Help with maths question

[Recycle Bin]

How do you do a)ii)?

 

 

 

 

 

Edited November 6, 2016 by Recycle Bin

[mac117]

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! 

[Recycle Bin]

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.

[Recycle Bin]

I don't get this question either :/ help would be appreciated. 

 

 

 

[kw0573]

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.

 

[Recycle Bin]

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.