Recycle Bin Posted November 6, 2016 Report Share Posted November 6, 2016 (edited) How do you do a)ii)? Edited November 6, 2016 by Recycle Bin Reply Link to post Share on other sites More sharing options...
mac117 Posted November 6, 2016 Report Share Posted November 6, 2016 In a (i) you found out that r = 6/(m-1) and r = (m+4)/6 Now to get the equation, you have to set the two equal to each other. This will look like this: (m+4)/6 = 6/(m-1) To get rid of the denominators, you have to multiply each side by both denominators. One of them will cancel out for each side, so you will get this: (m+4)(m-1) = 36 Now you just multiply the brackets out to get: m^2 - m +4m -4 = 36 m^2 +3m - 40 = 0 Hope it helps! 3 Reply Link to post Share on other sites More sharing options...
Recycle Bin Posted November 6, 2016 Author Report Share Posted November 6, 2016 1 hour ago, mac117 said: In a (i) you found out that r = 6/(m-1) and r = (m+4)/6 Now to get the equation, you have to set the two equal to each other. This will look like this: (m+4)/6 = 6/(m-1) To get rid of the denominators, you have to multiply each side by both denominators. One of them will cancel out for each side, so you will get this: (m+4)(m-1) = 36 Now you just multiply the brackets out to get: m^2 - m +4m -4 = 36 m^2 +3m - 40 = 0 Hope it helps! thank you so much! you explained it really nicely. Reply Link to post Share on other sites More sharing options...
Recycle Bin Posted November 6, 2016 Author Report Share Posted November 6, 2016 I don't get this question either :/ help would be appreciated. Reply Link to post Share on other sites More sharing options...
kw0573 Posted November 6, 2016 Report Share Posted November 6, 2016 I recently answered the simpler version, to simply evaluate such binomial expansion, http://www.ibsurvival.com/topic/42140-binomial-expansion/#comment-271213 Note that the expanded expression has 9 as a constant, which is due to 1n32 = 9, does not really help us in any way. The important term is 84x. Similar to the above post, we first find some r and s such that (x)r(x)s = x1, only two options are r = 0, s = 1, and r = 1, s = 0,we find both. 84x = (n choose 0) * (1)n - 0 * (2/3 x)0 * (2 choose 1) * (3)2-1 * (nx)1 + [this is case for r = 0, s = 1] ..........(n choose 1) * (1)n - 1 * (2/3 x)1 * (2 choose 0) * (3)2-0 * (nx)0 [this is case for r = 1, s = 0] 84x = 1 * 1 * 1 * 2 * 3 * nx + ..........n * 1 * 2/3 x * 1 * 9 * 1 = 6nx + 6nx = 12nx 84x = 12nx, If x =/= 0, n = 7. 1 Reply Link to post Share on other sites More sharing options...
Recycle Bin Posted November 7, 2016 Author Report Share Posted November 7, 2016 12 hours ago, kw0573 said: I recently answered the simpler version, to simply evaluate such binomial expansion, http://www.ibsurvival.com/topic/42140-binomial-expansion/#comment-271213 Note that the expanded expression has 9 as a constant, which is due to 1n32 = 9, does not really help us in any way. The important term is 84x. Similar to the above post, we first find some r and s such that (x)r(x)s = x1, only two options are r = 0, s = 1, and r = 1, s = 0,we find both. 84x = (n choose 0) * (1)n - 0 * (2/3 x)0 * (2 choose 1) * (3)2-1 * (nx)1 + [this is case for r = 0, s = 1] ..........(n choose 1) * (1)n - 1 * (2/3 x)1 * (2 choose 0) * (3)2-0 * (nx)0 [this is case for r = 1, s = 0] 84x = 1 * 1 * 1 * 2 * 3 * nx + ..........n * 1 * 2/3 x * 1 * 9 * 1 = 6nx + 6nx = 12nx 84x = 12nx, If x =/= 0, n = 7. ahh thank you so much! so well explained. Reply Link to post Share on other sites More sharing options...
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