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May 2nd, 2019, 01:17 AM   #1
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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.
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May 2nd, 2019, 02:50 AM   #2
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$\displaystyle a\leq 0$.
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May 2nd, 2019, 05:38 AM   #3
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Is that all?
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May 2nd, 2019, 07:02 AM   #4
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Quote:
Originally Posted by skipjack View Post
Is that all?
I need for a >0, here's more difficult

please have a look ...
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Last edited by aurel5; May 2nd, 2019 at 07:06 AM.
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May 2nd, 2019, 07:10 AM   #5
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Quote:
Originally Posted by aurel5 View Post
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?
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May 2nd, 2019, 07:28 AM   #6
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Quote:
Originally Posted by skeeter View Post
$(x-2)(x-3)\sqrt{2^x-a} = 0$

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
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May 2nd, 2019, 08:48 AM   #7
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Quote:
Originally Posted by idontknow View Post
$\displaystyle a\leq 0$.
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.
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May 2nd, 2019, 09:37 AM   #8
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Quote:
Originally Posted by skeeter View Post
... am I missing something?
No. You found the additional values, but didn't mention the values already found again.
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May 2nd, 2019, 10:58 AM   #9
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Quote:
Originally Posted by skeeter View Post
$(x-2)(x-3)\sqrt{2^x-a} = 0$

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.
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May 2nd, 2019, 11:21 AM   #10
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x = 2 is now outside the domain of existence
Why?
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