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 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 $\displaystyle (x^2-5x+6)\sqrt{2^x-a}=0$ 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? Thanks from idontknow May 2nd, 2019, 07:02 AM   #4
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Quote:
 Originally Posted by skipjack Is that all?
I need for a >0, here's more difficult

Last edited by aurel5; May 2nd, 2019 at 07:06 AM. May 2nd, 2019, 07:10 AM   #5
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Quote:
 Originally Posted by aurel5 Determine the real values of a, for which equation $\displaystyle (x^2-5x+6)\sqrt{2^x-a}=0$ admits exactly two real solutions.
$(x-2)(x-3)\sqrt{2^x-a} = 0$

Two real solutions are $x \in \{2,3\}$

$x=2 \implies a = 4$

$x=3 \implies a = 8$

... am I missing something? May 2nd, 2019, 07:28 AM   #6
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 Originally Posted by skeeter $(x-2)(x-3)\sqrt{2^x-a} = 0$ Two real solutions are $x \in \{2,3\}$ $x=2 \implies a = 4$ $x=3 \implies a = 8$ ... am I missing something?
oh, ya, of course...!

For a>0 it's very insidious, it seems May 2nd, 2019, 08:48 AM   #7
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 Originally Posted by idontknow $\displaystyle a\leq 0$. I Just gave a hint ..., Start by avoiding the positive values of a .
Then solve the quadratic equation and plug the solutions to the general equation to find the other values of a. May 2nd, 2019, 09:37 AM   #8
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 Originally Posted by skeeter ... am I missing something?
No. You found the additional values, but didn't mention the values already found again. May 2nd, 2019, 10:58 AM   #9
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 Originally Posted by skeeter $(x-2)(x-3)\sqrt{2^x-a} = 0$ Two real solutions are $x \in \{2,3\}$ $x=3 \implies a = 8$ ... am I missing something?

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
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Quote:
 Originally Posted by aurel5 x = 2 is now outside the domain of existence
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