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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

( 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.

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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##### help with a question please
« 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.

#### Freitag

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##### help with a question please
« 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 )