Answer:
C) x = −5 and x = 1
Step-by-step explanation:
I took the test and I got it right
The solutions to the quadratic equation [tex]4(x + 2)^2 = 36[/tex] are as x = -5 and x = 1.
Suppose that we've a function y = f(x) such that f(x) is quadratic.
When y = 0, then the values of x for which f(x) = 0 is called solution of quadratic equation f(x) = 0
This solution gives values of x, and when we plot x and f(x), we'd see that the graph intersects the x-axis at its solution points.
We have the given quadratic equation
[tex]4(x + 2)^2 = 36[/tex]
solving for x
[tex]4(x + 2)^2 = 36\\\\(x + 2)^2 = 36/4\\\\(x + 2)^2 = 9\\\\(x + 2)^2 = 3^2\\\\(x + 2) = 3\\\\x = 1\\or \\\\x + 2 = -3\\\\x = -5[/tex]
Learn more about quadratic equations;
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