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 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 $\displaystyle (x^2-5x+6)\sqrt{2^x-a}=0$ 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: 534 Thanks: 81 $\displaystyle a\leq 0$.
 May 2nd, 2019, 04:38 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 Is that all? Thanks from idontknow
May 2nd, 2019, 06:02 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 06:06 AM.

May 2nd, 2019, 06:10 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, 06:28 AM   #6
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Quote:
 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, 07:48 AM   #7
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Quote:
 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, 08:37 AM   #8
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Quote:
 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, 09:58 AM   #9
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Quote:
 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 10:00 AM.

May 2nd, 2019, 10:21 AM   #10
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Quote:
 Originally Posted by aurel5 x = 2 is now outside the domain of existence
Why?

 Tags equation

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