Solve (8x)^2=1

$$( 8 x ) ^ { 2 } = 1$$

Solve for x

$$x=\frac{1}{8}=0.125$$

$$x=-\frac{1}{8}=-0.125$$

Steps Using Difference of Squares

Expand \left(8x\right)^{2}.

$$8^{2}x^{2}=1$$

Calculate 8 to the power of 2 and get 64.

$$64x^{2}=1$$

Subtract 1 from both sides.

$$64x^{2}-1=0$$

Consider 64x^{2}-1. Rewrite 64x^{2}-1 as \left(8x\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).

$$\left(8x-1\right)\left(8x+1\right)=0$$

To find equation solutions, solve 8x-1=0 and 8x+1=0.

$$x=\frac{1}{8} x=-\frac{1}{8}$$

Steps by Finding Square Root

Expand \left(8x\right)^{2}.

$$8^{2}x^{2}=1$$

Calculate 8 to the power of 2 and get 64.

$$64x^{2}=1$$

Divide both sides by 64.

$$x^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$x=\frac{1}{8} x=-\frac{1}{8}$$

Steps Using the Quadratic Formula

Expand \left(8x\right)^{2}.

$$8^{2}x^{2}=1$$

Calculate 8 to the power of 2 and get 64.

$$64x^{2}=1$$

Subtract 1 from both sides.

$$64x^{2}-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 64 for a, 0 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 64\left(-1\right)}}{2\times 64}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 64\left(-1\right)}}{2\times 64}$$

Multiply -4 times 64.

$$x=\frac{0±\sqrt{-256\left(-1\right)}}{2\times 64}$$

Multiply -256 times -1.

$$x=\frac{0±\sqrt{256}}{2\times 64}$$

Take the square root of 256.

$$x=\frac{0±16}{2\times 64}$$

Multiply 2 times 64.

$$x=\frac{0±16}{128}$$

Now solve the equation x=\frac{0±16}{128} when ± is plus. Reduce the fraction \frac{16}{128} to lowest terms by extracting and canceling out 16.

$$x=\frac{1}{8}$$

Now solve the equation x=\frac{0±16}{128} when ± is minus. Reduce the fraction \frac{-16}{128} to lowest terms by extracting and canceling out 16.

$$x=-\frac{1}{8}$$

The equation is now solved.

$$x=\frac{1}{8} x=-\frac{1}{8}$$