What is the answer to this extremely hard equation?

For example 32=8K K=4

show your working

I tried the Wolfram Alpha Calcultor and it gave me this:

http://www.wolframalpha.com/input/?i=%282a+%2B+3%29%5E2+-+%282a+-+3%29%5E2+%3D+8K+

Scroll down for anser i think...

Oh and K was the multiple ?

Just a wild guess but was it

prove (2a + 3)^2 - (2a - 3)^2 is a multiple of 8 for all integer values of a ?

(where ^2 means squared)

so, (2a + 3)^2 - (2a - 3)^2 = 8K
4a^2 +12a +9 -(4a^2 -12a+9)=8k
24a=8k
Therefore k=3


This was posted from The Student Room's iPhone/iPad App

The question came up in the Edexcel mathematics exam. It asked:

Prove that the following equation is a multiple of 8, given that the value of a is a positive integer. (2a + 3)^2 - (2a - 3)^2

It didn;t have anything about a K, though.

But can it represent a number? Like the multiple we are looking for?

That's fine up to the therefore

what happened to the a ?

Thanks a lot Guys

Silly mistake...

so, (2a + 3)^2 - (2a - 3)^2 = 8K
4a^2 +12a +9 -(4a^2 -12a+9)=8k
24a=8k
8(3a)=8k
K=3a


This was posted from The Student Room's iPhone/iPad App

So what the answer is 24a/k=8 :/

A better approach would have been to realize that it represents the difference of two squares.

$(2a + 3)^2 - (2a - 3)^2 = 8k \implies [(2a+3)-(2a-3)] [(2a+3)+(2a-3)] = 8k \\ 6(4a)=8k \implies 24a = 8k \implies \boxed{ k=3a}$

Still should stress how I think this question is being grossly misunderstood by a lot of being posting here. The OP has done a very bad job of explaining, but I believe the teacher has asked for them to prove that the LHS is a multiple of 8 for all integers a. Obviously another way to show this is by showing the LHS = 8K for some integer K.


So while doing it this way and getting K = 3a is fairly valid, it doesn't really answer the question unless you clearly state 'since 3a is an integer then K is an integer, and so the LHS is a multiple of 8 for all integers a'..