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The roots of the equation $x^2+3x-1=0$ are also the roots of quartic equation $x^4+ax^2+bx+c=0$. Find $a+b+4c$.

This problem is from yesterday's Bangladesh National Math Olympiad 2017. I tried this using Vieta Root Jumping but no luck. After the contest my friend laughed at me "One doesn't simply try a 10 point problem with Vieta Root Jumping".
How to solve this problem?

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  • $\begingroup$ Is division allowed? $\endgroup$ – lab bhattacharjee Feb 12 '17 at 8:56
  • $\begingroup$ @labbhattacharjee This is the exact same question... I dont know $\endgroup$ – Rezwan Arefin Feb 12 '17 at 8:57
  • $\begingroup$ Sean Bean would like your friend. :) $\endgroup$ – Deepak Feb 12 '17 at 9:23
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Note that if $x^2+3x-1=0$, then roots of the quadratic $x^2+3x-1$ are also roots of $$(x^2+3x-1)(x^2-px+q)$$ Which follows from polynomial long divison.

Since the coefficient of $x^3$ is $0$, we have that $p=3$. Now note that $$(x^2+3x-1)(x^2-3x+q)=x^4+(q-10)x^2+(3q+3)x-q$$ So the value of $a+b+4c=q-10+3q+3-4q=-7$. So the answer is $-7$ no matter the value of $q$.

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  • $\begingroup$ How I missed this :| :| My friend became Champion of the Champions of the Olympiad .. and I am out of any places :| :| :'( $\endgroup$ – Rezwan Arefin Feb 12 '17 at 9:01
  • $\begingroup$ @RezwanArefin Was there a restriction that $a,b,c$ are rational? $\endgroup$ – S.C.B. Feb 12 '17 at 9:02
  • $\begingroup$ nope.. I gave the exact same question $\endgroup$ – Rezwan Arefin Feb 12 '17 at 9:03
  • $\begingroup$ @RezwanArefin Then the problem is incorrect.... $\endgroup$ – S.C.B. Feb 12 '17 at 9:03
  • $\begingroup$ I guess so :( But how to find other possibilities? $\endgroup$ – Rezwan Arefin Feb 12 '17 at 9:04
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$$(x^2+3x-1)(x^2-3x+3)=x^4-7x^2+12x-3$$ has the same real roots as $x^2+3x-1$. In this case $a=-7,b=12,c=-3$ so $$a+b+4c=-7.$$

Moreover $$(x^2+3x-1)(x^2-3x+d)=x^4+(d-10)x^2+(3d+3)x-d$$ for $$d>\frac{9}{4}$$ has the same real roots as $x^2+3x-1$. In this case $a=d-10,b=3d+3,c=-d$ so $$a+b+4c=-7.$$

So if the statement of the problem says the same real roots then it seems to be that $a+b+4c=-7$ in all cases.

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  • $\begingroup$ Why do you need $d>\frac{9}{4}$? Note the fact that it did not say that the roots of the given quartic were equal to the roots of the quadratic. it merely said that all of the roots of the quadratic were roots of the quartic. $\endgroup$ – S.C.B. Feb 12 '17 at 9:23
  • $\begingroup$ @S.C.B. For those values of $d$ the discriminant is negative so it doesn't have any more roots. It says "are the roots.." so it shouldn't have any more roots. $\endgroup$ – Test123 Feb 12 '17 at 9:25
  • $\begingroup$ The fact that it says that it is the roots does not imply that those are the only roots. For example we do say that $x=1$ is a root of $x^2-3x+2$ . Or we do say that $x=1,2$ are roots of $$(x-1)(x-2)(x-3)(x-4)$$ We just don't say that it is all the roots. $\endgroup$ – S.C.B. Feb 12 '17 at 9:26
  • $\begingroup$ @Test123 No .. I think it can have roots other than those two $\endgroup$ – Rezwan Arefin Feb 12 '17 at 9:26
  • $\begingroup$ @Test123 A more clear statement "All roots of ..... are also roots of .... " but not "Roots of ..... also all roots of .... " .. Maybe you misunderstood :| $\endgroup$ – Rezwan Arefin Feb 12 '17 at 9:28
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For the equation $x^2+3x-1=0$, the sum of the roots is $-3$ and the product is $-1$. For the equation $x^4+ax^2+bx+c=0$, the sum of the roots is $0$ and the product is $c$. If the roots of the first equation are roots of the second, the remaining two roots have sum $3$ and product $-c$. Therefore, we have

$$x^4+ax^2+bx+c=(x^2+3x-1)(x^2-3x-c)$$

The coefficient of the $x^2$ term is $-c-9-1=-c-10$. The coefficient of $x$ is $-3c+3$. Therefore,

$$a+b+4c=-c-10-3c+3+4c=-7$$

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