Solve 8q^2-2=0

$$8 q ^ { 2 } - 2 = 0$$

Solve for q

$$q=\frac{1}{2}=0.5$$

$$q=-\frac{1}{2}=-0.5$$

Steps Using Difference of Squares

Divide both sides by 2.

$$4q^{2}-1=0$$

Consider 4q^{2}-1. Rewrite 4q^{2}-1 as \left(2q\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).

$$\left(2q-1\right)\left(2q+1\right)=0$$

To find equation solutions, solve 2q-1=0 and 2q+1=0.

$$q=\frac{1}{2} q=-\frac{1}{2}$$

Steps by Finding Square Root

Add 2 to both sides. Anything plus zero gives itself.

$$8q^{2}=2$$

Divide both sides by 8.

$$q^{2}=\frac{2}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$q^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$q=\frac{1}{2} q=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$8q^{2}-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 0 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{0±\sqrt{0^{2}-4\times 8\left(-2\right)}}{2\times 8}$$

Square 0.

$$q=\frac{0±\sqrt{-4\times 8\left(-2\right)}}{2\times 8}$$

Multiply -4 times 8.

$$q=\frac{0±\sqrt{-32\left(-2\right)}}{2\times 8}$$

Multiply -32 times -2.

$$q=\frac{0±\sqrt{64}}{2\times 8}$$

Take the square root of 64.

$$q=\frac{0±8}{2\times 8}$$

Multiply 2 times 8.

$$q=\frac{0±8}{16}$$

Now solve the equation q=\frac{0±8}{16} when ± is plus. Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$q=\frac{1}{2}$$

Now solve the equation q=\frac{0±8}{16} when ± is minus. Reduce the fraction \frac{-8}{16} to lowest terms by extracting and canceling out 8.

$$q=-\frac{1}{2}$$

The equation is now solved.

$$q=\frac{1}{2} q=-\frac{1}{2}$$