MHB Prove that A Real Root Exists in [-1, 1]

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The discussion centers on proving that the polynomial function $$f(x)=5tx^4+sx^3+3rx^2+qx+p$$ has a real root in the interval [-1, 1] under the condition that $r+t=-p$. By evaluating the function at the endpoints, $f(1)$ and $f(-1)$, and applying the Intermediate Value Theorem, the participants explore the implications of the product $f(1) \cdot f(-1) < 0$. The conversation also introduces an alternative polynomial, $$p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$, which has roots at both endpoints, suggesting the existence of an extremum in the interval. Overall, the discussion emphasizes various approaches to demonstrate the existence of a real root for the given polynomial.
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Given $$f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.
 
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Some ideas:

We rewrite the polynomial as
$$f(x)=5tx^4+sx^3+3rx^2+qx-r-t=0,$$
where we are setting it equal to zero. We evaluate $f(1)$ and $f(-1)$:
\begin{align*}
f(1)&=5t+s+3r+q-r-t=4t+2r+s+q\\
f(-1)&=5t-s+3r-q-r-t=4t+2r-s-q.
\end{align*}
Idea: if $f(1)\cdot f(-1)<0$, then by the Intermediate Value Theorem, we would have shown there is a root in $[-1,1]$. Now
$$f(1)\cdot f(-1)=(4t+2r)^{2}-(s+q)^{2}.$$
Not seeing where to go with this. There's nothing stopping $s=q=0$, with $4t+2r\not=0$, in which case I have not proved what I want to prove.
 
anemone said:
Given $$f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.

Consider the polynomial $$p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.

Then $$p(1)=p(-1)=0$$ and so $$p$$ has an extremum in $$(-1,1)$$, so $$p'(x)$$ has a root in $$(-1,1)$$ ...

.
 
Last edited:
Thanks to both, Ackbach and zzephod for participating...

zzephod said:
Consider the polynomial $$p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.

Then $$p(1)=p(-1)=0$$ and so $$p$$ has an extremum in $$[-1,1]$$, so $$p'(x)$$ has a root in $$[-1,1]$$ ...

.

WoW! What an elegant way to approach this problem! Well done, zzephod!(Clapping)

And there is another quite straightforward and beautiful way to tackle it as well...therefore I'll wait for the inputs from other members for now...(Sun)
 
Another method proposed by other to solve this challenge problem is by using the integration method:

Hint:
$$\int_{-1}^1 p(x) dx=0$$
 
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