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

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== Problem ==
 
== Problem ==
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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 irratioanl roots.
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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 ==
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Note that the square root of the [[discriminant]] of the quadratic is:
  
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<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>
  
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Since <math>\nu</math> is a positive integer:
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<math>(\nu+1)^2 < \nu^2+4\nu < (\nu+2)^2</math>.
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So, <math>\nu^2+4\nu</math> is a non-perfect square, and thus, <math>\sqrt{\nu^2+4\nu}</math> is irrational.
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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.
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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 $\kappa, \lambda, \nu$ 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. $\boxed{\mathbb{Q.E.D.}}$

See also

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