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

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== Problem == | == Problem == | ||

− | Prove that if <math>\kappa, \lambda, \nu</math> are positive integers, then the equation <math>x^2-(\nu +2)\kappa\lambda x+\kappa^2\lambda^2 = 0</math> has | + | Prove that if <math>\kappa, \lambda, \nu</math> are positive integers, then the equation <math>x^2-(\nu +2)\kappa\lambda x+\kappa^2\lambda^2 = 0</math> has irrational roots. |

== Solution == | == Solution == | ||

+ | Note that the square root of the [[discriminant]] of the quadratic is: | ||

+ | <math>\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}</math> | ||

+ | Since <math>\nu</math> is a positive integer: | ||

+ | |||

+ | <math>(\nu+1)^2 < \nu^2+4\nu < (\nu+2)^2</math>. | ||

+ | |||

+ | So, <math>\nu^2+4\nu</math> is a non-perfect square, and thus, <math>\sqrt{\nu^2+4\nu}</math> is irrational. | ||

+ | |||

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

+ | |||

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

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## Latest revision as of 22:54, 22 May 2009

## Problem

Prove that if are positive integers, then the equation has irrational roots.

## Solution

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

Since is a positive integer:

.

So, is a non-perfect square, and thus, is irrational.

Since and are integers, the square root of the discriminant: is irrational.

Therefore, the quadratic: has irrational roots.