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melanie.dee

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help with a question please
« on: November 04, 2007, 06:52:09 pm »
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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?

Freitag

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help with a question please
« Reply #1 on: November 04, 2007, 06:57:37 pm »
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The two distinct solutions generally comes up when using the quadratic formula, with det > 0, but I don't think that'll come up in this question =/

It's more like, for what value of k will two different values of x satisfy the equation.

I saw the question too, wasn't sure how to approach it.

I'm interested to see Ahmad nail it :P

( EDIT ) Oh, and a real constant means a number without any complex numbers.

Check http://en.wikipedia.org/wiki/Real_number
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()

melanie.dee

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help with a question please
« Reply #2 on: November 04, 2007, 07:01:57 pm »
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ahah that makes me feel better than its not just me that didnt know how to approach it.. normally its just because im so shit at methods lol.

anyway.. coblin, ahmad?

eta; thanks for the real constant thing.. i thought it'd be something simple like that, but my complete lack of homework or listening all year means i have no idea about all the terminology and such lol

Freitag

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« Reply #3 on: November 04, 2007, 07:02:54 pm »
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Which exam is this from? I know I've done it, but I've done a few.
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()

melanie.dee

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help with a question please
« Reply #4 on: November 04, 2007, 07:04:47 pm »
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its from the mav cas 2007 exam.. are doing cas? its pretty possible it'd be on the normal mav 07 exam too though or something

Freitag

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help with a question please
« Reply #5 on: November 04, 2007, 07:06:09 pm »
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Yes, I'm doing CAS.
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()

cara.mel

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help with a question please
« Reply #6 on: November 04, 2007, 07:13:07 pm »
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Yay for CAS ^_^

I can't work out how to do it either. Strangly, this doesnt actually bother me, or my accompanied realisation I can't do half the methods exponential/log stuff anymore xD

Freitag

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help with a question please
« Reply #7 on: November 04, 2007, 07:15:50 pm »
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There's no way VCAA would give you such a question.. In my opinion anyways.

By graphing the function (plug in random x values and form a graph) you can see that the graph will have two distinct solution when  -6.25  < k < 0 . (I used my calculator).
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()

Collin Li

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help with a question please
« Reply #8 on: November 04, 2007, 07:16:08 pm »
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It's basically asking how do we ensure that:
4^x - 5(2^x) = k, has two distinct solutions?

Revision
If you recall from year 11, there was the concept of the "discriminant" (Δ).

Δ = b^2 - 4ac, where ax^2 + bx + c = 0

If Δ < 0, then there are no solutions
If Δ = 0, there is one solution
If Δ > 0, then there are two solutions.

This comes from the quadratic formula, where we have +/- root(b^2 - 4ac) = +/- root(Δ).

If Δ is negative, we can't take the square root
If Δ is zero, we get zero: one solution
If Δ is positive, we get two different solutions.

Okay, so lets actually do this:

4^x - 5(2^x) = k
=> (2^x)^2 - 5(2^x) = k [Here you must recognise that 4^x = (2^x)^2]

Let a = 2^x: a quadratic equation is evident:
=> a^2 - 5a = k
=> a^2 - 5a - k = 0

To fix 2 solutions: Δ > 0
Δ = (-5)^2 - 4(-k) = 25 + 4k
=> 25 + 4k > 0
=> 4k > -25
=> k > -25/4

This ensures there are 2 solutions to a^2 - 5a - k = 0.
(i.e.: a = u, and a = v are solutions to the equation, where u and v are some real numbers)

=> 2^x = u, and 2^x = v
=> x = log2(u), x = log2(v)

This means that we must ensure the two solutions, u and v are greater than zero (because you can't log negative numbers, or zero).

Use the quadratic formula: a^2 - 5a - k = 0

a = [ 5 +/- root(25 + 4k) ] / 2

We need to ensure:
=> [ 5 +/- root(25 + 4k) ] / 2 > 0
=> 5 - root(25 + 4k) > 0
[I removed the positive root solution, because that will always be larger than the negative root solution: remember, we just want to make sure both of the solutions are positive, so if the smallest solution is positive, both will be]

=> root(25 + 4k) < 5
=> root(25/4 + k) < 5/2

Sketch graph, with the root(x) function translated 25/4 units to the left. Convince yourself that for the function to be less than 5/2, k must be less than some value, which we will now find.

Solve: root(25/4 + k) = 5/2
=> 25/4 + k = 25/4
=> k = 0

So, k < 0.

This means the boundary is: k for (-25/4, 0)

melanie.dee

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« Reply #9 on: November 04, 2007, 07:19:03 pm »
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this makes me feel so much better, i thought it'd just be me who couldnt do it haha!

i hate the non calc or notes exam. hmph. not that i really know how to use the cas properly anyway haha, but i always manage to fiddle around on the calc and find a solutioin someway ha

enwiabe

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Re: help with a question please
« Reply #10 on: November 04, 2007, 07:19:13 pm »
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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?


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

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help with a question please
« Reply #11 on: November 04, 2007, 07:19:34 pm »
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beaten to the punch by 3 minutes. :P

Freitag

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« Reply #12 on: November 04, 2007, 07:19:37 pm »
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Well, my det > 0 proved to be useful ^^. I just wasn't sure how to apply it.
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()

melanie.dee

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« Reply #13 on: November 04, 2007, 07:21:27 pm »
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ahhh thanks both of you, i shall now proceed to read that and let it sink in aha

Freitag

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« Reply #14 on: November 04, 2007, 07:22:33 pm »
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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 ><
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()