# Problem Of The Week #392 Nov 15th, 2019

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#### anemone

##### MHB POTW Director
Staff member
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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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### anemone

##### MHB POTW Director
Staff member
Congratulations to the following members for their correct solution! 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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