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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! :(

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  • $\begingroup$ You do realize that if the expression in the absolute value equals -28 that would also be a solution, right? $\endgroup$ – JB King Jul 2 '14 at 2:27
  • $\begingroup$ When you factored $x^2 - 3x - 28$, you should have obtained $(x - 7)(x \color{red}{+} 4)$, as you can verify by multiplying the factors. $\endgroup$ – N. F. Taussig May 6 '18 at 21:43
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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.

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  • $\begingroup$ OK, but I've just put an edit in the original question. I found x=7, x=-4 and x=7, x=4. The answer is 7 and -4, 4 can't work. Does that happen when you check with your initial problem and you just rule out the ones that don't work? Thank you $\endgroup$ – Dani Jul 2 '14 at 2:45
  • $\begingroup$ The incorrect solution $x=4$ is coming up because you're factoring wrong. $x^2-3x-28=0$ factors as $(x-7)(x+4)=0$, as you originally had it. In your edit you seem to be factoring it as $(x-7)(x-4)=0$, which is incorrect. And what are you doing about the second possibility, $x^2-3x=-28$? That should turn into $x^2-3x+28=0$. How are you factoring that? $\endgroup$ – mweiss Jul 2 '14 at 2:54
  • $\begingroup$ Contrary to what the other answerer has said, there should not be any need to check your solutions and throw out the ones that don't work; if you have solved the two cases correctly, any solutions that you find will work in the original problem. (Of course checking your answers is still useful as a means of catching algebra mistakes, like the one you have made here.) $\endgroup$ – mweiss Jul 2 '14 at 2:56
  • $\begingroup$ I factored x^2 - 3x = 20 as (x-7)(x+4)=0. Sorry my edit is a bit confusing I realise! I factored the x^2 - 3x = -28 as (x-7)(x-4). Because that would have gone to x^2 - 3x + 28. Thats clearly wrong!! Thats where the positive 4 came from. I can't factorise that expression.. This is the part that is confusing me! $\endgroup$ – Dani Jul 2 '14 at 3:06
  • $\begingroup$ If it can't be factored, there are two possibilities to consider: (1) Maybe it has some real solutions, but they can't be found by simple factoring; do you have any other methods (e.g. the quadratic formula) available? (2) On the other hand maybe it doesn't have any real solutions. Not every quadratic equation does. If that turns out to be the case, then the only solutions to the problem are the ones that come from considering the solutions to $x^2-3x=28$. Which means that in this problem (but not usually!) the absolute value bars don't actually change the answer. $\endgroup$ – mweiss Jul 2 '14 at 3:20
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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 $3x-x^2=28$.

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  • $\begingroup$ Thank you!! Can you please just explain why we have to break it down into >= 0 and < 0? Sorry I had a hard time understanding this section!! $\endgroup$ – Dani Jul 2 '14 at 2:22
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    $\begingroup$ @Dani. We use the definition of the absolute value: $|y|=y$ if $y\ge 0$, and $|y|=-y$ if $y<0$. In your case $y=x^2-3x$. By breaking it into two cases we can get rid of the pesky absolute value signs. $\endgroup$ – Peter Woolfitt Jul 2 '14 at 2:32
  • $\begingroup$ Great. I remember seeing that now!! Thank you for your help! $\endgroup$ – Dani Jul 2 '14 at 2:49
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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.

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