Please help with absolute value $|x^2 - 3x| = 28$

Question

Just a question about solving an absolute value equation:

$$|x^2 - 3x| = 28$$

Do I just solve this as if the absolute value brackets weren't even there?

$$x^2 - 3x - 28 = 0$$

$$(x+4)(x-7) = 0$$

So $x=-4$ ; $x=7$

But I'm still confused why the absolute value signs would be there in the first place :(

EDIT:

So, I've found that $x=7, x=4, x=-4$

Just not $100\%$ now if they are correct as I've had a look at a few online abs value calculators to check my answers and only $x=-4$ and $x=7$ come up as answers.

Am I correct?

EDIT 2

Ok, $x=4$ can't work out. I found it by: \begin{align*} x^2 - 3x & = 28\\ x^2 - 3x - 28 & = 0\\ (x-7)(x-4) & = 0 \end{align*} My answers here are $7$ and $4$.

So I'm lost as to why I got that answer! :(

Answer 1

Hint: you have to break it up into two cases - case 1: $x^2-3x\ge0$ and case 2: $x^2-3x<0$.

In case 1, your equation becomes $x^2-3x=28$ and in case 2, your equation becomes $3x-x^2=28$.

Answer 2

Whenever you have $$| \rm{something} | = \rm{number}$$ you should think of this as a shorthand way of listing two possibilities at once:

1. something = number, or
2. something = $-$number

So in this case, the equation you are given really includes two separate equations to solve:

1. $x^2-3x=28$, or
2. $x^2-3x=-28$

The first equation is the one you already solved by just ignoring the absolute value signs. The second equation is the one you have not considered yet.

Answer 3

This expression $|x^2 - 3x| = 28$ not has only 2 roots because if you remove mod it becomes $x^2 - 3x = 28$ and $x^2 - 3x = -28$

The first equation becomes $x^2 - 3x - 28=0$ or $x=7$ and $x=-4$ are the roots and second becomes $x^2 - 3x + 28=0$ and this second equation if you solve will give you 2 another.