Question

what are the solution(s) to the quadratic equation $9x^2 = 4$? $x = \frac{4}{9}$ and $x = -\frac{4}{9}$; $x = \frac{2}{3}$ and $x = -\frac{2}{3}$; $x = \frac{3}{2}$ and $x = -\frac{3}{2}$; no real solution

Explanation:

Step1: Isolate $x^2$ term

Divide both sides by 9:
$$x^2 = \frac{4}{9}$$

Step2: Take square root of both sides

Recall square roots have two solutions:
$$x = \pm\sqrt{\frac{4}{9}} = \pm\frac{2}{3}$$

Answer:

$x = \frac{2}{3}$ and $x = -\frac{2}{3}$