Question

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

Explanation:

Step1: Isolate \(x^2\)

Divide both sides of \(9x^2 = 4\) by 9:
\(x^2=\frac{4}{9}\)

Step2: Take square roots

Take square root of both sides. Remember square root has two solutions (positive and negative):
\(x = \pm\sqrt{\frac{4}{9}}\)

Step3: Simplify the square root

Simplify \(\sqrt{\frac{4}{9}}\):
\(\sqrt{\frac{4}{9}}=\frac{\sqrt{4}}{\sqrt{9}}=\frac{2}{3}\)
Thus, \(x=\frac{2}{3}\) or \(x = -\frac{2}{3}\).

Answer:

B. \(x=\frac{2}{3}\) and \(x = -\frac{2}{3}\)