How to find the absolute value.?

Question

First problem:
(x-5)^2=16
Answer is x=1 and x=9, but how do I get to this answer?

Answer 1

(x - 5)^2 = 16

To solve this, you have to take the square root of both sides. Whenever you take the square root of both sides, you _always_ have to include a "plus or minus" sign, so this becomes

x - 5 = +/- sqrt(16)

And the square root of 16 is 4, so

x - 5 = +/- 4

Move the -5 to the right hand side, and we get

x = 5 +/- 4

So our two solutions are

x = {5 + 4, 5 - 4}
x = {9, 1}

Answer 2

If you REALLY want to use the concept of "absolute value" (### --- see footnote) to solve this, here's how you do it:

If (x - 5)^2 = 16, the ABSOLUTE VALUE of (x - 5), which is written as
| x - 5 |, is 4 (the POSITIVE square root of 16).

So, | x - 5 | = 4.

There are TWO cases:

1.) (x - 5) is +ve; then (x - 5) = 4, so x = 9.

2.) (x - 5) is -ve; then (5 - x) = 4, so x = 1.

You now have the two solutions.

Live long and prosper.


### Footnote : I don't necessarily RECOMMEND using the absolute value function as THE way to solve this, but that IS what you appear to have wanted. (No-one else seems to have given your particular request to involve the "absolute value" serious consideration.)

Answer 3

don't see an absolute value, but if you're trying to solve for x

(x-5)^2 = 16
expand
x^2 - 10x + 25 = 16
bring 16 to other side
x^2 - 10x + 9 = 0
factor
(x-9)(x-1) = 0

x = 9 and x = 1

edit: Puggy's method is probably the better approach

Answer 4

take square root of both sides
x-5 = +/- 4

add 5 to both sides
x = 5 +/- 4

the two solutions are then
x= 5+4 = 9, x = 5 - 4 = 1