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help with a question please

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enwiabe:

--- Quote from: "melanie.dee" ---im sure this is really simple, i just have no idea what to do, and i have no clue with methods lol

its on the non calc exam, so how to do it wthout a calc please

For what values of k, where k is a real constant, does the equation 4^x - 5(2^x) = k, have two distinct solutions?

a) what does real constant mean haha :oops:
b) how do i even do this haha

im sure its ridiculously simple and you're thinking what an idiot haha, but i have no idea.

ps. if i get about 24/40 on the first exam (just did one and got that. everything i attempt i get right, i just dont know how to do half of it ahha) and same sort of thing for the second exam, what study score am i looking at?
--- End quote ---


For two disctinct solutions, you need a discriminant greater than 0, meaning b^2 - 4ac > 0

Now, let A = 2^x,

so A^2 - 5A = k
A^2 - 5A - k = 0

So you need b^2 - 4ac > 0

meaning 25 - 4(-k)(1) > 0

25 + 4k > 0
k > -25/4

So 4^x - 5(2^x) = k has two distinct real solutions for {k: k > -25/4}

enwiabe:
beaten to the punch by 3 minutes. :P

Freitag:
Well, my det > 0 proved to be useful ^^. I just wasn't sure how to apply it.

melanie.dee:
ahhh thanks both of you, i shall now proceed to read that and let it sink in aha

Freitag:
Coblin, i graphed the function and it appears to me the solutions for k are never for k > 0 ( only one solution exists). Wtf am i doing wrong ><

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