Solve 18x^2-2=0

$$18 x ^ { 2 } - 2 = 0$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=-\frac{1}{3}\approx -0.333333333$$

Steps Using Difference of Squares

Divide both sides by 2.

$$9x^{2}-1=0$$

Consider 9x^{2}-1. Rewrite 9x^{2}-1 as \left(3x\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).

$$\left(3x-1\right)\left(3x+1\right)=0$$

To find equation solutions, solve 3x-1=0 and 3x+1=0.

$$x=\frac{1}{3} x=-\frac{1}{3}$$

Steps by Finding Square Root

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

$$18x^{2}=2$$

Divide both sides by 18.

$$x^{2}=\frac{2}{18}$$

Reduce the fraction \frac{2}{18} to lowest terms by extracting and canceling out 2.

$$x^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$x=\frac{1}{3} x=-\frac{1}{3}$$

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.

$$18x^{2}-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, 0 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 18\left(-2\right)}}{2\times 18}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 18\left(-2\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{0±\sqrt{-72\left(-2\right)}}{2\times 18}$$

Multiply -72 times -2.

$$x=\frac{0±\sqrt{144}}{2\times 18}$$

Take the square root of 144.

$$x=\frac{0±12}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{0±12}{36}$$

Now solve the equation x=\frac{0±12}{36} when ± is plus. Reduce the fraction \frac{12}{36} to lowest terms by extracting and canceling out 12.

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{0±12}{36} when ± is minus. Reduce the fraction \frac{-12}{36} to lowest terms by extracting and canceling out 12.

$$x=-\frac{1}{3}$$

The equation is now solved.

$$x=\frac{1}{3} x=-\frac{1}{3}$$