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January 22nd, 2012, 08:16 PM   #1
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Roots of the equation x2 3x 5 = 0

Hi I have recently received some work to complete on a certain question that contains three problems. I can manage to complete the first of the problems although not the second, and without the second I cannot complete the third. The question written verbatim is as follows:

Question 3
If ? and ? are the roots of the equation x^23x5 = 0, find the values of the following expression, without actually finding the roots:
a) 1/? + 1/? (express as a single fraction first)
b) ?^2 + ?^2 (use: (a+b)^2=a^2+2ab+b^2 and rearrange)
c) 1/?^2 + 1/?^2 (combine a) and b) above)

The help would be so very much appreciated. I have been dwelling on these questions since I got them and nothing has seemed to come to me on how to do b) and c). Thanks.
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January 22nd, 2012, 08:44 PM   #2
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Re: Roots of the equation x2 3x 5 = 0 help need

If are the roots, then we may write:





Thus, we have:

and

a)

b)

c)
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January 22nd, 2012, 08:46 PM   #3
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Re: Roots of the equation x2 3x 5 = 0 help need

Thank you for the reply.
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January 23rd, 2012, 01:26 PM   #4
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Re: Roots of the equation x2 3x 5 = 0

Darn you, Mark! I was hoping to get to typeset the beautiful Greek letters. :P
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January 23rd, 2012, 03:18 PM   #5
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Re: Roots of the equation x2 3x 5 = 0

You could demonstrate how to find the desired quantities by using and ...
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January 23rd, 2012, 04:11 PM   #6
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Re: Roots of the equation x2 3x 5 = 0

I have something that is clever and ridiculous:
Let there be two inverse functions:


By virtue of inverse functions, we have:


Thus,






Jokes aside. . .

I've kinda been pondering on precisely what you had in mind, Mark. I can't see anything worthwhile beside finding the value of alpha and beta directly from the quadratic formula. My imagination is being very bland. D:

Due to time constraints, it seems I'll have to get back to you later.
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January 23rd, 2012, 05:15 PM   #7
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Re: Roots of the equation x2 3x 5 = 0

This is what I had in mind:

(i)





Since the discriminant of the given equation is not zero, the roots are distinct, and we may divide through by :



(ii)















Now the questions can be found as in my first post.
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January 24th, 2012, 12:20 AM   #8
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1/? and 1/? are the roots of (1/x) - 3(1/x) - 5 = 0, i.e., x + (3/5)x - 1/5 = 0, so 1/? + 1/? = -3/5,
and by [color=#00AC00]MarkFL[/color]'s method for (b), 1/? + 1/? = (3/5) - 2(-1/5) = 9/25 + 10/25 = 19/25.
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