what are the solutions to the quadratic equation 4(x + 2)^2 = 36\no x = -11 and x = 7\no x = -7 and x =…
what are the solutions to the quadratic equation 4(x + 2)^2 = 36\no x = -11 and x = 7\no x = -7 and x = 11\no x = -5 and x = 1\no x = -1 and x = 5
Answer
Explanation:
Step1: Divide both sides by 4
Divide the equation $4(x + 2)^2=36$ by 4. We get $(x + 2)^2=\frac{36}{4}=9$.
Step2: Take square - root of both sides
Taking the square - root of both sides of $(x + 2)^2 = 9$, we have $x+2=\pm\sqrt{9}=\pm3$.
Step3: Solve for x in two cases
Case 1: When $x + 2=3$, subtract 2 from both sides: $x=3 - 2=1$. Case 2: When $x + 2=-3$, subtract 2 from both sides: $x=-3 - 2=-5$.
Answer:
$x=-5$ and $x = 1$