Question

what are the solutions to the quadratic equation $3(x - 4)^2 = 75$?
$\bigcirc$ $x = -9$ and $x = 1$
$\bigcirc$ $x = -5$ and $x = 5$
$\bigcirc$ $x = -4$ and $x = 4$
$\bigcirc$ $x = -1$ and $x = 9$

Explanation:

Step1: Divide both sides by 3

To isolate the squared term, divide each side of the equation \( 3(x - 4)^2 = 75 \) by 3.
\( \frac{3(x - 4)^2}{3} = \frac{75}{3} \)
Simplifying gives \( (x - 4)^2 = 25 \).

Step2: Take square root of both sides

Take the square root of both sides. Remember that taking the square root of a number gives two solutions: positive and negative.
\( \sqrt{(x - 4)^2} = \pm\sqrt{25} \)
Simplifying, we get \( x - 4 = \pm5 \).

Step3: Solve for x (two cases)

Case 1: \( x - 4 = 5 \)

Add 4 to both sides:
\( x = 5 + 4 \)
\( x = 9 \).

Case 2: \( x - 4 = -5 \)

Add 4 to both sides:
\( x = -5 + 4 \)
\( x = -1 \).

Answer:

x = -1 and x = 9 (the fourth option: \( \boldsymbol{x = -1} \) and \( \boldsymbol{x = 9} \))