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    O-Level Additional Math

    Scheduled Pinned Locked Moved Secondary Schools - Academic Support
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    • A Offline
      ADoc
      last edited by

      red rose:
      Would appreciate help on part (b) of this question. I typed out the whole question for clarity.


      The equation x^2/2-x-3/2=0 has roots a and b.

      (a) Without solving for a and b, find the value(s) of
      (i) a-b
      (ii) 1/b^2-1/a^2

      (b) If m=a-b, n=1/b^2 - 1/a^2, form all the possible quadratic equations with roots m and n (4 marks)

      Thanks in advance! šŸ™‚
      Hi. It's been a while since I last posted on KS.
      Guess this may be too late for your question. Nevertheless...

      (b) Simply substituting all the possible values from a (i) & (ii)
      which are (i) = +/- 4 (since (a-b)^2 = 16, therefore a - b = +/-4);
      (ii) = +/- 8/9

      so you'll obtain 2^2 = 4 possibilities in total. I'll save some cyber space by not typing them out. šŸ˜„

      By the way, just some useful identities when solving questions involving product and sum of roots that are suitable for Sec level especially for some of the IP schools such as DHS in particular.

      a^2 + b^2 = (a + b)^2 - 2ab

      (a - b)^2 = (a + b)^2 - 4ab

      a^4 - b^4 = (a^2 + b^2)(a + b)(a - b)

      a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2

      a^3 - b^3 = (a - b) [(a + b)^2 - ab]

      a^3 + b^3 = (a + b) [(a + b)^2 - 3ab]

      cheers
      Eugene
      Full-time tutor / Part-time Lecturer
      Specialising in IP group tuitions for DHS & VS.

      1 Reply Last reply Reply Quote 0
      • A Offline
        ADoc
        last edited by

        JadeDry:
        Hello,


        I would appreciate assistance with the following question:

        3 divided by (x-3) + (1 divided by (x+1)) divided by (2x divided by (x-3))

        Is the answer:

        (7(x squared) +9)/((2x)(x+1)(x-3))

        or

        2x/(x+1) ?

        Thanks in advance.
        hihi! First answer too, assuming the correct order of operations from what you typed above.

        cheers
        eugene

        1 Reply Last reply Reply Quote 0
        • MathIzzzFunM Offline
          MathIzzzFun
          last edited by

          koguma:
          Herbie:

          How to solve 499^2


          Assume 499^2 = 499 * 499,
          Since 499 = (500-1)
          so solve (500-1)^2 using (x-y)^2 formulae in algebra topic

          Hi

          it is faster to use (x+1)(x-1) = x^2-1 --> x^2 = (x+1)(x-1)+1
          499^2 = 500x 498 +1

          cheers.

          1 Reply Last reply Reply Quote 0
          • B Offline
            bran36
            last edited by

            Hi, as the exams are near, I'm preparing for my papers. However, while doing my assessment book, there's this question I do not understand.

            It's an inequality question. I don't understand why must the sign change. Perhaps anyone can enlighten me? Many thanks!!

            My work:
            http://i44.tinypic.com/2gxfqdv.jpg

            The Answer:
            http://i44.tinypic.com/35irdbp.jpg

            1 Reply Last reply Reply Quote 0
            • M Offline
              mum_sugoku
              last edited by

              bran36:
              Hi, as the exams are near, I'm preparing for my papers. However, while doing my assessment book, there's this question I do not understand.

              It's an inequality question. I don't understand why must the sign change. Perhaps anyone can enlighten me? Many thanks!!

              My work:
              http://i44.tinypic.com/2gxfqdv.jpg

              The Answer:
              http://i44.tinypic.com/35irdbp.jpg
              (Actually it's quite hard to explain without using drawings but let me try šŸ˜„ )

              First of all, the eqn you have is a quadratic eqn (ie eqn in the form of ax^2 + bx + c) which, if you were to draw a graph of the eqn, it's either a U shape curve (if a is +ve), or an inverted U (if a is -ve).

              Next, if we look at your eqn x^2 -4x +3 >=0, or (x-3)(x-1)>=0
              since a (ie coef of x^2) is = +ve, it's a U shape curve which cuts the x-axis (ie y=0) at x=3 and x=1

              What it means is that when x is between 1 and 3 (ie 1<x<3) the curve falls below the x-axis (ie eqn is -ve), and when x is <1 and when x is >3, the curve rises above the x-axis (ie eqn is +ve).

