2
$\begingroup$

Determine how many real solutions has the following equation:

$$x^2(|x|-6)=-15$$

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!

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.