Determine how many real solutions has the following equation:


I noticed that $|x|-6$ should be negative because $x^2$ is always a positive value. Thus, $x\in(-6;6)$. I made a substitution: $|x|=t$, hence $x^2=t^2$, and got the equation:$$t^3-6t^2+15=0$$

Now the problem is I cannot find any solution for this equation. Hope you'll give me the right explanation for this exercise. Thank you very much!

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    $\begingroup$ try to study the variations of the function you have in the last equation. $\endgroup$ – Denis Apr 24 '14 at 10:20
  • $\begingroup$ Thank you, but what do you mean by this? $\endgroup$ – I.Gandakov Apr 24 '14 at 10:20
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    $\begingroup$ You don't need to find the solutions. Only their number. $\endgroup$ – evil999man Apr 24 '14 at 10:21
  • $\begingroup$ compute the derivative, and see where the derivative is zero, it will tell you when this function reaches minimum and maximum. $\endgroup$ – Denis Apr 24 '14 at 10:21
  • $\begingroup$ So using the Descarte's rule I see there are two variations of signs for $P(t)=t^3-6t^2+15$. So does this mean I have two positive solution or none? And if so then what should I do? I think that if there are two positive solutions for the equation above this means we have four solutions in general when substituting in $|x|=t$. $\endgroup$ – I.Gandakov Apr 24 '14 at 10:47

One way to answer this question is to draw a graph of $y = x^2(|x| - 6)$, then draw a horizontal line at $y = -15$. The curve and the line obviously intersect in four places. QED.

Finding the solutions is a bit trickier, but that is not the question that was asked.

If you don't want to rely on drawing the graph, you can prove the result using $f(t)$ and $f'(t)$. As you showed, $f(t) = t^3 - 6t^2 + 15$, which gives $f'(t) = 3t^2 - 12t$.

Solving for $f'(t) = 0$ yields $t = \{0, 4\}.$ Thus, $f(t)$ has extrema at $t =0$ and $t = 4$. We only care about $t>0$, so we ignore that one. Calculating $f'(1) = -9$ and $f'(5) = 15$, we see that $f(t)$ must be strictly decreasing for $0 < t < 4$ and strictly increasing for $t > 4$.

Some key evaluations: $$ f(0) = 15 $$ $$ f(4) = -17 $$ $$ f(6) = 15 $$

Since $f(t)$ is strictly decreasing between $0$ and $4$, and $f(0) > 0$ and $f(4) < 0$, $f(t)$ must cross zero exactly once in that region. Likewise, since $f(t)$ is strictly increasing for $t > 4$ and $f(4) < 0$ and $f(6) > 0$, $f(t)$ must cross zero exactly once in that region.

Thus, there are two positive solutions to the initial equation. By symmetry, there must be two negative solutions, as well.


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