Solve x^2=81^-1

$$x ^ { 2 } = 81 ^ { - 1 }$$

Solve for x

$$x=\frac{1}{9}\approx 0.111111111$$

$$x=-\frac{1}{9}\approx -0.111111111$$

Steps Using Difference of Squares

Calculate 81 to the power of -1 and get \frac{1}{81}.

$$x^{2}=\frac{1}{81}$$

Subtract \frac{1}{81} from both sides.

$$x^{2}-\frac{1}{81}=0$$

Multiply both sides by 81.

$$81x^{2}-1=0$$

Consider 81x^{2}-1. Rewrite 81x^{2}-1 as \left(9x\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(9x-1\right)\left(9x+1\right)=0$$

To find equation solutions, solve 9x-1=0 and 9x+1=0.

$$x=\frac{1}{9} x=-\frac{1}{9}$$

Steps by Finding Square Root

Calculate 81 to the power of -1 and get \frac{1}{81}.

$$x^{2}=\frac{1}{81}$$

Take the square root of both sides of the equation.

$$x=\frac{1}{9} x=-\frac{1}{9}$$

Steps Using the Quadratic Formula

Calculate 81 to the power of -1 and get \frac{1}{81}.

$$x^{2}=\frac{1}{81}$$

Subtract \frac{1}{81} from both sides.

$$x^{2}-\frac{1}{81}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{1}{81} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{1}{81}\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{1}{81}\right)}}{2}$$

Multiply -4 times -\frac{1}{81}.

$$x=\frac{0±\sqrt{\frac{4}{81}}}{2}$$

Take the square root of \frac{4}{81}.

$$x=\frac{0±\frac{2}{9}}{2}$$

Now solve the equation x=\frac{0±\frac{2}{9}}{2} when ± is plus.

$$x=\frac{1}{9}$$

Now solve the equation x=\frac{0±\frac{2}{9}}{2} when ± is minus.

$$x=-\frac{1}{9}$$

The equation is now solved.

$$x=\frac{1}{9} x=-\frac{1}{9}$$