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IGCSE MATHS Doubts

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Vin:

y=(m-2n) (m+2n)
substitute

=[(2x + 3) -2(x - 1)] [(2x + 3) +2(x + 1)]
expand

= (2x + 3 - 2x + 2) ( 2x + 3 + 2x - 2)
multiply each number and variable from the first bracket to the second ..

= (2x + 3 - 2x + 2) ( 2x + 3 + 2x - 2)
(2x + 3 - 2x + 2) ( 2x + 3 + 2x - 2)
(2x + 3 - 2x + 2) ( 2x + 3 + 2x - 2)
(2x + 3 - 2x + 2) ( 2x + 3 + 2x - 2)
u get

= 4x2 + 6x + 4x2 - 4x + 6x + 9 + 6x - 6 - 4x2 - 6x - 4x2  + 4x + 4x + 6 + 4x - 4
cancel out similar +ve and -ve numbers
 
4x2 + 6x + 4x2 - 4x + 6x + 9 + 6x - 6 - 4x2 - 6x - 4x2  + 4x + 4x + 6 + 4x - 4
rearranging

= 6x + 6x + 4x + 4x + 9 - 4

= 20x + 5

$H00t!N& $t@r:
Thanks for the answer. Actually what i did was that i substituted the m and n values into y = m2 ? 4n2. This is also correct but my mistake was the i worked it out wrongly. I tried again and it worked.

Vin:

you can do it that way too .. ultimately u hav to factorise in the form of (a + b) ( a - b) .. jus one more step adds up .. finally u get the same ans .. safer method to do is the one i mentioned .. ;)

$H00t!N& $t@r:

--- Quote from: ~VIN1094~ on May 17, 2010, 08:09:40 am ---you can do it that way too .. ultimately u hav to factorise in the form of (a + b) ( a - b) .. jus one more step adds up .. finally u get the same ans .. safer method to do is the one i mentioned .. ;)

--- End quote ---
k ill keep that in mind  ;)

key04:
Hi Everyone!!!!!!!!!!

I have a question 4 maths paper 4.............
how to do ques 4 of may june 2006???????
anyone have any idea to do so??????
plz i need help urgently...........

Thanks

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