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How do you solve # (x + 2)^2 = 16#?

Answer 1

Hazel Anglin

#x=2# or #x=-6#

Get rid of the square by square rooting both sides: #sqrt((x+2)^2)# = #± sqrt(16)#

Squares are eliminated by the square root:

#x+2# = #± sqrt(16)#

#± sqrt(16)=+4 or -4#

So you have to solve for both #+4# and #-4#

#x+2 = 4#

#x=2#

and

#x+2=-4#

#x=-6#

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Answer 2

Jameson Allen

To solve the equation (x + 2)^2 = 16, follow these steps:

Expand the left side of the equation: (x + 2)(x + 2) = x^2 + 4x + 4.
Set the expanded expression equal to 16: x^2 + 4x + 4 = 16.
Subtract 16 from both sides: x^2 + 4x + 4 - 16 = 0, which simplifies to x^2 + 4x - 12 = 0.
To solve the quadratic equation x^2 + 4x - 12 = 0, you can use factoring, completing the square, or the quadratic formula.
Factoring the quadratic expression: (x + 6)(x - 2) = 0.
Set each factor equal to zero and solve for x: x + 6 = 0 and x - 2 = 0.
Solve for x in each equation: x = -6 and x = 2.

Therefore, the solutions to the equation (x + 2)^2 = 16 are x = -6 and x = 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.