## A Wrong Solution

I found this cute problem in the Russian book *Sharygin Geometry Olympiad* by Zaslavsky, Protasov and Sharygin.

Find numbers

pandqthat satisfy the equation:x^{2}+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 andp= 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 pm## Roman:

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 pm## Roman:

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 pm## Joseph:

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 am## Christ 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 am## Gregory 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 am## Andrei 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 am## Austin:

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 am## vamsi:

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 am## Anurag:

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