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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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    Is division allowed? – 2017-02-12
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    @labbhattacharjee This is the exact same question... I dont know – 2017-02-12
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    Sean Bean would like your friend. :) – 2017-02-12

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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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    How I missed this :| :| My friend became Champion of the Champions of the Olympiad .. and I am out of any places :| :| :'( – 2017-02-12
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    @RezwanArefin Was there a restriction that $a,b,c$ are rational? – 2017-02-12
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    nope.. I gave the exact same question – 2017-02-12
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    @RezwanArefin Then the problem is incorrect.... – 2017-02-12
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    I guess so :( But how to find other possibilities? – 2017-02-12
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    But the solution must be unique :| This problem used in national Olympiad – 2017-02-12
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    @RezwanArefin The problem was wrong. Unfortunately. – 2017-02-12
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    @RezwanArefin If you multiply $x^2+3x-1$ by a polynomial of degree 2 without real roots such that coeff. of $x^3$ is 0 (i.e. take $x^2-3x+d$ for appropriate value of $d$) you get another possibility if you don't have any conditions on $a,b,c$ – 2017-02-12
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    @Test123 There is a restriction that the coefficient of $x^3$ is $0$. – 2017-02-12
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    Is there any chance that the value of $a+b+4c$ will remain the same? – 2017-02-12
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    @RezwanArefin No, if you see my edited answer you can see that is not true. – 2017-02-12
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    @S.C.B. Yes, but again it's possible to do that – 2017-02-12
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    @S.C.B. Yes, thank you, I removed that example. – 2017-02-12
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    @S.C.B. This is my friends solution... Could you please look at this? -http://i.imgur.com/lqVrgxh.jpg I think his proof is complete .... a+b+4c remains same – 2017-02-12
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    @RezwanArefin Oh, did it say the roots of the function, as in both roots? – 2017-02-12
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    @RezwanArefin So do you mean both roots? – 2017-02-12
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    @S.C.B. yes ... the problem says "roots of the ... " – 2017-02-12
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    @RezwanArefin I think it would have been more clearer if it had said all roots. However, since that is the case I will edit my answer. – 2017-02-12
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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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    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. – 2017-02-12
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    @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. – 2017-02-12
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    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. – 2017-02-12
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    @Test123 No .. I think it can have roots other than those two – 2017-02-12
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    @Test123 A more clear statement "All roots of ..... are also roots of .... " but not "Roots of ..... also all roots of .... " .. Maybe you misunderstood :| – 2017-02-12
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    @S.C.B. When saying "a root" or "the roots" is different. We need to write correct statements! "The roots" means all the roots. Also, assuming the problem is correctly statement, the set of polynomials that I consider is a subset so if $a+b+4c=fixed$ then my solution is in any case correct. Moreover, I posted this answer before the discussion you had in the comment section and before you edited your answer! – 2017-02-12
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    @Test123 There was only a $30$ second gap between your answer and my edit. I had not seen your answer. – 2017-02-12
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    @RezwanArefin The word "the" makes the difference. We need to be careful when writing a mathematical statement. – 2017-02-12
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    @Test123 I am not American, nor have I ever seen an American math textbook, so I do not know what "the roots" mean. However, could you show me a reference where it says that "the roots" means "all the roots"? – 2017-02-12
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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,

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