1. ## 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...)

Thanks!

2. ## Re: Find the roots.

Originally Posted by eleventhhour
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...)

Thanks!
you sort of had the right idea but didn't quite pull it off.

what you want to do is let

$u=2^t$

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

$(2u-1)(8u-1)=0$

$u=\dfrac 1 2 \bigvee u=\dfrac 1 8$ (not sure how you got 1 as a root)($\bigvee$ means OR)

now

$u=2^t \Rightarrow t=\ln_2(u)$ so

$t=\ln_2\left(\dfrac 1 2\right)=-1 \bigvee t=\ln_2\left(\dfrac 1 8\right)=-3$

so $t=-1 \bigvee t=-3$