Question

someone help me please :c

Solve the following quadratic equation: (x + 3)^2 = 16.


A. x = -3, -4
B. x = 1
C. x = -1, 7
D. x = 1, -7

Answer 1

(x + 3)^2 = 16
x + 3 = ± 4
x = ± 4 - 3

x = -7
x = 1

Answer 2

thank u!

Answer 3

Do you understand the steps?

Answer 4

not at all. im trying to understand bt it just doesn't click...

Answer 5

You start with (x + 3)^2 = 16.
Then you take the square root of both sides:
\[\sqrt{(x + 3)^2} = \sqrt{16}\]

Remember that square root and square are inverses of each other so they cancel
\[x + 3 = \pm 4\]

The plus or minus sign results from the fact that the square root of 16 is 4 and -4 because \((-4)^2 = 16\) and \(4^2 = 16\)

So now you split \(x + 3 = ±4\) into two equations:

\(x + 3 = 4\)
\(x + 3 = -4\)

From there you solve for x