2
$\begingroup$

Find all $a \in \mathbb R$ such that $3x^5 - 5x^3 + a = 0$ has exactly two different real roots.

My attempt: This equation has to have one double root and one single root. I have no idea how to move on from this.

$\endgroup$
  • 1
    $\begingroup$ The double root must be a root of the derivative as well. What could it be? $\endgroup$ – dxiv Apr 18 '18 at 23:19
3
$\begingroup$

Consider the graph of $f(x) = 3x^5 - 5x^3$. Then $f'(x) = 15x^4 - 15x^2 = 15x^2(x^2-1)$. Thus $f'(\pm 1) = 0$. Also $f''(x) = 60x^3-30x = 30x(2x^2-1).$ Thus $f''(\pm 1/\sqrt{2}) = 0.$

Thus $f(x)$ increases on $(-\infty, -1)$, $(1,\infty)$, and decreases on $(-1, 1)$. And $f(x)$ has a local maximum and minimum at $x = \pm 1$.

Therefore, $a = f(\pm 1) = 2$ or $-2$.

$\endgroup$
1
$\begingroup$

Hint The complex roots come in pair.

So if you are counting multiplicity, the number of real roots is always odd.

If you are not counting the multiplicity, the only way there are two real roots is if one of the rots is multiple. This is the case if and only if there exists an $r$ such that $$f(r)=0 \\ f'(r)=0$$

Note that the second equation is easy to solve.

$\endgroup$
1
$\begingroup$

Hint:

The derivative of $p_a(x)=3x^5 - 5x^3 + a$ is $p'_a(x)=15x^2(x^-1)$, so $p_a(x)$ is increasing on $(-\infty,-1]$ and on $[1,+\infty)$, decreasing on $[-1,1]$, so, supposing $a\ne 0$, we have these cases - $p_a(x)=0$ has a single real root if its minimum (attained at $1$) and its maximum (attained at $-1$) have the same sign; - $p_a(x)=0$ has three real roots if its minimum and maximum have different signs; - L $p_a(x)=0$ has three real roots if $1$ or $-1$ is a root.

The case $a=0$ is to be considered apart: the equation is then $$3x^5-5x^3=x^3(3x^2-5)=0,$$ and it has three real roots.

$\endgroup$
1
$\begingroup$

All the answers thus far involve taking the derivative. But the problem can be solved without using derivatives.

To have an even number of real roots the quintic equation gas to have a double root, meaning there is a squared factor. Thus

$3x^3-5x^3+a=(x-r)^2(bx^3+cx^2+dx+e)$

Then expanding the product and matching like degree terms gives:

Degree 5: $b=3$

Degree 4: $-2rb+c=-6r+c=0, c=6r$

Degree 3: $r^2b-2rc+d=3r^2-12r^2+c=-5, d=9r^2-5$

Degree 2: $r^2c-2rd+e=6r^3-18r^3+10+e=0, e=12r^3-10$

Degree 1: $r^2d-2re=-15r^4+15r^2=0$

We have rederived the usual derivarive equation up to a constant factor which is, of course, irrelevant. In general, when we require a polynomial function $P(x)$ to have a maximum or minimum at $(x_0,y_0)$, we impose the requirement that $x_0$ be a double root for $P(x)-y_0=0$, and without calculus an analysis like the one above gives the same equation as $P'(x)=0$ for $x_0$.

Then in this case, $r=0$ or $\pm 1$ whereupon, as in other answers, these values substituted for $x$ give $a=0$ or $\pm 2$. These must still be checked, because the squared factor is only a necessary condition; it turns out insufficient in the case of one root in the original problem.

$\endgroup$
0
$\begingroup$

$ f^{'}(x)=15x^2(x^2-1) $, $ f(x) $ has a local maximum at $ x=-1 $, and local minimum at $ x=1 $. To ensure $ f $ has exactly two real roots, one need to let $ f(-1)=0 $ or $ f(1)=0 $, which implies $ a=2 $ or $ a=-2 $.

$\endgroup$

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.