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 January 22nd, 2012, 08:16 PM #1 Newbie   Joined: Jan 2012 Posts: 10 Thanks: 0 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^2–3x–5 = 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.
 January 22nd, 2012, 08:44 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Roots of the equation x2 –3x –5 = 0 help need If $\alpha,\,\beta$ are the roots, then we may write: $$$x-\alpha$$$$x-\beta$$=x^2-3x-5$ $x^2-$$\alpha+\beta$$x+\alpha\beta=x^2-3x-5$ Thus, we have: $\alpha+\beta=3$ and $\alpha\beta=-5$ a) $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\bet a}{\alpha\beta}=\frac{3}{-5}=-\frac{3}{5}$ b) $\alpha^2+\beta^2=$$\alpha+\beta$$^2-2\alpha\beta=3^2-2(-5)=9+10=19$ c) $\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{\alpha^ 2+\beta^2}{$$\alpha\beta$$^2}=\frac{19}{25}$
 January 22nd, 2012, 08:46 PM #3 Newbie   Joined: Jan 2012 Posts: 10 Thanks: 0 Re: Roots of the equation x2 –3x –5 = 0 help need Thank you for the reply.
 January 23rd, 2012, 01:26 PM #4 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: Roots of the equation x2 –3x –5 = 0 Darn you, Mark! I was hoping to get to typeset the beautiful Greek letters. :P
 January 23rd, 2012, 03:18 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Roots of the equation x2 –3x –5 = 0 You could demonstrate how to find the desired quantities by using $f$$\alpha$$=f$$\beta$$=0$ and $f$$\alpha$$+f$$\beta$$=0$...
 January 23rd, 2012, 04:11 PM #6 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: Roots of the equation x2 –3x –5 = 0 I have something that is clever and ridiculous: Let there be two inverse functions: $f^{a}(x)=\frac{1}{2}\left(3-\sqrt{29-4x}\right) \text{ and } f^{b}(x)=\frac{1}{2}\left(3+\sqrt{29-4x}\right)$ By virtue of inverse functions, we have: $\{\alpha, \beta\}=\left\{\left[f^{a}(f(\alpha)),f^{b}(f(\beta))\right] \text{ or } \left[f^{b}(f(\alpha)),f^{a}(f(\beta))\right]\right\}$ Thus, $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{f^{a}(f(\al pha)))+f^{b}(f(\beta))}{f^{a}(f(\alpha)))f^{b}(f(\ beta))}$ $\alpha^2+\beta^2=f^{a}(f(\alpha)))^2+f^{b}(f(\beta ))^2$ $\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{f^{a}(f (\alpha)))^2+f^{b}(f(\beta))^2}{f^{a}(f(\alpha)))f ^{b}(f(\beta))}$ 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.
 January 23rd, 2012, 05:15 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Roots of the equation x2 –3x –5 = 0 This is what I had in mind: (i) $f$$\alpha$$=f$$\beta$$$ $\alpha^2-3\alpha-5=\beta^2-3\beta-5$ $\alpha^2-\beta^2=3$$\alpha-\beta$$$ Since the discriminant of the given equation is not zero, the roots are distinct, and we may divide through by $\alpha-\beta$: $\alpha+\beta=3$ (ii) $f$$\alpha$$+f$$\beta$$=0$ $\alpha^2-3\alpha-5+\beta^2-3\beta-5=0$ $$$\alpha^2+\beta^2$$-3$$\alpha+\beta$$=10$ $\alpha^2+\beta^2=10+3$$3$$=19$ $$$\alpha+\beta$$^2=3^2$ $$$\alpha^2+\beta^2$$+2\alpha\beta=9$ $2\alpha\beta=9-19=-10$ $\alpha\beta=-5$ Now the questions can be found as in my first post.
 January 24th, 2012, 12:20 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,388 Thanks: 2012 1/? and 1/? are the roots of (1/x)² - 3(1/x) - 5 = 0, i.e., x² + (3/5)x - 1/5 = 0, so 1/? + 1/? = -3/5, and by [color=#00AC00]MarkFL[/color]'s method for (b), 1/?² + 1/?² = (3/5)² - 2(-1/5) = 9/25 + 10/25 = 19/25.

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