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The equations $x^2 + x + a = 0$ and $x^2 + ax+ 1 = 0$

a) Cannot have a common real root for any value of a b) have common real root for exactly one value of a c) have a common real root for exactly two values of a d) have a common real root for all three values of a

My attempt : I used to quadratic formula to find the roots of the first two equations and then equated any two of them , which after simplification lead me to solve a cube equation in a. Since only one of the roots of the given cubic equation gave a real value of x as an answer , the correct option was b)

I would, however, like to know if there is a more shorter/neater way of doing the same question .

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Taking the difference of the two equations we obtain $(x-1)(a-1)=0$. Suppose first that $a=1$. Then the two equations coincide and we have two solutions of $x^2+x+1$, namely $x_{1,2}=\frac{\pm \sqrt{-3}-1}{2}$. In the other case we have $x=1$, which gives $a=-2$. These are all possible solutions. Now we can answer all questions.

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