Solve sqrt{1-x^2/10}=1-x/3

$$\sqrt { 1 - \frac { x ^ { 2 } } { 10 } } = 1 - \frac { x } { 3 }$$

Solve for x

$$x=0$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{1-\frac{x^{2}}{10}}\right)^{2}=\left(1-\frac{x}{3}\right)^{2}$$

Calculate \sqrt{1-\frac{x^{2}}{10}} to the power of 2 and get 1-\frac{x^{2}}{10}.

$$1-\frac{x^{2}}{10}=\left(1-\frac{x}{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1-\frac{x}{3}\right)^{2}.

$$1-\frac{x^{2}}{10}=1+2\left(-\frac{x}{3}\right)+\left(-\frac{x}{3}\right)^{2}$$

Express 2\left(-\frac{x}{3}\right) as a single fraction.

$$1-\frac{x^{2}}{10}=1+\frac{-2x}{3}+\left(-\frac{x}{3}\right)^{2}$$

Calculate -\frac{x}{3} to the power of 2 and get \left(\frac{x}{3}\right)^{2}.

$$1-\frac{x^{2}}{10}=1+\frac{-2x}{3}+\left(\frac{x}{3}\right)^{2}$$

To raise \frac{x}{3} to a power, raise both numerator and denominator to the power and then divide.

$$1-\frac{x^{2}}{10}=1+\frac{-2x}{3}+\frac{x^{2}}{3^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3^{2}}{3^{2}}.

$$1-\frac{x^{2}}{10}=\frac{3^{2}}{3^{2}}+\frac{-2x}{3}+\frac{x^{2}}{3^{2}}$$

Since \frac{3^{2}}{3^{2}} and \frac{x^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

$$1-\frac{x^{2}}{10}=\frac{3^{2}+x^{2}}{3^{2}}+\frac{-2x}{3}$$

Combine like terms in 3^{2}+x^{2}.

$$1-\frac{x^{2}}{10}=\frac{9+x^{2}}{3^{2}}+\frac{-2x}{3}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3^{2} and 3 is 9. Multiply \frac{-2x}{3} times \frac{3}{3}.

$$1-\frac{x^{2}}{10}=\frac{9+x^{2}}{9}+\frac{3\left(-2\right)x}{9}$$

Since \frac{9+x^{2}}{9} and \frac{3\left(-2\right)x}{9} have the same denominator, add them by adding their numerators.

$$1-\frac{x^{2}}{10}=\frac{9+x^{2}+3\left(-2\right)x}{9}$$

Do the multiplications in 9+x^{2}+3\left(-2\right)x.

$$1-\frac{x^{2}}{10}=\frac{9+x^{2}-6x}{9}$$

Divide each term of 9+x^{2}-6x by 9 to get 1+\frac{1}{9}x^{2}-\frac{2}{3}x.

$$1-\frac{x^{2}}{10}=1+\frac{1}{9}x^{2}-\frac{2}{3}x$$

Multiply both sides of the equation by 90, the least common multiple of 10,9,3.

$$90-9x^{2}=90+10x^{2}-60x$$

Subtract 90 from both sides.

$$90-9x^{2}-90=10x^{2}-60x$$

Subtract 90 from 90 to get 0.

$$-9x^{2}=10x^{2}-60x$$

Subtract 10x^{2} from both sides.

$$-9x^{2}-10x^{2}=-60x$$

Combine -9x^{2} and -10x^{2} to get -19x^{2}.

$$-19x^{2}=-60x$$

Add 60x to both sides.

$$-19x^{2}+60x=0$$

Factor out x.

$$x\left(-19x+60\right)=0$$

To find equation solutions, solve x=0 and -19x+60=0.

$$x=0 x=\frac{60}{19}$$

Substitute 0 for x in the equation \sqrt{1-\frac{x^{2}}{10}}=1-\frac{x}{3}.

$$\sqrt{1-\frac{0^{2}}{10}}=1-\frac{0}{3}$$

Simplify. The value x=0 satisfies the equation.

$$1=1$$

Substitute \frac{60}{19} for x in the equation \sqrt{1-\frac{x^{2}}{10}}=1-\frac{x}{3}.

$$\sqrt{1-\frac{\left(\frac{60}{19}\right)^{2}}{10}}=1-\frac{\frac{60}{19}}{3}$$

Simplify. The value x=\frac{60}{19} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{19}=-\frac{1}{19}$$

Equation \sqrt{-\frac{x^{2}}{10}+1}=-\frac{x}{3}+1 has a unique solution.

$$x=0$$