Show that all real roots of the polynomial $P (x) = x^5 − 10x + 35$ are negative.

Question

I got this problem out of Andreescu's Putnam and Beyond. I solved it differently from the given solution and could not understand the solution. Can you explain what is happening in the last step of the solution?

Because P (x) has odd degree, it has a real zero r. If r > 0, then by the AM–GM inequality

$P (r)$

$ = r^5 + 1 + 1 + 1 + 2^5 − 5 · 2 · r $

$≥ 0.$ (why?)

And the inequality is strict since $1 \neq 2$. Hence r < 0, as desired.

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Also, here is my own (edit: incorrect) solution:

Let p be a positive root. Then $p^5-10p+35=0$

$10p > p^5 + 35 \implies 10 > p^4 + 35/p > 2 \sqrt{35p^3} > 10p \implies p < 1$

$10p > 35 \implies p > 3.5$

Contradiction!

Answer 1

That should say $P(r) = r^5+1+1+1+2^5-5\cdot 2 \cdot r$.

The AM-GM inequality applied to $\{r^5,1,1,1,2^5\}$ gives us:

$\dfrac{r^5+1+1+1+2^5}{5} \ge \sqrt[5]{r^5 \cdot 1 \cdot 1 \cdot 1 \cdot 2^5}$

$\dfrac{r^5+35}{5} \ge 2r$

$r^5+35 \ge 10r$

$r^5-10r+35 \ge 0$

Equality only holds if $r^5 = 1 = 1 = 1 = 2^5$, which gives $r = 1 = 2$, which is impossible.

Thus, $r^5-10r+35 > 0$ for any number $r > 0$. Hence $x^5-10x+35$ cannot have a positive root.

One error in your solution is that if $p$ is a root, then $10p = p^5+35$, so you can't say $10p > p^5+35$. Also, $2\sqrt{35p^3} > 10p$ is only true for $p > \dfrac{5}{7}$

Answer 2

We have $f(x) = x^5 - 10x + 35 = x^5 - x^2 + (x - 5)^2 + 10 \geq x^2 (x^3 - 1) + 10$. Thus $f(x) \geq 10$ if $x\geq 1$ and $f(x)\geq 1(-1) + 10 = 9$ if $0\leq x\leq 1$.

Answer 3

Your proof that there are no positive roots can be easily fixed, keeping all of the same ideas.

Suppose that $10p=p^5+35$, with $p$ positive. Then $p\gt 3.5$.

However, if $10p=p^5+35$, then $$10=p^4+\frac{35}{p}\ge 2\sqrt{35p^3}.$$ It follows that $35p^3\le 25$, which contradicts $p\gt 3.5$.

Answer 4

Using calculus:

If we take the derivative of $P(x)=x^5-10x+35$, we get $\frac{dP}{dx}=5x^4-10=0$. This will give critical points of $x=\pm\sqrt[4]{2}$.

Notice that if $x > \sqrt[4]{2}$, then $\frac{dP}{dx} > 0$, so $P(x)$ will be increasing, hence always positive, at that interval.

At $x=\sqrt[4]{2}$, we have $P(x) > 0$, so the local minimum there is positive. (You may wish to verify this by evaluating $P(\sqrt[4]{2})$).

Also, $\frac{dP}{dx} < 0$ on the interval $-\sqrt[4]{2} < x < \sqrt[4]{2}$, so the slope of $P(x)$ will be negative on that interval. But the local minimum at $x=\sqrt[4]{2}$ is positive. Hence, $P(x)$ is positive as well on the interval $-\sqrt[4]{2} < x < \sqrt[4]{2}$.

Therefore, we can conclude for the interval $-\sqrt[4]{2} < x < \infty$ that $P(x)$ is always positive. Hence, $P(x)$ will never cross the $x$-axis for any positive value of $x$, which means $P(x)$ has no positive real roots.

However, $P(x)$ does indeed have a negative real root because $P(-2) > 0$ and $P(-3) < 0$, and $P(x)$ is increasing and differentiable on the interval $-3 < x < -2$ (because $\frac{dP}{dx}$ exists and is positive on that interval).

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For a visual illustration, here is a graph of $P(x)$ from WolframAlpha:

Answer 5

One way is to look at the two equations $$y = x^5 + 35 \tag{1}$$ $$y = 10x \tag{2}$$

Both are (not-strictly) increasing, their intersection determines the x-value of the zeroes.

For $0 \le x \le 2$, (1) has a range $35 \le y \le 2^5 + 35$ and (2) has a range of $0 \le y \le 20$ so there is no intersection.

For $2 < x$, (1) is always increasing faster than (2) and is larger at $x=2$ so there is no intersection there either.

So any root must be in $x < 0$.

Answer 6

To see what can be done without calculus, we can use a little bit from the "theory of equations" to show that $ \ x^5 \ - \ 10x \ - \ 35 \ $ has either five, three, or one real roots (complex conjugate zeroes) and that either two or none of those are positive and just one is negative (Descartes' Law of Signs).

If we just work with $ \ x^5 \ - \ 10x \ = \ x \ (x^2 \ - \ 10^{1/2} ) \ (x^2 \ + \ 10^{1/2} ) $ , we have an even function with three real zeroes. From what we know about the form of such a curve ("negative tail" goes to negative infinity, "positive tail" to positive infinity), between $ \ x \ = \ -(10^{1/4}) \ $ and $ \ x \ = \ 0 \ $ lies the local maximum for this function and the local minimum lies between $ \ x \ = \ 0 \ $ and $ \ x \ = \ +(10^{1/4}) \ $ This latter value of $ \ x \ $ is about $ \ 3.2^{1/2} \ \approx \ 1.8 \ $ (since $ \ \sqrt{10} \ \approx \ 3.2 \ $ and $ \ \sqrt{3.24} \ = \ 1.8 \ $ ) .

We can now estimate the minimal value of the function, using values of $ \ x \ $ between $ \ 1 \ $ and $ \ 1.8 \ $ , by noting that

$$ P \ (1) \ = \ 1^5 \ - \ 10 \ = \ -9 \ \ , $$

$$ P \ ( \ \frac{3}{2} \ ) \ = \ \left( \frac{3}{2} \right)^5 \ - \ 10 \ \left( \frac{3}{2} \right) \ = \ \left( \frac{243}{32} \right) \ - \ 15 \ \sim \ 8 \ - \ 15 \ \sim \ -7 \ \ , $$

$$ P \ ( \ \frac{4}{3} \ ) \ = \ \left( \frac{4}{3} \right)^5 \ - \ 10 \ \left( \frac{4}{3} \right) \ = \ \left( \frac{1024}{243} \right) \ - \ \frac{40}{3} \ \sim \ 4 \ - \ 13 \ \sim \ -9 \ \ , $$

where small integer ratios are chosen to keep the calculations simple. It is clear that the local minimal value is probably not significantly different from $ \ -10 \ $ (in fact, it's about $ \ -9.51 \ $ ) .

Shifting the curve for $ \ x^5 \ - \ 10x \ $ "vertically" by 35 units then guarantees that $ \ P \ (0) \ $ and the local minimum of $ \ P(x) \ $ are positive. So only the single negative $ \ x-$ intercept remains. (The other four zeroes are in complex conjugate pairs.)