Given $x^4+x^3+5x^2-x+M=0$ , find $M$ such that $X_1 +X_2=X_3\cdot X_4$.

Given the polynomial equation $x^4+x^3+5x^2-x+M=0$, I am looking for the value of $M$ such that $X_1 +X_2=X_3\cdot X_4$, where $X_1,X_2,X_3,X_4$ are the complex roots.

I have tried doing it with Vieta's formulas: $$X_1+X_2+X_3+X_4 = -1$$ $$X_1X_2+X_1X_3+X_1X_4+X_2X_3+X_2X_4+X_3X_4=5$$ $$X_1X_2X_3+X_1X_2X_4+X_1X_3X_4+X_2X_3X_4=1$$ $$X_1X_2X_3X_4=M$$ and of course $$X_1+X_2=X_3X_4$$ but I can't solve it... How do I go on from here? Any tips ?

• Here's a tutorial in MathJax – Sahiba Arora Jun 8 '17 at 13:04
• What are the capital $X$s ? – onurcanbektas Jun 8 '17 at 13:20
• Would OPs PLEASE write their questions the way they want it to be AT ONCE! The original formulation had the condition $X1 +X2=X1*X2$ so I was hiding away to some quiet place to spend time on it (and I solved it), only to find later that the condition was changed meanwhile to $X_1 + X_2 = X_3 X_4$. Please do not waste my time. – Andreas Jun 8 '17 at 13:49
• @RobertZ: Your edit makes the question consistent, but I cannot see whether that is the originally intended version. (See comment by @Andreas) – ccorn Jun 8 '17 at 14:04
• @ccorn Yes, I hope that the OP will confirm my editing. Let's wait for a response form Simon Jachson – Robert Z Jun 8 '17 at 14:11

You are on the right track. Let $p=cd$ and $s=c+d$, then from the Vieta's formulas and the given equation $a+b=p$ we obtain $$p+s=-1,\quad\frac{M}{p}+p+sp=5,\quad \frac{Ms}{p}+p^2=1.$$ After eliminating $s$, we get $$\frac{M}{p}-p^2=5,\quad-\frac{M}{p}-M+p^2=1\implies M=-1-5=-6.$$
• Since this deduction is $\implies$ only, a verification step should be added. With $M=-6$ the polynomial is $(x^2+x+6)(x-1)(x+1)$ and indeed the sum of the quadratic roots equals $(+1)\cdot(-1)$. – ccorn Jun 8 '17 at 14:43
• Graphically, I found that $M = -6$ is correct. – Felix Marin Jun 9 '17 at 4:08