Rational Roots

[Guest]

My homework question is

Which of the following is NOT a possible rational root of \(\displaystyle 2x^3\, +\, 7x\, +\, 5\, =\, 0\)?

A. -5
B. -1/2
C. 1
D. -5/2

When I used Descartes' Rule to find the number of positive, negative and imaginary roots, I got:

0 positive
1 negative
2 imaginary.

Automatically I assumed that since there are no positive roots and the only positive answer is C, that the answer IS C, however, when listing out all possible roots, 1 is possible.

When I listed out roots, I got

1/2, 1, 5, 5/2

Those are the possible roots. Are these possible roots positive AND negative or are these possible roots ONLY positive because the original equation has no sign changes.

Or, I guess the easier question would be, how do I solve this?

 

[galactus]

You're correct about the nature of the roots. There are 1 neg. and 2 non-real.

If c/d is a rational root of f(x) and c and d have no common factors, then

the numerator, c, of the zero is a factor of the constant term and the

denominator, d, of the zero is a factor of the leading coefficient.

So, youre possible roots are:

Choices for numerator c: \(\displaystyle \L\\\pm{1}, \;\ \pm{5}\)

Choices for numerator d: \(\displaystyle \L\\\pm{1}, \;\ \pm{2}\)

Choices for c/d:\(\displaystyle \L\\\pm{1}, \;\ \pm{\frac{1}{2}}, \;\ \pm{5}, \;\ \pm{\frac{5}{2}}\)

Upon entering these possibilities we find \(\displaystyle f(x)\neq{0}\)

Therefore, no rational roots. Ol' Descartes pretty much told us that, though.

 

[Guest]

...Therefore, no rational roots. Ol' Descartes pretty much told us that, though.

Hmm...I'm a little bit confused at this part. If there are no rational roots, then my homework question would be impossible to answer, correct?

I don't think I would have an impossible-to-answer homework question, so where are we going wrong?

:?

 

[galactus]

The roots of your cubic are:

\(\displaystyle \L\\0.319773262272+1.95109323657i; \;\ \\
-0.319773262272-1.95109323657i;\\ \;\
-0.639546524544\)

What do you think?. Are we wrong?.

 

[Guest]

Hm. No. You are right. There are no rational roots, but I guess the answer would be C since, in the first step, we can eliminate any positive answers.

How would you answer this question? I'd have to say this was a bad example since the problem doesn't even contain any rational roots...lol.

 

[Mrspi]

My homework question is

Which of the following is NOT a possible rational root of \(\displaystyle 2x^3+7x+5=0\)?

A. -5
B. -1/2
C. 1
D. -5/2

...how do I solve this?

Did you (or any of the other responders) LOOK at the question?

You are asked: Which of these ARE NOT possible rational roots of
2x<SUP>3</SUP> + 7x + 5 = 0

You've been reminded of the rational roots theorem...and that's helpful here.

Look at the possible answers given. Which (if any) could not be rational roots for the equation?

If you want me to get more specific, I will....but GEEZ, guys! :wink:

 

[Guest]

Did you (or any of the other responders) LOOK at the question?

You are asked: Which of these ARE NOT possible rational roots of
2x<SUP>3</SUP> + 7x + 5 = 0

You've been reminded of the rational roots theorem...and that's helpful here.

Look at the possible answers given. Which (if any) could not be rational roots for the equation?

If you want me to get more specific, I will....but GEEZ, guys!

Obviously I don't know how to solve this problem...if I did I wouldn't be coming here to seek help from other forum members.

When using the rational roots theorem, the roots I came up with were

As far as I can tell, all the possible solutions that I came up with are covered in the answer choices. This is why I took it a step further and used Descartes' rule to find the number of positive, negative, and imaginary roots. Since I came up with ZERO positive roots, I assumed that answer choice C is the correct answer. My question is: is this correct? If not, why not and what should I have done differently?

----------------------------

Mrspi, I am not a retired math teacher, I am an uninformed student. That is why I am asking for help. Being a teacher, I am sure you appreciate it when your students come prepared for their classes, and that is merely what I am trying to do. Hopefully you can either put aside your condescending tone or ignore my plea for help.

 

[stapel]

Obviously I don't know how to solve this problem...if I did I wouldn't be coming here to seek help from other forum members.

You're missing the point of Mrs. Pi's reply:

You don't need to find the roots. You don't need to find the number of roots. You don't need to find the signs of the roots. You only need to apply the Rational Roots Test to say which of the listed options could NOT be the roots.

That is all that the question asks. That is all that the solution requires. And understanding this was almost certainly the whole point of this exercise.

You started this thread by asking, "How do I solve this?" Mrs. Pi's reply was (in my opinion) friendly, correct, and exactly on target. She got the thread back to the question at hand, and provided the solution. It is to be regretted if this offends you.

Eliz.

 

[galactus]

Sorry for the apparent overkill. Yes, I read the question. I was just trying to show the poster what it was about.

 

[stapel]

No prob! You posted some good info!

Mrs. Pi was just helpng us all get back on track.

Eliz.