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I'm certain there must be a certain formula/theorem involved but I haven't learnt about it. Rather than simply being given the answer, could I be given short and concise useful tips for questions like these in the future? Much thanks in advance.

emacs drives me nuts
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Rae
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    What have you tried ? Where are you stuck ? MathJax tutorial and How to ask a good question ? – Surb Jul 08 '22 at 12:36
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    Hint : First solve $x^2+mx+1=x^2+x+m$. It will give you only one candidate to be the common root. Then you will be able to see on which condition on $m$ this candidate is a root of $x^2+x+m=0$. – TheSilverDoe Jul 08 '22 at 12:38
  • @Surb i've tried the quadratic formula but it seemed to get me nowhere – Rae Jul 08 '22 at 12:42
  • You looked at the hint and solved $x^2+mx+1 = x^2+x+m$? – aschepler Jul 08 '22 at 13:29
  • I think here you don't have to say m ≠ 1 , because if m =1 two quadratic equations are identical and both roots are common but in your problem you want to find m in order to have a common root. Therefore by problem itself you can say m can not be 1 even if you don't specifically mention it. – Janaka Rodrigo Jul 08 '22 at 13:30
  • Consider this as the system $x^2+mx+1=0$, $x^2+x+m=0$ with respect to $x$ and $m$. Subtracting gives $(m-1)x+(1-m)=0 \Rightarrow (m-1)(x-1)=0$. $m\neq 1$, then $x=1$, then $0=x^2+mx+1=1+m+1\Rightarrow m=-2$. – Ivan Kaznacheyeu Jul 11 '22 at 12:32

2 Answers2

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Let the common root be $x$ then

$ x^2 + m x + 1 = 0 $ and $ x^2 + x + m = 0 $

Hence, by subtracting

$ m x + 1 - (x + m) = 0 $

$ (m - 1) x + (1 - m) = 0 $

Since $ m \ne 1 $ then $ x = 1 $ is the common root.

Substituting $x = 1$ in either equation gives $ m = -2 $

Hosam Hajeer
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There are a set of formulas named Viête formulas: You can use those.

In this case, if $x_1$ and $x_2$ are the roots of the first equation and $x_1$ and $x_3$ the roots of the second one (note that $x_1 is the common solution of both equations).

For the first equation the Viête formulas are $x_1+x_2=-m$ and $x_1x_2=1$. For the second equations are $x_1+x_3=-1$ and $x_1x_3=m$.

In other words, you can solve this system of equations for $x_1$, $x_2$ and $x_3$.

Does this help? :D

James Garrett
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