Question

gwen was asked to solve $3x^2 - 20 = 7$ by factoring using the difference of squares pattern. what is/are the correct solution(s)? (1 point)
$\bigcirc$ $x = -3, x = 3$
$\bigcirc$ $x = -3$
$\bigcirc$ $x = 3$
$\bigcirc$ $x = 9, x = -9$

Explanation:

Step1: Simplify the equation

First, we need to get the equation in the form \(a^2 - b^2 = 0\). Start with the given equation \(3x^2 - 20 = 7\). Add 20 to both sides:
\(3x^2 = 7 + 20\)
\(3x^2 = 27\)

Then divide both sides by 3:
\(x^2 = \frac{27}{3}\)
\(x^2 = 9\)

Step2: Rewrite as difference of squares

The equation \(x^2 = 9\) can be rewritten as \(x^2 - 9 = 0\). Notice that \(x^2 - 9\) is a difference of squares, since \(x^2 = (x)^2\) and \(9 = (3)^2\). The difference of squares formula is \(a^2 - b^2 = (a - b)(a + b)\), so:
\(x^2 - 9 = (x - 3)(x + 3) = 0\)

Step3: Solve for \(x\)

Set each factor equal to zero:

1. \(x - 3 = 0\) gives \(x = 3\)
2. \(x + 3 = 0\) gives \(x = -3\)

Answer:

\(x = -3, x = 3\) (corresponding to the first option: \(x = -3, x = 3\))