looking at this proof: http://www.proofwiki...ence_of_Squares
I dont understand how it gets from (k+1)(2k2+7k+6)6
to
(k+1)(k+2)(2(k+1)+1)6
if someone can enlighten me that would be great. thanks in advance
Induction help
Page 1 of 13 Replies  1861 Views  Last Post: 07 December 2013  01:47 PM
Replies To: Induction help
#2
Re: Induction help
Posted 07 December 2013  11:36 AM
Well, it's been twenty years, but:
(2(k+1)+1) = 2k + 2 + 1 = 2k + 3
(k+2)(2k+3) = 2k^2 + 3k + 4k + 6 = 2k^2 + 7k + 6
(2(k+1)+1) = 2k + 2 + 1 = 2k + 3
(k+2)(2k+3) = 2k^2 + 3k + 4k + 6 = 2k^2 + 7k + 6
#3
Re: Induction help
Posted 07 December 2013  01:16 PM
yeah but where did they pull the (k+2) from? I just feel like theres steps missing.
#4
Re: Induction help
Posted 07 December 2013  01:47 PM
andrew showed you how to work backwards. We basically just reverse the process. With nice numbers, you can use a trial and error process to derive factors. You may have learned from algebra that (ax+b )(cx+d) = acx^2+(ad+bc)x+bd. So, when given an expression of the form nx^2+mx+y you need to come up with numbers for a,b,c and d such that
In your equation, n is 2, m is 7, and y is 6. The only nice numbers for ac=2 are a=1 and c=2. Now just solve the system of equations
So, bd is either (1 and 6) or (2 and 3). When we test d=3 and b=2, it satisfies both equations. So, the factored equation is (x+2)(2x+3).
ac=n ad+bc=m bd=y
In your equation, n is 2, m is 7, and y is 6. The only nice numbers for ac=2 are a=1 and c=2. Now just solve the system of equations
1*2=2 d+2b=7 bd=6
So, bd is either (1 and 6) or (2 and 3). When we test d=3 and b=2, it satisfies both equations. So, the factored equation is (x+2)(2x+3).
This post has been edited by mojo666: 07 December 2013  01:47 PM
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