Polynomials help

[Robert G]

Hello everybody,

I have some problems to do and could use some help. It's all about polynomials and complex numbers, and I basically just started the topic and don't have too much of a clue about it. Any help would be appreciated, but if you could give me some explanations of what to do, how to do it and why to do it, that would be really great.

#1 |(x/(8-x)] <= 2 I need to find the exact values for x. Unfortunately I never had inequalities. Please help me with this.

#2 This one is very hard, please help me with it. x³-x+1=0 has the roots a and b, I need to show that x³+x²-1=0 has the root ab. I also got the hint that I should use x³-x+1=(x-a)(x-b)(x-c) but that doesn't help me too much.

#3 |z+16| = 4|z+1| I need to find |z| if z is a complex number

#4 I need to find the cube roots of -27.

Thank you very very much.

Regards,

Robert

 

[jwpaine]

\(\displaystyle \frac{x}{8-x} \,\leq\, 2\) Is solved the same way as an equality

\(\displaystyle x \,\leq\, 2(8-x)\)

\(\displaystyle x \,\leq\, 16 - 2x\)

\(\displaystyle 3x \,\leq\, 16\)

\(\displaystyle x \,\leq\, \frac{16}{3}\)

 

[o_O]

2. Well find the roots of x[sup:esb9ibpq]3[/sup:esb9ibpq] - x + 1, which will be a and b, and see if their product ab is a root of x[sup:esb9ibpq]3[/sup:esb9ibpq] + x[sup:esb9ibpq]2[/sup:esb9ibpq] - 1

3. I never dealt with complex numbers before but from what I read, if z = a + bi and \(\displaystyle |z| = \sqrt{a^{2} + b^{2}}\), then wouldn't
\(\displaystyle |z+16|= \sqrt{(a + 16)^{2} + b^{2}}\)
And something similar with |z + 1|? Then solve for b in terms of a and solve for a? Correct me if I'm wrong.

4. What cubed equals to -27? Do you know what the cube root of 27 is? That is kind of a start ...

 

[galactus]

Try setting your absolute value up as:

\(\displaystyle -2\leq\frac{x}{8-x}\leq{2}\)

\(\displaystyle -2\leq\frac{8}{8-x}-1\leq{2}\)

Proceed. You will get two solutions. JW gave you one.

 

[stapel]

\(\displaystyle \frac{x}{8-x} \,\leq\, 2\) Is solved the same way as an equality

Um... no, it's not.... :shock:

What's the rule for multiplying (or dividing) through an inequality by a negative?

Do you know (before you find the final solution) whether or not 8 - x is positive or negative?

Eliz.

 

[o_O]

I think he missed the absolute value signs ... I sure did.

And he probably meant solving it in the sense that you don't have to test around the roots as you would with quadratic inequalities.

 

[stapel]

I think he missed the absolute value signs ... I sure did.

There's a "bar" in front, but a standard square bracket at the end. Are we sure this is an absolute-value inequality?

Eliz.

 

[o_O]

Guess not

At least jwpaine and galactus covered both possibilities :wink:

I really need to do something else with my spare time ...

 

[Unco]

I think he missed the absolute value signs ... I sure did.

And he probably meant solving it in the sense that you don't have to test around the roots as you would with quadratic inequalities.

You are missing the point Eliz is making. Even if there were no absolute value signs, as in Jwpaine's post, the work that followed was assuming that 8-x>0, and consequently did not find all possible values for x. Similarly, following Galactus's post, one has to consider the two cases 8-x<0 and 8-x>0.

With regards to problem #4, Robert should consider solving z^3 = -27, where z is complex.

 

[jwpaine]

Sorry
How about this:

\(\displaystyle \frac{x}{8-x} <= 2\)

\(\displaystyle \frac{x}{8-x} -2 <= 0\)

\(\displaystyle \frac{x-2(8-x)}{8-x} <= 0\)

\(\displaystyle \frac{3x-16}{8-x} <= 0\)

<-----( - )----(16/3)-----( + )-------(8)-------( - )----->

So: \(\displaystyle (-\infty \,\,..\,\, \frac{16}{3}] \,\,U\,\, (8 \,\,..\,\, +\infty)\) Is the set of numbers which satisfies the inequality.

John

 

[stapel]

How about this:

\(\displaystyle \frac{x}{8-x} <= 2\)

\(\displaystyle \frac{x}{8-x} -2 <= 0\)

Exactly right! :wink:

Eliz.

 

[Robert G]

Hey, thanks for your help. But one thing: I don't think that the bars are supposed to mean absolute value, I think they stand for the modulus, because that's part of what the chapter is about. That's what's confusing me. Can you help me with it?

 

[stapel]

I don't think that the bars are supposed to mean absolute value, I think they stand for the modulus, because that's part of what the chapter is about.

Ah. Providing the assignment's context, correct formatting, and full instructions can be very helpful in expressing your meaning. We've been having to make many assumptions, some of which were apparently in error...? Sorry!

If you mean "modulus", are we assuming "x" is any complex-valued number? Also, what is the definition your book gives for the "modulus" of a number or expression? What techniques have you been taught for doing these computations?

Please be complete in your reply, including in showing what you have tried so far, and how far you got.

Thank you!

Eliz.

 

[Robert G]

Providing the assignment's context, correct formatting, and full instructions can be very helpful in expressing your meaning. We've been having to make many assumptions, some of which were apparently in error...? Sorry! .

Sorry, my mistake, I apologize.

Well, I got 4 as a solution, and that fits to what my book says.

Thanks to all, you helped me significantly!