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help with a question please

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Freitag:
Yes, I'm doing CAS.

cara.mel:
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:
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).

Collin Li:
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:
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

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