Solve x+sqrt{1-x^2}=2/3

$$x + \sqrt { 1 - x ^ { 2 } } = \frac { 2 } { 3 }$$

Solve for x

$$x=-\frac{\sqrt{14}}{6}+\frac{1}{3}\approx -0.290276231$$

Solution Steps

Subtract x from both sides of the equation.

$$\sqrt{1-x^{2}}=\frac{2}{3}-x$$

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{2}{3}-x\right)^{2}.

$$1-x^{2}=\frac{4}{9}-\frac{4}{3}x+x^{2}$$

Subtract \frac{4}{9} from both sides.

$$1-x^{2}-\frac{4}{9}=-\frac{4}{3}x+x^{2}$$

Subtract \frac{4}{9} from 1 to get \frac{5}{9}.

$$\frac{5}{9}-x^{2}=-\frac{4}{3}x+x^{2}$$

Add \frac{4}{3}x to both sides.

$$\frac{5}{9}-x^{2}+\frac{4}{3}x=x^{2}$$

Subtract x^{2} from both sides.

$$\frac{5}{9}-x^{2}+\frac{4}{3}x-x^{2}=0$$

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

$$\frac{5}{9}-2x^{2}+\frac{4}{3}x=0$$

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.

$$-2x^{2}+\frac{4}{3}x+\frac{5}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, \frac{4}{3} for b, and \frac{5}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{4}{3}±\sqrt{\left(\frac{4}{3}\right)^{2}-4\left(-2\right)\times \frac{5}{9}}}{2\left(-2\right)}$$

Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{4}{3}±\sqrt{\frac{16}{9}-4\left(-2\right)\times \frac{5}{9}}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\frac{4}{3}±\sqrt{\frac{16}{9}+8\times \frac{5}{9}}}{2\left(-2\right)}$$

Multiply 8 times \frac{5}{9}.

$$x=\frac{-\frac{4}{3}±\sqrt{\frac{16+40}{9}}}{2\left(-2\right)}$$

Add \frac{16}{9} to \frac{40}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{4}{3}±\sqrt{\frac{56}{9}}}{2\left(-2\right)}$$

Take the square root of \frac{56}{9}.

$$x=\frac{-\frac{4}{3}±\frac{2\sqrt{14}}{3}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-\frac{4}{3}±\frac{2\sqrt{14}}{3}}{-4}$$

Now solve the equation x=\frac{-\frac{4}{3}±\frac{2\sqrt{14}}{3}}{-4} when ± is plus. Add -\frac{4}{3} to \frac{2\sqrt{14}}{3}.

$$x=\frac{2\sqrt{14}-4}{-4\times 3}$$

Divide \frac{-4+2\sqrt{14}}{3} by -4.

$$x=-\frac{\sqrt{14}}{6}+\frac{1}{3}$$

Now solve the equation x=\frac{-\frac{4}{3}±\frac{2\sqrt{14}}{3}}{-4} when ± is minus. Subtract \frac{2\sqrt{14}}{3} from -\frac{4}{3}.

$$x=\frac{-2\sqrt{14}-4}{-4\times 3}$$

Divide \frac{-4-2\sqrt{14}}{3} by -4.

$$x=\frac{\sqrt{14}}{6}+\frac{1}{3}$$

The equation is now solved.

$$x=-\frac{\sqrt{14}}{6}+\frac{1}{3} x=\frac{\sqrt{14}}{6}+\frac{1}{3}$$

Substitute -\frac{\sqrt{14}}{6}+\frac{1}{3} for x in the equation x+\sqrt{1-x^{2}}=\frac{2}{3}.

$$-\frac{\sqrt{14}}{6}+\frac{1}{3}+\sqrt{1-\left(-\frac{\sqrt{14}}{6}+\frac{1}{3}\right)^{2}}=\frac{2}{3}$$

Simplify. The value x=-\frac{\sqrt{14}}{6}+\frac{1}{3} satisfies the equation.

$$\frac{2}{3}=\frac{2}{3}$$

Substitute \frac{\sqrt{14}}{6}+\frac{1}{3} for x in the equation x+\sqrt{1-x^{2}}=\frac{2}{3}.

$$\frac{\sqrt{14}}{6}+\frac{1}{3}+\sqrt{1-\left(\frac{\sqrt{14}}{6}+\frac{1}{3}\right)^{2}}=\frac{2}{3}$$

Simplify. The value x=\frac{\sqrt{14}}{6}+\frac{1}{3} does not satisfy the equation.

$$\frac{1}{3}\times 14^{\frac{1}{2}}=\frac{2}{3}$$

Equation \sqrt{1-x^{2}}=\frac{2}{3}-x has a unique solution.

$$x=-\frac{\sqrt{14}}{6}+\frac{1}{3}$$