Solve sqrt{1-x^2}+x=1/5

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

Solve for x

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

Solution Steps

Subtract x from both sides of the equation.

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

Square both sides of the equation.

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

Calculate \sqrt{1-x^{2}} to the power of 2 and get 1-x^{2}.

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

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

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

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

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

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

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

Add \frac{2}{5}x to both sides.

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

Subtract x^{2} from both sides.

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

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

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

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}.

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

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

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

Multiply -4 times -2.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

The equation is now solved.

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

Substitute -\frac{3}{5} for x in the equation \sqrt{1-x^{2}}+x=\frac{1}{5}.

$$\sqrt{1-\left(-\frac{3}{5}\right)^{2}}-\frac{3}{5}=\frac{1}{5}$$

Simplify. The value x=-\frac{3}{5} satisfies the equation.

$$\frac{1}{5}=\frac{1}{5}$$

Substitute \frac{4}{5} for x in the equation \sqrt{1-x^{2}}+x=\frac{1}{5}.

$$\sqrt{1-\left(\frac{4}{5}\right)^{2}}+\frac{4}{5}=\frac{1}{5}$$

Simplify. The value x=\frac{4}{5} does not satisfy the equation.

$$\frac{7}{5}=\frac{1}{5}$$

Equation \sqrt{1-x^{2}}=\frac{1}{5}-x has a unique solution.

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