Sec 3 A Math Question. Pls help.

• ProjectsMates

  6 Feb 11, 15:52

  

• SBS n SMRT

  6 Feb 11, 16:47

   

  Den...... what you expect us to do

  Ans is 6 (squareroot 3) and 4 (squareroot 2) if you want it.

  Tips to solve

   

  • Rationalize denominator by multiplying it with the conjugate surds
  • squareroot 6 times 4 squareroot 2 equals to 4 squareroot 12 (rules of indices), after which bring out the factor 4 (4X3=12), will become 2 outside as it is squareroot 4. Do likewise for the other one (3 squareroot 3, except now is factor 9 (9X2 = 18)
  • Combine the entire equation together

  Btw, I am afraid of this kind of question from Sec 3, but after I get it, it is now quite easy to solve

  (Btw, i had not use any calculator for this qns and did it in 2 mins - practice more and it will be easy) :)

   

   

• ProjectsMates

  6 Feb 11, 18:02

  I still can't get the answer. This is getting frustrating.

  I got (22 square root 6 minus four) / (5 square root 2 minus 2 square root 3)

• qdtimes2

  6 Feb 11, 18:27

  you missed out the step: Rationalize denominator by multiplying it with the conjugate surds

• ProjectsMates

  6 Feb 11, 18:44

  

• ProjectsMates

  6 Feb 11, 19:40

  I managed to simplify till: (212 square root 3 plus 112 square root 2) / 38

   

  How do you get 6 (squareroot 3) and 4 (squareroot 2)??

• SBS n SMRT

  6 Feb 11, 23:32

  

  This may help, I had tried a long time to upload, is there any easier way to upload and type?
• ProjectsMates

  7 Feb 11, 00:05

  Hi all, really appreciate your all assistance. I finally managed to solve it after taking a stroll outside. Guess my mind is too congested to think clearly.

  To SBS n SMRT: You can use imageshack to upload the images. It is easy and fast.

   

   

   

• Nowories-wart

  7 Feb 11, 23:37

  Hi everybody..

  I too have questions in AMATH..

  Below are my qns

  1. If x = 1, √x = 1.
  
  √x = -1 => x = 1 is invalid. In fact, the expression √x = -1 is invalid (rejected); root of anything cannot be negative!

  Is that statement true?

   

  2. . Solve the equations: 64 (4)^y = 16^y and (4)^(3x-2) - 1.
  
  I can find the solution x=2 and y=1 but cannot find x=1 or x=-1

  :(.. Please find the other solution!

   

  3. 

   I know the answer but can you explain me how to do it...

  Thanks:)...
  

• wee_ws

  9 Feb 11, 11:02

  Hi Nowories-wart,

    1. sqrt(1) = -1 does not make sense in the first place.

    2. The equations you gave do not look right.

    3. The steps are:

    a. Deal with the complex expression part by part. Convert each part to index form, i.e.,

    (r^2)/4 becomes r^2 . 2^(-2),

    (3x)^r becomes 3^r . x^r,

    (2 / (9x^2))^(6 - r) becomes 2^(6 - r) . 9^(-6 + r) . x^(-12 + 2r).

    b. Compare the indices of x, i.e.,

    -12 + 3r = -3,

    from which r can be found.

    c. Find the value of k given that r has been found.

    Thanks.

  Cheers,
  Wen Shih