**Source:**Very interesting problem taken from Tanya Khovanova’s Math Blog

**Problem:**

*Sharygin Geometry Olympiad*by Zaslavsky, Protasov and Sharygin.

Find numbersThe book asks you to find a mistake in the following solution:pandqthat satisfy the equation:x^{2}+px+q= 0.

By Viète’s formulae we get a system of equationsWhat is wrong with this solution?p + q = − p,pq = q. Solving the system we get two solutions:p = q= 0 andp= 1,q= −2.

Update (26-05-2011)

**Solution:**

Posted by Sudeep Kamath (EECS PhD Student, UC Berkeley, EE IITB 2008 Alumnus) in comments and of course on Tanya Khovanova’s Math Blog

here, we are looking for the coefficients "p" & "q" most probably,

ReplyDeletei think the question DOES NOT that

P & q ARE THE SOLUTIONS OF GIVEN QUADRATIC..

Very nice problem.

ReplyDeleteThis approach misses out the case when p=q, so that you only need to enforce one root of the equation to be p. This gives the additional solution p=q=-1/2.

@Shaan. I did not understand your comment.

ReplyDelete@Sudeep. \m/ \m/ I could not solve it, but was happy_max when I read the solution :) Thanks.

but when you put p=q=-1/2 in the equation the roots of the quadratic are coming out to be -1/2 and 1 which are distinct and we need the equal roots..

ReplyDelete@satpal..

ReplyDeleteWhen you put p = q = -1/2,

The equation becomes 2x^2 - x - 1 = 0 and -1/2 is a root of this equation. The question never says that p and q are the only roots of the equation. :)

Its just a terminology difference. I would not have put this if this was not in Russian Olympiad book :)