May 2nd, 2019, 01:17 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 01:19 AM. 
May 2nd, 2019, 02:50 AM  #2 
Senior Member Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math 
$\displaystyle a\leq 0$. 
May 2nd, 2019, 05:38 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 21,124 Thanks: 2332 
Is that all?

May 2nd, 2019, 07:02 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 07:06 AM. 
May 2nd, 2019, 07:10 AM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677  
May 2nd, 2019, 07:28 AM  #6 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  
May 2nd, 2019, 08:48 AM  #7 
Senior Member Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math  
May 2nd, 2019, 09:37 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 21,124 Thanks: 2332  
May 2nd, 2019, 10:58 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 11:00 AM.  
May 2nd, 2019, 11:21 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 21,124 Thanks: 2332  

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