tag:blogger.com,1999:blog-3045676214936462080.post2695221623572865492..comments2010-06-07T05:10:25.982-07:00Comments on J(p)=s: S.A.TrickyJ Functionhttp://www.blogger.com/profile/00753970401479983613noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-3045676214936462080.post-80594683961365619832008-06-20T07:48:00.000-07:002008-06-20T07:48:00.000-07:00I guess John's was smarter ???I guess John's was smarter ???Varunnoreply@blogger.comtag:blogger.com,1999:blog-3045676214936462080.post-70388439614040591482008-05-20T18:54:00.000-07:002008-05-20T18:54:00.000-07:00You thought that was tricky? It took me like, 30 ...You thought that was tricky? It took me like, 30 seconds.<BR/><BR/>a-b = 10<BR/><BR/>a^2 - b^2 = 50<BR/>(a-b)(a+b)=50<BR/>10(a+b)=50<BR/>a+b = 5<BR/><BR/>now you have 2 first degree equations:<BR/><BR/>a-b=10<BR/>a+b=5<BR/><BR/>no sweat.Johnhttp://www.blogger.com/profile/10876775111703252840noreply@blogger.comtag:blogger.com,1999:blog-3045676214936462080.post-69207450883641705512008-05-20T01:22:00.000-07:002008-05-20T01:22:00.000-07:00Missed a (-) in there, too: b = -2.5.This one is f...Missed a (-) in there, too: b = -2.5.<BR/><BR/>This one is fun to think about geometrically, too. At first, think about a situation in which both numbers are positive lengths with corresponding squares. If <I>a</I> is longer than <I>b</I> (both are positive, remember), then <I>a² - b²</I> is the area between the edges of the squares when <I>b²</I> is inside <I>a².</I> <A HREF="http://lh4.ggpht.com/tom.marley/SDJet8Q6VgI/AAAAAAAAATI/xB0Ee8wCTeg/geometric%21.jpg" REL="nofollow">**See image.**</A> We can represent this area in another way and discover: <I>(10a + 10b) = a² - b² = 50</I>. If we try to solve the problem by adding a modified version of this new equation, <I>a + b = 5</I>, to our existing <I>a - b = 10</I>, we find that <I>2a = 15 --> a = 7.5</I>. Solving for <I>b</I> (and yes, we could have found <I>b</I> without solving for <I>a</I>, but that wouldn't have been as fun!), we get <I>b = -2.5</I>. <BR><BR/>If you aren't scratching your head right now, think about why you should be for a good minute and then continue reading. (Hint: How did we start this <I>geometric</I> approach?)<BR><BR/>We assumed in the beginning that both values were <I>positive</I> lengths, but our answer is negative. Are we even allowed to have negative lengths? Let's explore a bit further.<BR/>We can check that the math works out if we let <I>a = 7.5</I> and <I>b = -2.5</I> by simple substitution, and <A HREF="http://lh4.ggpht.com/tom.marley/SDJnZ8Q6VhI/AAAAAAAAATQ/M_mT5SHgjm4/geometric2%21.jpg" REL="nofollow">this image</A> shows x and y axes with squares with sides of lengths <I>a</I> and <I>b</I>, where <I>a</I> is positive and <I>b </I> is negative. Subtracting a negative length (as in our <I>a - b = 10</I>) is essentially the same as adding a positive length, and the areas are the same as they would be if both numbers were positive because the areas are always positive, regardless of position on the coordinate system.tomhttp://www.blogger.com/profile/10680796021596355761noreply@blogger.comtag:blogger.com,1999:blog-3045676214936462080.post-78000235839739462992008-05-05T10:00:00.000-07:002008-05-05T10:00:00.000-07:00You're right, joe, thanks. That should have been 2...You're right, joe, thanks. That should have been 20b instead of 2b when I expanded the square.Jacqueline Granchehttp://www.blogger.com/profile/00753970401479983613noreply@blogger.comtag:blogger.com,1999:blog-3045676214936462080.post-60473266221368693192008-05-04T10:08:00.000-07:002008-05-04T10:08:00.000-07:00are you sure than answer is right?b= -2.5are you sure than answer is right?<BR/><BR/>b= -2.5joenoreply@blogger.com