# Difference between revisions of "Specimen Cyprus Seniors Provincial/2nd grade/Problem 3"

## Problem

Prove that if $\kappa, \lambda, \nu$ are positive integers, then the equation $x^2-(\nu +2)\kappa\lambda x+\kappa^2\lambda^2 = 0$ has irrational roots.

## Solution

Note that the square root of the discriminant of the quadratic is:

$\sqrt{(\nu + 2)^2\kappa^2\lambda^2 - 4\kappa^2\lambda^2} = \sqrt{(\kappa\lambda)^2(\nu^2 + 4\nu + 4 - 4)} = \kappa\lambda\sqrt{\nu^2+4\nu}$

Since $\nu$ is a positive integer:

$(\nu+1)^2 < \nu^2+4\nu < (\nu+2)^2$.

So, $\nu^2+4\nu$ is a non-perfect square, and thus, $\sqrt{\nu^2+4\nu}$ is irrational.

Since $\kappa$ and $\lambda$ are integers, the square root of the discriminant: $\kappa\lambda\sqrt{\nu^2+4\nu}$ is irrational.

Therefore, the quadratic: $x^2-(\nu +2)\kappa\lambda x+\kappa^2\lambda^2 = 0$ has irrational roots. $\boxed{\mathbb{Q.E.D.}}$