Question

Absolute Equations - Is my textbook wrong?

3|x+2| -1 = 8

I did the problem along wit the book and we both go: x = 1 or x = -5

Then we checked it
3|1+2| -1 ? 8 For which we BOTH got 8
and
3|-5 + 2| -1 ? 8 However here they got 8 and I got -10

Am I wrong and if so how??

Answer 1

You want to show that your book is wrong?? :P

Answer 2

For, this you must be aware of MODULUS.. |..| this has some meaning..

Answer 3

\[|x| = |-x| = x\]

Answer 4

Whether the value inside the modulus is positive or negative, the final result after removing Modulus brackets, will be Positive..

Answer 5

For example: \(|3| = 3\) and also \(|-3| = 3\)

Answer 6

This I knoooow but this was the equation before the brackets came off 3|-3| -1 how can that possibly come out to positive 8 or even -8 it didn't even come out to negative 8 (which I already know is equal to positive 8) @waterineyes

Answer 7

I did not get you.. :(

Answer 8

See :

\[3 |-5 +2| -1 \implies 3|-3| - 1 \implies 3 \color{green}{(3)} -1 = ??\]

Answer 9

I did this:
3|-5 +2| -1
3|-3| -1
-9 -1
-10

So wait @waterineyes once I get to 3|-3|-1 the 3 becomes positive. Oooh wait I get it because the absolute value of -3 is 3 thank you!

Answer 10

yep.. :)

Answer 11

Use absolute value first and then multiply it with \(3\).. :)

Answer 12

\(|x|\) = \(\begin{cases} x, && x > 0 \\-x, && x < 0\end{cases}\)

Answer 13

So, whenever you deal with Mod, then you have to divide it into two parts, one greater than 0 and other less than 0.

Answer 14

\(|x+2|\) = \(\begin{cases} (x+2), && (x+2) > 0 \\-(x+2) , && (x+2) < 0\end{cases}\)

Answer 15

So, one time use (x+2) and solve the equation, you will get one value..

And next time : use -(x+2) and solve, you will get another value. :)

Answer 16

That's the best explanation thank you. :)

Answer 17

you are welcome dear.. :)