3 Problems

Danforth

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Joined
Jan 29, 2006
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1) Find two numbers whose sum is 55 and whose product is 684.

2) A farmer has a rectangular garden plot surrounded by 200 ft of fence. Find the length and width of the garden if its area is 2400 square feet.


I'm not sure how to enter special characters on here, so this might be difficult to understand:

3) Find all solutions, real and complex, of the equation.
x^6 - 9x^3 + 8 = 0



Answers:
1) 19 & 36
2) 60 ft x 40 ft
3) 1, 2, (-1 + or - i square root of 3)/2, and -1 + or - i square root of 3



Sorry if it got confusing since I couldn't figure out how to use symbols.
Anyway, for #3, I have it set up like this:

(x^3 - 8)(x^3 - 1)=0 ---> x = cube root of 8 ; x = cube root of 1

So I end up with 1 and 2. I don't understand how to get the two complex solutions.


Thank you very much in advance for any help that any of you are able to provide.
 
Hello, Danforth!

I can explain #3 . . .

3) Find all solutions, real and complex: \(\displaystyle \:x^6\,-\,9x^3\,+\,8\:=\:0\)
What you did was correct . . . but you didn't factor completely.

\(\displaystyle \;\;x^6\,-\,9x^3\,+\,8\:=\:0\;\;\Rightarrow\;\;(x^3\,-\,1)(x^3\,-\,8)\:=\:0\)

You can factor a "difference of cubes", \(\displaystyle a^3\,-\,b^3\) . . . remember?

So we have: \(\displaystyle \:(x\,-\,1)(x^2\,+\,x\,+\,1)(x\,-\,2)(x^2\,+2x\,+\,4)\:=\:0\)

Now solve the four equations: \(\displaystyle \;\begin{array}{cccc}x\,-\,1\:=\:0\\ x\,+\,1\:=\:0\\x^2\,+\,x\,+\,1\:=\:0\\ x^2\,+\,2x\,+\,4\:=\:0\end{array}\)
 
1) the numbers are a & b
a+b=55
a*b=684
Solve one for a
Substitute that for a in the other.

2) It is l long and w wide.
lw=2400
2l+2w=200
Same method.
 
First, thank you both very much for the help. I understand perfectly how to work out all three of them now. I really like that you don't work out the whole thing for me, but rather get me started, and then let me do it myself. It definitely helps me learn better that way.


Denis said:
What is your problem with #1 and #2, Danforth?

Anyway, to answer your question (sorry if it's lengthy):

1 & 2) These seem quite simple now that I understand what I'm doing. Unfortunately, I had never been taught this. I knew how to set it up if it read something like this: "The length is 20 ft. more than the width," with the rest reading exactly the same as #2. Then I set it up as x(x+20)=2400. I was just unsure of how to determine what both the length and width would be without being given one of the two (like my example). I just needed to know how to get started.

3) Sometimes I just forget about finding the difference of cubes, especially when one of those cubes is 1. That always throws me off.


Anyway, once again, thank you for helping me out. I do understand these problems (and solutions) perfectly now.
 
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