Solve 81x^2=25

$$81 x ^ { 2 } = 25$$

Solve for x

$$x=\frac{5}{9}\approx 0.555555556$$

$$x=-\frac{5}{9}\approx -0.555555556$$

Steps Using Difference of Squares

Divide both sides by 81.

$$x^{2}=\frac{25}{81}$$

Subtract \frac{25}{81} from both sides.

$$x^{2}-\frac{25}{81}=0$$

Multiply both sides by 81.

$$81x^{2}-25=0$$

Consider 81x^{2}-25. Rewrite 81x^{2}-25 as \left(9x\right)^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(9x-5\right)\left(9x+5\right)=0$$

To find equation solutions, solve 9x-5=0 and 9x+5=0.

$$x=\frac{5}{9} x=-\frac{5}{9}$$

Steps by Finding Square Root

Divide both sides by 81.

$$x^{2}=\frac{25}{81}$$

Take the square root of both sides of the equation.

$$x=\frac{5}{9} x=-\frac{5}{9}$$

Steps Using the Quadratic Formula

Divide both sides by 81.

$$x^{2}=\frac{25}{81}$$

Subtract \frac{25}{81} from both sides.

$$x^{2}-\frac{25}{81}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{25}{81} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{25}{81}\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{25}{81}\right)}}{2}$$

Multiply -4 times -\frac{25}{81}.

$$x=\frac{0±\sqrt{\frac{100}{81}}}{2}$$

Take the square root of \frac{100}{81}.

$$x=\frac{0±\frac{10}{9}}{2}$$

Now solve the equation x=\frac{0±\frac{10}{9}}{2} when ± is plus.

$$x=\frac{5}{9}$$

Now solve the equation x=\frac{0±\frac{10}{9}}{2} when ± is minus.

$$x=-\frac{5}{9}$$

The equation is now solved.

$$x=\frac{5}{9} x=-\frac{5}{9}$$