Find the roots.
4) Consider the equation H(t) = 16(2)^2t - 10(2)^t + 1. What are its roots? (HINT: Does this look like a quadratic? Perhaps, at least at first, it should be treated like one).
5) Do the same for Y(x) = 2sin^2x - 3sinx - 2. What is wrong with your solutions?
So, this is what I've done so far (which may be completely wrong!)
16(2t)^2 − 10(2)^t + 1 = 0
u = 2t
16u^2 − 10u + 1 = 0
Factored it: (8u-1)(2u-1) = 0
and then found the zeros: 1, 1/8 and 1/2
And then I'm not sure about the next part. Do I substitute 2^t for u?
[8(2^t)-1][2(2^t)-1] = 0
(16^t - 1)(4^t-1)
and then how would you find the roots of that? (assuming that I did that correctly...)
Re: Find the roots.
you sort of had the right idea but didn't quite pull it off.
Originally Posted by eleventhhour
what you want to do is let
then you can write your original equation as
$H(u) = 16 u^2 -10 u+1$
and to find the roots we solve
$H(u) = 0$
factoring we get
$u=\dfrac 1 2 \bigvee u=\dfrac 1 8$ (not sure how you got 1 as a root)($\bigvee$ means OR)
$u=2^t \Rightarrow t=\ln_2(u)$ so
$t=\ln_2\left(\dfrac 1 2\right)=-1 \bigvee t=\ln_2\left(\dfrac 1 8\right)=-3$
so $t=-1 \bigvee t=-3$