Finding Roots of Quadratic Equations & Sinusoidal Functions

[eleventhxhour]

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?

Even with the hint, I'm not really sure how to start this, especially with the variables in the exponents...Could someone help? Thanks

 

[MarkFL]

For the first equation, first write it as:

\(\displaystyle 16\left(2^t\right)^2-10(2)^t+1=0\)

Next, you may use a substitution such as:

\(\displaystyle u=2^t\)

And you may now write:

\(\displaystyle 16u^2-10u+1=0\)

Now solve for $u$, and once you have the roots, then back-substitute for $u$ and solve for $t$.

The other problem may be worked the same way.

 

[eleventhxhour]

This is what I did:
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 again!

 

[MarkFL]

You factored correctly, but from where did the root $u=1$ come? Let's look at the root:

\(\displaystyle u=\frac{1}{2}\)

Now back substitute for $u$:

\(\displaystyle 2^t=\frac{1}{2}=2^{-1}\)

Equate exponents:

$t=-1$

Can you now do the same for the root \(\displaystyle u=\frac{1}{8}\) ?

 

[eleventhxhour]

You factored correctly, but from where did the root $u=1$ come? Let's look at the root:

\(\displaystyle u=\frac{1}{2}\)

Now back substitute for $u$:

\(\displaystyle 2^t=\frac{1}{2}=2^{-1}\)

Equate exponents:

$t=-1$

Can you now do the same for the root \(\displaystyle u=\frac{1}{8}\) ?

Hm...would it be 2^-3 for the root u = 1/8?

 

[MarkFL]

Recall, we used the substitution \(\displaystyle u=2^t\), so now we are putting $2^t$ back in, in place of $u$.

Also recall:

\(\displaystyle \frac{1}{a^b}=a^{-b}\)

Hence:

\(\displaystyle \frac{1}{2}=\frac{1}{2^1}=2^{-1}\)

 

[eleventhxhour]

Recall, we used the substitution \(\displaystyle u=2^t\), so now we are putting $2^t$ back in, in place of $u$.

Also recall:

\(\displaystyle \frac{1}{a^b}=a^{-b}\)

Hence:

\(\displaystyle \frac{1}{2}=\frac{1}{2^1}=2^{-1}\)

Sorry, I edited my previous reply. I think I understand now. Would 2^-3 be correct?

 

[MarkFL]

Yes, good! :D

\(\displaystyle \frac{1}{8}=\frac{1}{2^3}=2^{-3}\)

 

[eleventhxhour]

Yes, good! :D

\(\displaystyle \frac{1}{8}=\frac{1}{2^3}=2^{-3}\)

So does that mean the roots are t=-1 and t=-3?

And for the second part of the question, this is what I've done so far:

5) Do the same for Y(x) = 2sin^2x - 3sinx - 2. What is wrong with your solutions?

let sinx = u, so...
y(x) = 2u^2 - 3u - 2
(u - 2)(2u + 1)
u = 2, u =-1/2

but I'm not sure what to do after that...

Thanks again! (:

 

[MarkFL]

What is the range of the sine function?

 

[eleventhxhour]

What is the range of the sine function?

Can't they only range from -1 to 1?

 

[MarkFL]

Correct, so you must discard the root:

\(\displaystyle \sin(x)=2\)

Because no real value for $x$ will satisfy that equation. So, you are left with:

\(\displaystyle \sin(x)=-\frac{1}{2}\)

In which quadrants will the solutions reside?

 

[eleventhxhour]

Correct, so you must discard the root:

\(\displaystyle \sin(x)=2\)

Because no real value for $x$ will satisfy that equation. So, you are left with:

\(\displaystyle \sin(x)=-\frac{1}{2}\)

In which quadrants will the solutions reside?

\(\displaystyle \sin(x)=\frac{y}{r}\)

$y = -1$, $r = 2$

So since $y$ is negative, it'd reside in either quadrant 3 or 4.

Is this correct?

 

[MarkFL]

Yes, the sine function is negative in the 3rd and 4th quadrants. I would try to write all solutions using one statement. In degrees, we should observe that we may write the two angles as:

\(\displaystyle \theta=270^{\circ}\pm60^{\circ}\)

Notice I picked the angle midway between the two solutions, and then used $\pm$ to pick both angles up with one statement. Now, what must we add to this to get all the solutions...Hint: think of the periodicity of the sine function. :D

 

[eleventhxhour]

Yes, the sine function is negative in the 3rd and 4th quadrants. I would try to write all solutions using one statement. In degrees, we should observe that we may write the two angles as:

\(\displaystyle \theta=270^{\circ}\pm60^{\circ}\)

Notice I picked the angle midway between the two solutions, and then used $\pm$ to pick both angles up with one statement. Now, what must we add to this to get all the solutions...Hint: think of the periodicity of the sine function. :D

Hm, I don't really understand how you got those angles?

 

[MarkFL]

Hm, I don't really understand how you got those angles?

When I think of $\sin(x)=-\frac{1}{2}$, I picture the unit circle and the line $y=-\frac{1}{2}$. I then see that the two points of intersection between the two occur at $\theta=210^{\circ},\,330^{\circ}$, and I can see that $210^{\circ}=270^{\circ}-60^{\circ}$ and $330^{\circ}=270^{\circ}+60^{\circ}$. This allows us to write both angles in one statement:

\(\displaystyle \theta=270^{\circ}\pm60^{\circ}\)

There are of course other ways to think of it, but this is how I do it. :D

 

[eleventhxhour]

When I think of $\sin(x)=-\frac{1}{2}$, I picture the unit circle and the line $y=-\frac{1}{2}$. I then see that the two points of intersection between the two occur at $\theta=210^{\circ},\,330^{\circ}$, and I can see that $210^{\circ}=270^{\circ}-60^{\circ}$ and $330^{\circ}=270^{\circ}+60^{\circ}$. This allows us to write both angles in one statement:

\(\displaystyle \theta=270^{\circ}\pm60^{\circ}\)

There are of course other ways to think of it, but this is how I do it. :D

Ohh okay, that makes sense. Though I'm not sure what to add to get all the solutions?

 

[MarkFL]

You were not given any restrictions on $x$, so use the following:

\(\displaystyle \sin\left(x+k\cdot360^{\circ}\right)=\sin(x)\) where $k\in\mathbb{Z}$ (this means $k$ can be any integer).

In other words, moving around the unit circle any number of complete revolutions in either direction will put you right back in the same place on the circle, and the sine of that new angle will be the same as the original.