Question

question
solve the following quadratic equation for all values of $x$ in simplest form.
$3(x-4)^2=27$
answer attempt 1 out of 2
additional solution $\bigcirc$ no solution
$x=$

Explanation:

Step1: Divide both sides by 3

$\frac{3(x-4)^2}{3}=\frac{27}{3}$
$(x-4)^2=9$

Step2: Take square roots of both sides

$x-4=\pm\sqrt{9}$
$x-4=\pm3$

Step3: Solve for x (positive case)

$x=3+4$
$x=7$

Step4: Solve for x (negative case)

$x=-3+4$
$x=1$

Answer:

$x=7$ and $x=1$