Solve left(8xright)^2+2x=0

$${ \left(8x \right) }^{ 2 } +2x=0$$

Solve for x

$$x=-\frac{1}{32}=-0.03125$$

$$x=0$$

Steps Using Factoring

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

$$8^{2}x^{2}+2x=0$$

Calculate 8 to the power of 2 and get 64.

$$64x^{2}+2x=0$$

Factor out x.

$$x\left(64x+2\right)=0$$

To find equation solutions, solve x=0 and 64x+2=0.

$$x=0 x=-\frac{1}{32}$$

Steps Using the Quadratic Formula

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

$$8^{2}x^{2}+2x=0$$

Calculate 8 to the power of 2 and get 64.

$$64x^{2}+2x=0$$

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

$$x=\frac{-2±\sqrt{2^{2}}}{2\times 64}$$

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\times 64}$$

Multiply 2 times 64.

$$x=\frac{-2±2}{128}$$

Now solve the equation x=\frac{-2±2}{128} when ± is plus. Add -2 to 2.

$$x=\frac{0}{128}$$

Divide 0 by 128.

$$x=0$$

Now solve the equation x=\frac{-2±2}{128} when ± is minus. Subtract 2 from -2.

$$x=-\frac{4}{128}$$

Reduce the fraction \frac{-4}{128} to lowest terms by extracting and canceling out 4.

$$x=-\frac{1}{32}$$

The equation is now solved.

$$x=0 x=-\frac{1}{32}$$

Steps for Completing the Square

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

$$8^{2}x^{2}+2x=0$$

Calculate 8 to the power of 2 and get 64.

$$64x^{2}+2x=0$$

Divide both sides by 64.

$$\frac{64x^{2}+2x}{64}=\frac{0}{64}$$

Dividing by 64 undoes the multiplication by 64.

$$x^{2}+\frac{2}{64}x=\frac{0}{64}$$

Reduce the fraction \frac{2}{64} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{32}x=\frac{0}{64}$$

Divide 0 by 64.

$$x^{2}+\frac{1}{32}x=0$$

Divide \frac{1}{32}, the coefficient of the x term, by 2 to get \frac{1}{64}. Then add the square of \frac{1}{64} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{32}x+\left(\frac{1}{64}\right)^{2}=\left(\frac{1}{64}\right)^{2}$$

Square \frac{1}{64} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{32}x+\frac{1}{4096}=\frac{1}{4096}$$

Factor x^{2}+\frac{1}{32}x+\frac{1}{4096}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{64}\right)^{2}=\frac{1}{4096}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{64}\right)^{2}}=\sqrt{\frac{1}{4096}}$$

Simplify.

$$x+\frac{1}{64}=\frac{1}{64} x+\frac{1}{64}=-\frac{1}{64}$$

Subtract \frac{1}{64} from both sides of the equation.

$$x=0 x=-\frac{1}{32}$$