May 2nd, 2019, 12:17 AM  #1 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  Equation
Determine the real values of a, for which equation $\displaystyle (x^25x+6)\sqrt{2^xa}=0$ admits exactly two real solutions. Thanks a lot !!! Last edited by aurel5; May 2nd, 2019 at 12:19 AM. 
May 2nd, 2019, 01:50 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 
$\displaystyle a\leq 0$. 
May 2nd, 2019, 04:38 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,968 Thanks: 2217 
Is that all?

May 2nd, 2019, 06:02 AM  #4 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  I need for a >0, here's more difficult please have a look ... Last edited by aurel5; May 2nd, 2019 at 06:06 AM. 
May 2nd, 2019, 06:10 AM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600  
May 2nd, 2019, 06:28 AM  #6 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  
May 2nd, 2019, 07:48 AM  #7 
Senior Member Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91  
May 2nd, 2019, 08:37 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,968 Thanks: 2217  
May 2nd, 2019, 09:58 AM  #9  
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  Quote:
and now, what are the two solutions of the initial equation? x = 2 is now outside the domain of existence Last edited by aurel5; May 2nd, 2019 at 10:00 AM.  
May 2nd, 2019, 10:21 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,968 Thanks: 2217  

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