Solve (1/5-x)^2+x^2=1

$$( \frac { 1 } { 5 } - x ) ^ { 2 } + x ^ { 2 } = 1$$

Solve for x

$$x=-\frac{3}{5}=-0.6$$

$$x=\frac{4}{5}=0.8$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{5}-x\right)^{2}.

$$\frac{1}{25}-\frac{2}{5}x+x^{2}+x^{2}=1$$

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

$$\frac{1}{25}-\frac{2}{5}x+2x^{2}=1$$

Subtract 1 from both sides.

$$\frac{1}{25}-\frac{2}{5}x+2x^{2}-1=0$$

Subtract 1 from \frac{1}{25} to get -\frac{24}{25}.

$$-\frac{24}{25}-\frac{2}{5}x+2x^{2}=0$$

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.

$$2x^{2}-\frac{2}{5}x-\frac{24}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -\frac{2}{5} for b, and -\frac{24}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{2}{5}\right)±\sqrt{\left(-\frac{2}{5}\right)^{2}-4\times 2\left(-\frac{24}{25}\right)}}{2\times 2}$$

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{2}{5}\right)±\sqrt{\frac{4}{25}-4\times 2\left(-\frac{24}{25}\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-\frac{2}{5}\right)±\sqrt{\frac{4}{25}-8\left(-\frac{24}{25}\right)}}{2\times 2}$$

Multiply -8 times -\frac{24}{25}.

$$x=\frac{-\left(-\frac{2}{5}\right)±\sqrt{\frac{4+192}{25}}}{2\times 2}$$

Add \frac{4}{25} to \frac{192}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{2}{5}\right)±\sqrt{\frac{196}{25}}}{2\times 2}$$

Take the square root of \frac{196}{25}.

$$x=\frac{-\left(-\frac{2}{5}\right)±\frac{14}{5}}{2\times 2}$$

The opposite of -\frac{2}{5} is \frac{2}{5}.

$$x=\frac{\frac{2}{5}±\frac{14}{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{\frac{2}{5}±\frac{14}{5}}{4}$$

Now solve the equation x=\frac{\frac{2}{5}±\frac{14}{5}}{4} when ± is plus. Add \frac{2}{5} to \frac{14}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{16}{5}}{4}$$

Divide \frac{16}{5} by 4.

$$x=\frac{4}{5}$$

Now solve the equation x=\frac{\frac{2}{5}±\frac{14}{5}}{4} when ± is minus. Subtract \frac{14}{5} from \frac{2}{5} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{12}{5}}{4}$$

Divide -\frac{12}{5} by 4.

$$x=-\frac{3}{5}$$

The equation is now solved.

$$x=\frac{4}{5} x=-\frac{3}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{5}-x\right)^{2}.

$$\frac{1}{25}-\frac{2}{5}x+x^{2}+x^{2}=1$$

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

$$\frac{1}{25}-\frac{2}{5}x+2x^{2}=1$$

Subtract \frac{1}{25} from both sides.

$$-\frac{2}{5}x+2x^{2}=1-\frac{1}{25}$$

Subtract \frac{1}{25} from 1 to get \frac{24}{25}.

$$-\frac{2}{5}x+2x^{2}=\frac{24}{25}$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-\frac{2}{5}x=\frac{24}{25}$$

Divide both sides by 2.

$$\frac{2x^{2}-\frac{2}{5}x}{2}=\frac{\frac{24}{25}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{\frac{2}{5}}{2}\right)x=\frac{\frac{24}{25}}{2}$$

Divide -\frac{2}{5} by 2.

$$x^{2}-\frac{1}{5}x=\frac{\frac{24}{25}}{2}$$

Divide \frac{24}{25} by 2.

$$x^{2}-\frac{1}{5}x=\frac{12}{25}$$

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{12}{25}+\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{12}{25}+\frac{1}{100}$$

Add \frac{12}{25} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{49}{100}$$

Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{49}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{49}{100}}$$

Simplify.

$$x-\frac{1}{10}=\frac{7}{10} x-\frac{1}{10}=-\frac{7}{10}$$

Add \frac{1}{10} to both sides of the equation.

$$x=\frac{4}{5} x=-\frac{3}{5}$$