A Wrong Solution
I found this cute problem in the Russian book Sharygin Geometry Olympiad by Zaslavsky, Protasov and Sharygin.
Find numbers p and q that satisfy the equation: x2 + px + q = 0.
The book asks you to find a mistake in the following solution:
By Viète’s formulae we get a system of equations p + q = − p, pq = q. Solving the system we get two solutions: p = q = 0 and p = 1, q = −2.
What is wrong with this solution?
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Joshua Zucker:
I think it’s a perfectly good solution to a somewhat different question.
My answer would be q = -x^2 – xp, pick any value of p you want.
The question they’re answering, I think, is when are p and q the two values of x that satisfy the equation, which is an amusing question!
18 May 2011, 9:36 pmRoman:
if p and q are solutions simultaneously then the solution correct. but if we are asked to find, for instance, all numbers p which satisfy the equation then the second root can be any number (not necessary equal q). the same for q.
let’s say that X1, X2 are two roots. if p is a solution then we have the following system of equations:
X1(p,q)=p, X2(p,q)=’any number’ or X2(p,q)=p, X1(p,q)=’any number’
probably i did not get the condition correctly or it is too simple.
Roman
P.S. the “wrong” solution corresponds to X1(p,q)=p, X2(p,q)=q
18 May 2011, 10:02 pmRoman:
i was wrong… he he. p and q are two solutions of the same equation, but there is one more possibility (except p=q=0 and p=1, q=-2). the missed solution is p=-1/2, q=-1/2.
simple and fun problem. thank you. Roman
18 May 2011, 11:06 pmJoseph:
The problem is that p and q might be equal, and the two solutions of the equation might be p (which equals q) and something else.
If p is a solution of x^2+px+p=0, then we have p^2+p^2+p=0. From that, p could be 0 or -1/2. We already got the solution p=q=0, but p=q=-1/2 is a new possibility.
19 May 2011, 12:39 amChrist Schlacta:
it doesn’t take into account the value of x at all. for x=0, p=q=0 works, but if x is any other value, say, 1.. 1^2 = 1, 1*0 = 0, 0 = 0. 1+0+0 = 1. 1 != 0. the only viable solution is p = 0, q = -(x^2). if the solution doesn’t reference x^2, then it cannot be guaranteed to stay true when x changes.
19 May 2011, 3:09 amGregory Marton:
It feels like the bug is not so much in the solution as in the problem statement. p and q as numbers cannot, by themselves, satisfy the equation, because whatever you set them to, x might take on values where the equation doesn’t hold. So a better question is to find values for p and q where it is possible to satisfy the equation with, say, real numbers, and you’d come out with a description of the discriminant…? Or to pre-specify some x where you’d like this to be true (which would make it trivial). Or as Christ says, to make them functions of x rather than “numbers”. Any way you slice it, the problem statement is screwy.
19 May 2011, 6:35 amAndrei Zelevinsky:
I agree with Joseph’s comment: there is one more solution p=q=-1/2. And I don’t understand the other two comments.
19 May 2011, 8:11 amAustin:
I also agree with Joseph, but in defense of the other commenters, the problem statement is a bit unnatural. This would be clearer:
Find numbers p and q such that x=p and x=q satisfy x^2+px+q = 0.
19 May 2011, 9:12 amvamsi:
it should have been {a+b=-p} and {ab=q} and then solve for solutions of a,b
23 May 2011, 1:10 amhehe the author of this question definitely had his own style of sense of humour.
Pratik Poddar:
Nice problem. Awesome solution by Joseph. Thanks.
24 May 2011, 12:05 amAnurag:
p is a solution iff
p^2 + p^2 + q = 0
q is a solution iff
q^2 + pq + q = 0
taking q = 0 we get p = 0
24 May 2011, 1:22 amif q is not 0 then 2nd eqn becomes q + p + 1 = 0
using this and 1 we get
2p^2 – p – 1 = 0
which has 2 solutions p = 1, p = -1/2
and corresponding q’s
So this approach somehow misses p = -1/2, q = -1/2
Cheska:
I think it’s a perfectly good solution to a somewhat different question.
1 June 2011, 4:45 amMy answer would be q = -x^2 – xp, pick any value of p you want.This is a nice problem.It is not hard to solve.
estoyanov:
The mistake is that p and q are not given numbers!They are parameters!And eqution is not only one. And p can be a root with some other number diferent of q. Same for q.
9 November 2011, 3:43 am