[Math] Please help with absolute value $|x^2 – 3x| = 28$

absolute valuealgebra-precalculus

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! 🙁

Best Answer

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.

Related Question