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"Find the set of values of k for which |(x-4)(x+2)| = k has four solutions."

EDIT:

Ok so I thought I'd start with setting the modulus function equal to k and -k to get the two set of results.

Doing that I ended up with:

(x -4)(x +2) = k

and

(x - 4)(x +2) = -k .

After solving for x I got x = 4 or -2 OR (the other set of results) x = -4 and 2. -> not sure if my logic here is right, as I simply took the "negative" version of the function to get the second set of results...

I do not know how to progress with this question, the x-values don't really help me much here. Any help would be appreciated.

Posted this in the question so that it's easier to see :)

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  • $\begingroup$ Draw the function first. It is a parabola with zeros at $x=-2$ and $x=4$. You will have to flip over the lower part of the parabola with negative $y$ values. You will see for what values the parabola has four zeros $0<k<k_{max}$. $\endgroup$ – MrYouMath Jun 1 '16 at 16:35
  • $\begingroup$ I did draw the parabola, sadly it says that solutions done by accurate drawing will not be accepted! :/ Can I justify my drawings as the mathematics calculation? $\endgroup$ – Daniel F. Jun 1 '16 at 16:39
  • $\begingroup$ No don't use the drawing for your answer. It should give you an insight what value of $x$ you have to choose to find $k_{max}$. Look at the vertex of your parabola (wolframalpha.com/input/…) $\endgroup$ – MrYouMath Jun 1 '16 at 16:44
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Solving the equation $(x-4)(x+2)=k$ gives you $x_{1,2}=1\pm\sqrt{9+k}$; solving the equation $(x-4)(x+2)=-k$ gives you $x_{3,4}=1\pm\sqrt{9-k}$.

So in order to get four solutions, the following conditions have to be met:

  • $k>-9$ has to hold in order to get two solutions out of $x_{1,2}=1\pm\sqrt{9+k}$
  • $k<9$ has to hold in order to get two solutions out of $x_{3,4}=1\pm\sqrt{9-k}$
  • $k\neq 0$ has to hold, because otherwise $\{1\pm\sqrt{9+k}\}=\{1\pm\sqrt{9-k}\}$

Therefore the solution is (given $k$ is an integer): $k\in\{-8,-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,7,8\}$

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  • $\begingroup$ I do not see why you restricted your solution to the integers. That said, it is clear how to find all real values of $k$ that produce four solutions from your arguments. (+1). $\endgroup$ – N. F. Taussig Jun 2 '16 at 10:42

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