt{614}}{2}.

$$x=\frac{\frac{\sqrt{614}i}{2}+80}{\frac{17}{8}}$$

Divide 80+\frac{i\sqrt{614}}{2} by \frac{17}{8} by multiplying 80+\frac{i\sqrt{614}}{2} by the reciprocal of \frac{17}{8}.

$$x=\frac{640+4\sqrt{614}i}{17}$$

Now solve the equation x=\frac{80±\frac{\sqrt{614}i}{2}}{\frac{17}{8}} when ± is minus. Subtract \frac{i\sqrt{614}}{2} from 80.

$$x=\frac{-\frac{\sqrt{614}i}{2}+80}{\frac{17}{8}}$$

Divide 80-\frac{i\sqrt{614}}{2} by \frac{17}{8} by multiplying 80-\frac{i\sqrt{614}}{2} by the reciprocal of \frac{17}{8}.

$$x=\frac{-4\sqrt{614}i+640}{17}$$

The equation is now solved.

$$x=\frac{640+4\sqrt{614}i}{17} x=\frac{-4\sqrt{614}i+640}{17}$$

Steps for Completing the Square

Expand \left(\frac{1}{4}x\right)^{2}.

$$\left(\frac{1}{4}\right)^{2}x^{2}+\left(40-x\right)^{2}=58$$

Calculate \frac{1}{4} to the power of 2 and get \frac{1}{16}.

$$\frac{1}{16}x^{2}+\left(40-x\right)^{2}=58$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(40-x\right)^{2}.

$$\frac{1}{16}x^{2}+1600-80x+x^{2}=58$$

Combine \frac{1}{16}x^{2} and x^{2} to get \frac{17}{16}x^{2}.

$$\frac{17}{16}x^{2}+1600-80x=58$$

Subtract 1600 from both sides.

$$\frac{17}{16}x^{2}-80x=58-1600$$

Subtract 1600 from 58 to get -1542.

$$\frac{17}{16}x^{2}-80x=-1542$$

Divide both sides of the equation by \frac{17}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{17}{16}x^{2}-80x}{\frac{17}{16}}=-\frac{1542}{\frac{17}{16}}$$

Dividing by \frac{17}{16} undoes the multiplication by \frac{17}{16}.

$$x^{2}+\left(-\frac{80}{\frac{17}{16}}\right)x=-\frac{1542}{\frac{17}{16}}$$

Divide -80 by \frac{17}{16} by multiplying -80 by the reciprocal of \frac{17}{16}.

$$x^{2}-\frac{1280}{17}x=-\frac{1542}{\frac{17}{16}}$$

Divide -1542 by \frac{17}{16} by multiplying -1542 by the reciprocal of \frac{17}{16}.

$$x^{2}-\frac{1280}{17}x=-\frac{24672}{17}$$

Divide -\frac{1280}{17}, the coefficient of the x term, by 2 to get -\frac{640}{17}. Then add the square of -\frac{640}{17} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1280}{17}x+\left(-\frac{640}{17}\right)^{2}=-\frac{24672}{17}+\left(-\frac{640}{17}\right)^{2}$$

Square -\frac{640}{17} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1280}{17}x+\frac{409600}{289}=-\frac{24672}{17}+\frac{409600}{289}$$

Add -\frac{24672}{17} to \frac{409600}{289} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1280}{17}x+\frac{409600}{289}=-\frac{9824}{289}$$

Factor x^{2}-\frac{1280}{17}x+\frac{409600}{289}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{640}{17}\right)^{2}=-\frac{9824}{289}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{640}{17}\right)^{2}}=\sqrt{-\frac{9824}{289}}$$

Simplify.

$$x-\frac{640}{17}=\frac{4\sqrt{614}i}{17} x-\frac{640}{17}=-\frac{4\sqrt{614}i}{17}$$

Add \frac{640}{17} to both sides of the equation.

$$x=\frac{640+4\sqrt{614}i}{17} x=\frac{-4\sqrt{614}i+640}{17}$$