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I am working on a question which begins with

The number $\alpha$ is a common root of the equations $x^2+ax+b=0$ and $x^2+cx+d=0$. Given that $a\neq c$, show that $$\alpha=-\frac{b-d}{a-c}$$

This was easy enough to show algebraically by assuming that alpha does satisfy the above two equations. Am I correct in saying that, because I made this assumption, it is sufficient that the root alpha takes on the above form, but it is not necessary? This doesn't seem to fit with the question though, because that would imply that if alpha takes on that value, then the two equations automatically have a common root, however it is not necessary for alpha to take on this value for them to have a common root. In that case, I have not shown that alpha being a common root of the equations implies that alpha must take on this value, I have shown that if alpha does take. On this value, then the two equations have a common root... I'm sorry if this is very confusing! I am trying to explain myself as best as I can...

The second part of the question is

Hence, or otherwise, show that the equations have at least one common root if and only if $$(b-d)^2-a(b-d)(a-c)+b(a-c)^2=0$$

Since I am not sure whether the first statement for the value of alpha is necessary or sufficient, I am having trouble deciding how far I can deal with the second part of the question using the result from the first part...

Thank you in advance for any guidance. If anyone has a link to any particularly good resource that will help with using necessary/sufficient and if/iff I would be very grateful.

N. F. Taussig
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Meep
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1 Answers1

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Assume for now that $a\ne c$. Then you’ve shown that if if the two equations have a common root $\alpha$, then that common root must be

$$-\frac{b-d}{a-c}\;.\tag{1}$$

You’ve not shown, however, that the expression in $(1)$ always is a common root of the two equations. That is, you’ve shown that having $\alpha$ equal to $(1)$ is necessary, but not that it’s sufficient. And in fact it’s not sufficient. Consider the equations $x^2=0$ and $x^2+x+1=0$, with $a=b=0$ and $c=d=1$: then

$$-\frac{b-d}{a-c}=-\frac{0-1}{0-1}=-1\;,$$

but $-1$ isn’t a root of either equation.

To find a sufficient condition, substitute the expression $(1)$ for $x$ in each of the equations; in the first you get

$$\frac{(b-d)^2}{(a-c)^2}-\frac{a(b-d)}{a-c}+b$$

on the left-hand side, so in order for $(1)$ to be a root of $x^2+ax+b=0$, we must have

$$\frac{(b-d)^2}{(a-c)^2}-\frac{a(b-d)}{a-c}+b=0\;.\tag{2}$$

Similarly, from the second equation we find that we must have

$$\frac{(b-d)^2}{(a-c)^2}-\frac{c(b-d)}{a-c}+d=0\;.\tag{3}$$

Multiplying $(2)$ through by $(a-c)^2$ yields

$$(b-d)^2-a(b-d)(a-c)+b(a-c)^2=0\;,\tag{4}$$

which should look familiar. Doing the same thing to $(3)$ yields

$$(b-d)^2-c(b-d)(a-c)+d(a-c)^2=0\;;\tag{5}$$

with a bit of algebra you can show that $(4)$ and $(5)$ are equivalent, in the sense that each is true if and only if the other is true. (HINT: Solve them for $\frac{(b-d)^2}{a-c}$.) Thus, we need only consider $(4)$.

If $(4)$ holds, and $a\ne c$, then $(2)$ and $(3)$ hold, and $(1)$ is a root of the first equation. What does $(4)$ imply when $a=c$? Does it still ensure that the two equations have a common root?

Brian M. Scott
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  • Thank you for your reply! I have been thinking over this for some time now but I am still a bit confused. My first difficulty is with double headed arrows in the first part of the argument. – Meep Mar 31 '15 at 09:56
  • @21joanna12: The first part of the problem is just a one-way implication: if there is a common root it must be $(1)$. In your discussion you exactly inverted what you'd proved. You don't have to deal with a two-way implication until the second part of question. – Brian M. Scott Mar 31 '15 at 10:05
  • Apologies- I accidentally submitted my comment before I fully write out what I was having trouble with. I wrote: "Assume f(x) and g(x) (the two functions) have a common roots alpha $\leftrightarrow$ $\alpha^2+a\alpha+b=0$ and $\alpha^2+c\alpha+d$..." and the rest was just algebraic manippulation so it had double headed arrows all the way through. I thought that if you can use double headed arrows all the way through your argument, then what you have at each step is necessary and sufficient? – Meep Mar 31 '15 at 10:10
  • I am just unsure of how II used the double headed arrows wrong, because using them still allows me to go back up the argument from (1) to 'assume f(x) and g(x) have a common root alpha' which to me seems to say (1) is sufficient? – Meep Mar 31 '15 at 10:14
  • My second difficulty is with understanding why (4) is both necessary and sufficient for at least one common root alpha. I understand that if the common root alpha exists, it is necessary that it takes the form (1) and that is alpha exists and a is not equal to c, then (4) follows. But since (1) is necessary, but not sufficient, and I have used it to get (4), then it seems to me that (4) is also necessary, but not sufficient, for there to be at least one common root alpha. And II haven't shown any 'if' because I cannot say that (4) implies the common root in any way... – Meep Mar 31 '15 at 10:20
  • @21joanna12: The problem is that not all of your steps actually are reversible, so you have at least one arrow that should not be double-headed. Your entire calculation is contingent on the assumption that $\alpha$ really is a common root of the equations. As long as $a\ne c$, you can always calculate $(1)$, but there's no reason to expect that number to satisfy either equation. – Brian M. Scott Mar 31 '15 at 10:21
  • @21joanna12: Don't forget that $(4)$ implies $(2)$, which in turn implies that $(1)$ is a root of the first equation. I also sketched the argument that you need to show that $(4)$ implies $(3)$, making $(1)$ a root of the second equation as well. (I'll be offline for a few hours, but I'll certainly check for further questions when I get back.) – Brian M. Scott Mar 31 '15 at 10:25
  • I apologise for asking more questions and thank you for taking the time to answers them, but I was wondering which step is not reversible. From what I understand, all algebraic manipulation is reversible bar division (when I had to introduce the 'a not equal to c' constraint), so everything that followed after the "Assume f(x) and g(x) (the two functions) have a common roots alpha ↔ α2+aα+b=0 and α2+cα+d..." would be reversible provided the constraint on a and c. And I assumed from the beginning that alpha satisfies both equations because I substituted alpha into both equations... – Meep Mar 31 '15 at 10:28
  • @21joanna12: Yes, and it’s only that assumption that allows you to conclude that $-\frac{b-d}{a-c}$ satisfies both equations. As my example shows, if the equations do not have a common solution, then $-\frac{b-d}{a-c}$ need not satisfy either one of them. Your calculation is possible in the first place only because you assumed that there was a common solution. Remove that assumption, and the number $-\frac{b-d}{a-c}$ ceases to have any special significance. (I probably shouldn’t have said that one of your steps had to be irreversible; the real problem is that the whole calculation ... – Brian M. Scott Apr 01 '15 at 03:22
  • ... depends on the assumption that the equations have a common solution. You’ve shown only that if there is a solution, then it has to be a certain number. This says nothing about that number when the equations have no common solution.) – Brian M. Scott Apr 01 '15 at 03:23