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 the equation \(9x^2 = 4\) by 9.
\(x^2=\frac{4}{9}\)

Step2: Solve for \(x\)

Take the square root of both sides. Remember, when taking the square root of a positive number, we get two solutions: positive and negative.
\(x = \pm\sqrt{\frac{4}{9}}\)
Simplify the square root: \(\sqrt{\frac{4}{9}}=\frac{\sqrt{4}}{\sqrt{9}}=\frac{2}{3}\), so \(x=\frac{2}{3}\) or \(x = -\frac{2}{3}\).

Answer:

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