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Let $a$ and $b$ be distinct integers. If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then which of the following statements are true?

1) $c*(a+c)=-b$

2) $a+b=-1$

3) $a+b+c=0$

4) $c=0$

Update

I just tried to sub $c$ into both of the equations:

$c^2+cb+a=0$ and $c^2+ac+b=0$ which then gives us the equality

$c^2+cb+a=c^2+ac+b$

$ => cb+a=ac+b$

$=> b(c-1)=a(c-1)$

which gives me then a=b which is contradictory because the integers are supposed to be distinct.

Update #2 Ok it looks like 1) is true, 2) is true, 3) is true, and 4) is false .. right?

user2574069
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    Have you tried the simplest approach, for example substituting $;c;$ in both equations and seeing what happens?? – DonAntonio Apr 18 '14 at 22:21
  • That only works for multiple choice and I fail to see how that is good from a pedagogical point of view. – Christopher K Apr 18 '14 at 22:22
  • @ChrisK, whom are you addressing? My comment? Well, he has a multiple choice... – DonAntonio Apr 18 '14 at 22:22
  • Yes, he has multiple choice, but how often does one really have that luxury: high school contests, SAT, GRE and that's pretty much it. – Christopher K Apr 18 '14 at 22:31
  • Chris K , what exactly is wrong with DonAntonio's approach ? The problem asks us to evaluate the validity of a given set of sentences . How else would one take on this problem , other than just plugging in c ? – Victor Apr 18 '14 at 22:33
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    @ChrisK, I refer to the question asked as it is asked. I cannot begin guessing or assuming stuff, like "how ofter this or that". In this site I try to help people with his questions as much as I can, not to give them whole (or even partial) university courses. – DonAntonio Apr 18 '14 at 22:34
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    BTW, the question is either part of a much developed, complete question...or else it is a poorly worded and low-level one, as the correct option (1) is clear when one substitutes only in the second equation, and thus the first one is superfluous. – DonAntonio Apr 18 '14 at 22:36
  • Okay... fine. I retract my comment. Yes, indeed, more than one answer is valid. So, it would make sense to check on a case-by-case basis. – Christopher K Apr 18 '14 at 22:41
  • As for (4) it is clearly false, since then $a = b = 0$, but we already known that $a$ and $b$ are distinct. – Christopher K Apr 18 '14 at 22:42

3 Answers3

1

We know that $c^2+bc+a = 0$ and $c^2+ac+b = 0$. Subtract to get $c(b-a)-(b-a) = 0$ or $(b-a)(c-1) = 0$. This means that either $b = a$ or $c = 1$. If $c = 1$, then $1+a+b = c+a+b = 0$.

Sandeep Silwal
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Factor Theorem implies $x-c$ is a factor of $x^2+ax+b-(x^2+bx+a)$, and this is equal to $(a-b)(x-1)$, so $c=1$, let $r$ be the another root of $x^2+ax+b$ then $(x-1)(x-r)=x^2-(1+r)x+r=x^2+ax+b$, so $r=b$ and $a=-(1+r)=-1-b$. Therefore $a+b=-1$.

Ángel Mario Gallegos
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Express $$x^2+bx+a=(x-c)(x-d)$$ $$x^2+ax+b=(x-c)(x-f)$$ Matching the coefficients of the powers of $x$ for both equations, we have $$c+d=-b$$ $$c+f=-a$$ $$cd=a$$ $$cf=b$$ Solving for $d$ and $f$ in terms of $a$ and $b$ we have $$d=a,f=b$$ leading to $c$ having only one possible value, which is $1$.

Thus we have $$c+a=-b\Rightarrow a+b+c=0$$ as well as $$a+b+1=0\Rightarrow a+b = -1$$

Also, $$c(a+c)=a+1=-b$$

Thus, options (1), (2) and (3) are true.

Alijah Ahmed
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