MHB How to Calculate \(a+b+c\) for Equations Sharing Common Real Roots?

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The problem involves finding the sum \(a+b+c\) for two pairs of quadratic equations that share common real roots. The first pair is \(x^2 + ax + 1 = 0\) and \(x^2 + bx + c = 0\), while the second pair is \(x^2 + x + a = 0\) and \(x^2 + cx + b = 0\). Participants are encouraged to follow the provided guidelines for submitting solutions. Members Greg and Opalg successfully provided correct answers to the problem. The discussion emphasizes the importance of understanding common roots in quadratic equations.
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Here is this week's POTW:

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The equations $x^2+ax+1=0$ and $x^2+bx+c=0$ have a common real root and the equations $x^2+x+a=0$ and $x^2+cx+b=0$ have a common real root. Find $a+b+c$.

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Congratulations to the following members for their correct solution!(Cool)

1. Greg
2. Opalg

Solution from Opalg:
The condition for the first equation to have a real root is $a^2-4\geqslant0$. The condition for the third equation to have a real root is $1-4a\geqslant0$. Those two conditions imply that $a\leqslant-2$.

Let $u$ be a common real root of the first two equations, and let $v$ be a common real root of the other two equations. Then $$(1)\qquad u^2 + au + 1 = 0,$$ $$(2)\qquad u^2 + bu + c = 0,$$ $$(3)\qquad v^2 + v + a = 0,$$ $$(4)\qquad v^2 + cv + b = 0.$$ Subtract (2) from (1) to get $(a-b)u + 1-c = 0$, so that $u = \dfrac{c-1}{a-b}$ (provided that $a-b\ne0$).

Now subtract (3) from (4) to get $(c-1)v + b-a = 0$, so that $v = \dfrac{a-b}{c-1}$ (provided that $c-1\ne0$). Thus $uv = 1$. But the product of the roots of (1) is $1$, and one of those roots is $u$. Therefore the other root must be $v$, so that $$(5)\qquad v^2 + av + 1 = 0.$$ Subtract (3) from (5) to get $(a-1)v = a-1$. But $a-1\ne0$ (because $a\leqslant-2$), and so $v=1$. It follows from (3) that $a=-2$, and from (4) that $b+c=-1$. Therefore $\boxed {a+b+c=-3}$.

It remains to consider what happens when $a-b=0$ or $c-1=0$. If $a=b$ then the first two equations have no common root unless $c=1$, in which case they are the same equation and it hardly makes sense for them to have a common root. Similarly, if $c=1$ then the other pair of equations have no common root unless $a=b$, in which case they are the same equation and again it hardly makes sense for them to have a common root. But if you do allow this situation to satisfy the conditions of the problem, then there is a solution whenever $a=b\leqslant-2$ and $c=1$. In that case, $a+b+c = 2a-1$, which needs not be equal to $-3$.
 
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