tys e maths ques

• skythewood

  22 Oct 08, 14:22

  Originally posted by crucifiedx:

  
  then how to solve..?

  try to solve it. it's something you need to try bah. giving you the answer like that is not really helping you much.

• crucifiedx

  22 Oct 08, 14:23

  Originally posted by eagle:

  There are 2 values

  1) Consider 2 as the hypotenuse
  side will be sqrt(2^2 - 1^2) = sqrt(3)

  2) Consider final side as hypotenuse
  side will be sqrt (2^2 + 1^2) = sqrt(5)

  so a can be 3 or 5

  
  i understand le. thanks.

• crucifiedx

  22 Oct 08, 14:28

  (3-2x)^2 - x(2x-3) = 0

   

  for this ques must i expand everything then factorise or can i just write:

   

  (3-2x)^2=0 or x(2x-3)=0 ?

• eagle

  22 Oct 08, 14:32

  Originally posted by crucifiedx:

  (3-2x)^2 - x(2x-3) = 0

   

  for this ques must i expand everything then factorise or can i just write:

   

  (3-2x)^2=0 or x(2x-3)=0 ?

  Expand is too troublesome

  the 2nd one also wrong

  Must note that (3-2x)^2 = (2x-3)^2

  Then you can factorise 2x-3 out and solve...

• crucifiedx

  22 Oct 08, 14:37

  Originally posted by eagle:

  Expand is too troublesome

  the 2nd one also wrong

  Must note that (3-2x)^2 = (2x-3)^2

  Then you can factorise 2x-3 out and solve...

  
  (3-2x)^2 = (2x-3)^2

   

  huh? why are they the same?

• eagle

  22 Oct 08, 14:40

  Originally posted by crucifiedx:

  
  (3-2x)^2 = (2x-3)^2

   

  huh? why are they the same?

  First, you must know that (3-2x) = (-1)(2x-3)

  (3-2x)^2 = (3-2x)(3-2x)
  = (-1) (2x-3) (-1) (2x-3)
  = (2x-3)^2

• crucifiedx

  22 Oct 08, 15:10

  Originally posted by eagle:

  First, you must know that (3-2x) = (-1)(2x-3)

  (3-2x)^2 = (3-2x)(3-2x)
  = (-1) (2x-3) (-1) (2x-3)
  = (2x-3)^2

  
  don't understand.. =X

• eagle

  22 Oct 08, 15:11

  which part dun understand?

• crucifiedx

  22 Oct 08, 15:35

  Originally posted by eagle:

  which part dun understand?

  
  tml then ask u ba.. =)

• crucifiedx

  22 Oct 08, 16:27

  i forgot how to do this type of ques..

   

  5t^2 = 12t

• crucifiedx

  22 Oct 08, 16:29

  and how do i factorise

   

  y^2 - 6xy + 5x^2 ?

• eagle

  22 Oct 08, 16:36

  Originally posted by crucifiedx:

  i forgot how to do this type of ques..

   

  5t^2 = 12t

  5t^2 - 12t = 0
  t(5t - 12) = 0

  :D

• eagle

  22 Oct 08, 16:37

  Originally posted by crucifiedx:

  and how do i factorise

   

  y^2 - 6xy + 5x^2 ?

  y     -5x  |  -5xy
  y      -x   |  -xy

  so you get (y-5x)(y-x)

  tmr I teach u 1 more short cut magic eye method for these type questions

• crucifiedx

  22 Oct 08, 17:04

  ok. thanks.. =)

   

  so is it like that:

   

  t(5t - 12) = 0

   

  t=0

   

  or 5t-12=0

      5t=12

      t=12/5  ?

• eagle

  22 Oct 08, 17:05

  correct!

  you are one step closer to your A :D

• crucifiedx

  22 Oct 08, 17:12

  Originally posted by eagle:

  correct!

  you are one step closer to your A :D

  
  =D