Solve 0=4/9x^2-4

$$0 = \frac { 4 } { 9 } x ^ { 2 } - 4$$

Solve for x

$$x=3$$

$$x=-3$$

Steps Using Difference of Squares

Swap sides so that all variable terms are on the left hand side.

$$\frac{4}{9}x^{2}-4=0$$

Divide both sides by \frac{4}{9}.

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

Consider x^{2}-9. Rewrite x^{2}-9 as x^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

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

$$x=3 x=-3$$

Steps by Finding Square Root

Swap sides so that all variable terms are on the left hand side.

$$\frac{4}{9}x^{2}-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$\frac{4}{9}x^{2}=4$$

Multiply both sides by \frac{9}{4}, the reciprocal of \frac{4}{9}.

$$x^{2}=4\times \frac{9}{4}$$

Multiply 4 and \frac{9}{4} to get 9.

$$x^{2}=9$$

Take the square root of both sides of the equation.

$$x=3 x=-3$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$\frac{4}{9}x^{2}-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{4}{9} for a, 0 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times \frac{4}{9}\left(-4\right)}}{2\times \frac{4}{9}}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times \frac{4}{9}\left(-4\right)}}{2\times \frac{4}{9}}$$

Multiply -4 times \frac{4}{9}.

$$x=\frac{0±\sqrt{-\frac{16}{9}\left(-4\right)}}{2\times \frac{4}{9}}$$

Multiply -\frac{16}{9} times -4.

$$x=\frac{0±\sqrt{\frac{64}{9}}}{2\times \frac{4}{9}}$$

Take the square root of \frac{64}{9}.

$$x=\frac{0±\frac{8}{3}}{2\times \frac{4}{9}}$$

Multiply 2 times \frac{4}{9}.

$$x=\frac{0±\frac{8}{3}}{\frac{8}{9}}$$

Now solve the equation x=\frac{0±\frac{8}{3}}{\frac{8}{9}} when ± is plus.

$$x=3$$

Now solve the equation x=\frac{0±\frac{8}{3}}{\frac{8}{9}} when ± is minus.

$$x=-3$$

The equation is now solved.

$$x=3 x=-3$$