I don't know where I went wrong

S

Square Root of Scute

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The problem is 2(x +4)^2 -16(x +4) +24 = 0.

My work:
2(x^2 +4^2) -16x -64 +24 = 0

2x^2 +8^2 -16x -64 +24 = 0

2x^2 -16x = -8^2 +64 -24

x^2(2 -4) = 64 +64 -24

x^2(-2) = 128 -24

x^2(-2) = 104

(x^2(-2)) / -2 = 104 / -2

x^2 = 52

x = 2 times the square root of 13 (phone has no radicand in the keyboard)

This is the wrong answer. The correct answer has been reveal to me to be both 2 and -2. I do not know how this absolute value was reached for the answer, any help clarifying what I've done incorrect would be appreciated.
 
2*4^2 is NOT 8^2.
Before you multiply two numbers you MUST know the numbers that you are multiply. One number is to the left of the times symbol and another is to the right of the times symbol.

In 2*4^2 you are multiply 2 and 4^2. You are NOT multiplying 2 and 4!
2 is 2 and 4^2 is 16 so you need to compute 2*16 which is 32

Just to point out another way to approach this problem you can let y=(x+4) and solve 2y^2 -16y +24 = 0
Whether you do it this way or the way you attempted it is best to divide by 2 as your first step
Once you solve y^2 -18y +12 = 0 for y, THEN set x+4 to whatever you think y can be and then solve for x.
 
You would probably have avoided your mistakes had you used a so-called u-substitution.

[MATH]2(x + 4)^2 - 16(x + 4) + 24 = 0.[/MATH]
[MATH]\text {Let } u = x + 4.[/MATH]
[MATH]\therefore 2u^2 - 16u + 24 = 0 \implies u^2 - 8u + 12 = 0 \implies[/MATH]
[MATH](u - 6)(u - 2) = 0 \implies u = 6 \text { or } u = 2 \implies[/MATH]
[MATH]x + 4 = 6 \text { or } x + 4 = 2 \implies x = 6 - 4 = 2 \text { or } x = 2 - 4 = -\ 2.[/MATH]
I venture a guess that you could have done that error free.

Jomo pointed out this approach, but he did not stress that it can reduce the risk of error.
 
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