November 9th, 2016, 04:19 AM  #31 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,685 Thanks: 701 
Amen.

November 11th, 2016, 02:49 PM  #32 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 371 Thanks: 68  
November 12th, 2016, 12:07 AM  #33 
Global Moderator Joined: Dec 2006 Posts: 18,052 Thanks: 1395 
As x and y are the zeros of u²  4u + 5, 1/x and 1/y are the zeros of 5v²  4v + 1, and so their sum is 4/5.

November 15th, 2016, 10:00 PM  #34 
Banned Camp Joined: Nov 2016 From: St. Louis, Missouri Posts: 28 Thanks: 4 Math Focus: arithmetic, fractions 
Although this problem is impossible to solve, you can make things easier in the future by making it so that the solution of x+y is larger than that for x times y.

November 16th, 2016, 02:58 AM  #35 
Global Moderator Joined: Dec 2006 Posts: 18,052 Thanks: 1395 
That doesn't make sense.

November 16th, 2016, 04:12 AM  #36 
Senior Member Joined: May 2008 Posts: 258 Thanks: 50  
December 1st, 2016, 12:43 AM  #37 
Member Joined: Sep 2016 From: India Posts: 88 Thanks: 30 
$x+y=4...........(1)$ $xy=5.............(2)$ dividing eq(1) by $xy$ $\dfrac{x}{xy}+\dfrac{y}{xy}=\dfrac{4}{xy}$ $\dfrac{1}{y}+\dfrac{1}{x}=\dfrac{4}{xy}$ since $xy=5$ therefore, $\dfrac{1}{y}+\dfrac{1}{x}=\dfrac{4}{5}$ 

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