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Let $a, b, c, d$ be real number such that polynomial $ax^2 + (b+c)x + (d+e)$ has real roots greater than $1$. Prove that polynomial $ax^4+bx^3+cx^2+dx+e$ has at least one real root.

Is my work correct ?

Let $r$ be real root of $ax^2+(c+b)x+(e+d)$, so $ar^2+cr+e=(br+d)(-1)$.

Let $P(x) = ax^4+bx^3+cx^2+dx+e$

so $P(\sqrt{r}) = ar^2+cr+e + br\sqrt{r}+d\sqrt{r}= (br+d)(\sqrt{r}-1)$

$P(-\sqrt{r}) = ar^2+cr+e - br\sqrt{r}-d\sqrt{r}= (br+d)(-\sqrt{r}-1)$

Since $\sqrt{r}>1$, so $P(\sqrt{r})>0>P(-\sqrt{r})$

By Intermediate value theorem, $P(x) = ax^4+bx^3+cx^2+dx+e$ has at least one real root.

user403160
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1 Answers1

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Assume the roots are $r_1,r_2$. Then: $$a(x-r_1)^2(x-r_2)^2=ax^2+(-ar_1-ar_2)x+ar_1r_2=0.$$ Hence the second equation: $$f(x)=ax^4-ar_1x^3-ar_2x^2+(ar_1r_2-e)x+e=0.$$ Note: $$f(r_1)=-er_1+e$$ $$f(0)=e$$

Now IVT is applicable.

farruhota
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  • Thank you, faruhota. Please show me on how to apply IVT. – user403160 Aug 06 '17 at 17:12
  • Consider two cases: $e>0$ and $e<0$. In each case $f(r_1)$ and $f(0)$ have opposite signs. – farruhota Aug 06 '17 at 17:20
  • Also remember $r_1>1$. And of course, if $e=0$, then $0$ is the root of $P(x)=0$. – farruhota Aug 06 '17 at 17:31
  • Thank you. I'm clear now. Could you please point out the error in my work ? – user403160 Aug 07 '17 at 10:46
  • if $r>1$ is real, what is $P(\sqrt{-r})$? Isn't it a complex number you are operating with? – farruhota Aug 07 '17 at 10:52
  • I've edited my work, $P(\sqrt{-r})$ has been changed to $P(-\sqrt{r})$. Is it correct now ? BTW, why do you have two roots of $r_1$ and two roots of $r_2$ ,e.g., the polynomial may have three roots of $r_1$ and one roots of $r_2$ ? – user403160 Aug 07 '17 at 16:19
  • Referring to the comments, $P(\sqrt{r})>0$ depends on the sign of $br+d$, like in my solution on $e$. The quadratic equation has two roots $r_1,r_2$ and I am using only one of them, namely $r_1$ snd $0$. For the polynomial $r_2$ is not used, though it could also be used instead of $r_1$. – farruhota Aug 07 '17 at 16:41
  • Thank you. If $br+d$ is positive, we have $P(\sqrt{r})>0>P(-\sqrt{r})$. And if $br+d$ is negative, we have $P(-\sqrt{r})>0>P(\sqrt{r})$. In either case, we can apply IVT. Am I right ? – user403160 Aug 08 '17 at 03:15
  • Right. Now it is complete and in a sense advantageous to mine because you handled the problem with one root. Good luck. – farruhota Aug 08 '17 at 03:42
  • I appreciate for your kind help, farruhota. – user403160 Aug 08 '17 at 03:47