# To find number of real roots

Consider the equation $$x^5-5x=c$$ where c is a real number.

Determine all c such that this equation has exactly 3 real roots.

I know that between consecutive real roots of $$f$$ there is a real root of $$f'$$. Now $$f'$$ in this case is $$5x^4-5$$ which always has two real roots. So the claim should be true for all c.

But I KNOW IT IS NOT TRUE. Where am I messing up?

• Note that while it is necessary for $f'$ to have two real roots (in order for $f$ to have exactly three real roots), it is not sufficient. Of course an odd degree polynomial will always have one real root. – hardmath Dec 12 '19 at 16:12
• ok so how do I find all such c? – Angry_Math_Person Dec 12 '19 at 16:14
• Consider the graph of $p(x) = x^5 - 5x$. Changing the constant $c$ in your equation amounts to moving a horizontal line up or down across this graph. – hardmath Dec 12 '19 at 16:15

Yes, between any two roots of $$f$$, there is a root of $$f'$$. However, just because $$f'$$ has a root, that doesn't mean that $$f$$ has a root on either side. Consider $$f(x)=x^2+1$$.

As for solving this problem, the derivative has only two roots, so we can at most have three roots. For some values of $$c$$ we have three roots, for some values of $$c$$ we have a single root, and for exactly two values of $$c$$, there are two roots. The three-root region is exactly the interval between the two two-root values of $$c$$.

And finding the values of $$c$$ that gives two roots is easier than one might think. They happen exactly when one root of $$f$$ coincides with a root of $$f'$$. So find the roots of $$f'$$, and find the values of $$c$$ that make each of them a root of $$f$$, and you have found the interval of $$c$$-values that gives three roots.

As you know, the given equation has extrema at $$x=\pm1$$. These correspond to values of the polynomial

$$1-5-c$$ and $$1+5-c$$ (the RHS was moved to the left).

Hence the polynomial will grow from $$-\infty$$, reach the maximum, then the minimum and continue growing to $$\infty$$. There are three roots when $$0$$ is in the range $$(-4-c,6-c)$$.

Where am I messing up? Just look at a graph where it fails.

$$x^5-5 x -5$$

I know that between consecutive real roots of f there is a real root of f′. Now f′ in this case is 5x4−5 which always has two real roots. So the claim should be true for all c.

$$A \implies B$$ does not mean $$B \implies A$$.

The two real roots of $$5x^4 - 5$$ are the two roots at $$x = \pm 1$$.

If $$x^5 - 5x=c$$ has three roots then they will be at $$x < -1; -1 < x < 1;$$ and at $$x > 1$$ by your condition.

But there won't be three real roots if there is no root for any $$x< -1$$, or no root between $$-1$$ and $$1$$, or no root for any $$x < -1$$.

$$x=\pm 1$$ are extreme points and if one, the max, is $$>0$$ and the other $$<0$$ then there will be three real roots. But if both are "on the same side of $$0$$" there is no root between them and no root to "the other side".

$$x^5-5x -c|_{-1} = 4-c$$ and $$x^5 - 5x -c|1 = -4-c$$ so $$x =-1$$ is a max and $$x = 1$$ is a min.

If $$f(-1) = 4-c \le 0$$ is a max there will be no root for $$x < -1$$ or for $$-1 < x \le 1$$. If $$f(1) = -4-c\ge 0$$ is a min there will be no root for $$x > 1$$ or for $$-1 \le x < 1$$. So if either $$c \ge 4$$ or if $$c\le 4$$ then there are fewer than three real roots. But if $$-4 < c < 4$$ then there will be three.

Alternatively: we know what the shape of an odd polynomial $$x^5 -5x$$ looks like. It's that polynomial curve with a twisty bit in the middle, goes off to infinity as $$x \to \infty$$, goes to negative infinity as $$x \to -\infty$$. It has roots were the x-axis crosses it (or where it crosses the x-axis-- everything is relative). If we shift it up or shift it down by $$c$$ we can force the x-axis to avoid the twisty bits in the middle and have it have only one root. Or we can deliberately shift it so that the x-axis goes smack through the twisty bits and we have a maximum number of roots. So if $$c$$ is between the max and mins we maximize the number of roots and the x-axis goes through the twisty bits. If $$c$$ is beyond the max an mins we've shoved the twisty bits below or above the x-axis and there is only one root.