              Hence for eqn to be >= 0 (ie +ve), x<1 or x>3.

              If you can't understand the above explanation, alternatively, just bear in mind that for quadratic inequality eqn, say (x-3)(x-1)>0, ans is always either in the form

              1<x<3, or

              x<1, or x>3

              To find out which is the correct ans, one way is to pick a number between 1 and 3, say x=2, and insert into the eqn, and you'll get (2-3)(2-1) = -1, which is <0. Since question ask for >=0, correct ans has to be the second one, ie x<1 or x>3.

              1 Reply Last reply Reply Quote 0
              • B Offline
                bran36
                last edited by

                mum_sugoku:
                bran36:

                Hi, as the exams are near, I'm preparing for my papers. However, while doing my assessment book, there's this question I do not understand.

                It's an inequality question. I don't understand why must the sign change. Perhaps anyone can enlighten me? Many thanks!!

                My work:
                http://i44.tinypic.com/2gxfqdv.jpg

                The Answer:
                http://i44.tinypic.com/35irdbp.jpg

                (Actually it's quite hard to explain without using drawings but let me try šŸ˜„ )

                First of all, the eqn you have is a quadratic eqn (ie eqn in the form of ax^2 + bx + c) which, if you were to draw a graph of the eqn, it's either a U shape curve (if a is +ve), or an inverted U (if a is -ve).

                Next, if we look at your eqn x^2 -4x +3 >=0, or (x-3)(x-1)>=0
                since a (ie coef of x^2) is = +ve, it's a U shape curve which cuts the x-axis (ie y=0) at x=3 and x=1

                What it means is that when x is between 1 and 3 (ie 1<x<3) the curve falls below the x-axis (ie eqn is -ve), and when x is <1 and when x is >3, the curve rises above the x-axis (ie eqn is +ve).

                Hence for eqn to be >= 0 (ie +ve), x<1 or x>3.

                If you can't understand the above explanation, alternatively, just bear in mind that for quadratic inequality eqn, say (x-3)(x-1)>0, ans is always either in the form

                1<x<3, or

                x<1, or x>3

                To find out which is the correct ans, one way is to pick a number between 1 and 3, say x=2, and insert into the eqn, and you'll get (2-3)(2-1) = -1, which is <0. Since question ask for >=0, correct ans has to be the second one, ie x<1 or x>3.

                Hi thanks a lot for your help! However, how do you which value of x is > or <? I don't really understand this part. Thanks again!

                1 Reply Last reply Reply Quote 0
                • M Offline
                  mum_sugoku
                  last edited by

                  bran36:
                  Hi thanks a lot for your help! However, how do you which value of x is > or <? I don't really understand this part. Thanks again!

                  Just bear in mind that for a quadratic inequality eqn, (range of) x is either (1) in between the smaller no. and bigger no., or (2) x is less than the smaller no. or more than the bigger no.,

                  for eg, the eqn (x+3)(x-5) >0, ans must be either

                  (1) -3<x<5 (x in between -3 and 5) or
                  (2) x<-3; x>5. (x less than -3 or more than 5)

                  (PS. you cannot have something like x>-3; x>5 because if x>5 then of cos x must be >-3, likewise you also cannot have something like x<-3; x<5 since if x<-3 then of cos x must be <5)

                  Presuming that (sign of) x^2 is +ve, then

                  if question asks for \">0\" (eg (x+3)(x-5)>0), always choose (2). And if question asks for \"<0\" (eg (x+3)(x-5)<0), always choose (1). (The reason has already been given in the 1st part of my previous post šŸ˜„)

                  1 Reply Last reply Reply Quote 0
                  • H Offline
                    htn
                    last edited by

                    Hi, need help for the following:


                    A) 1/(x+2)-1/(2x+4)=3/2

                    B) 1/(x-1)+1/6=5/12


                    C) 7/(2x+4)-2/(3x+6)=3/4


                    TIA

                    1 Reply Last reply Reply Quote 0
                    • K Offline
                      Kafer
                      last edited by

                      John spwes ia x m/s.

                      Mary speed ia y m/s
                      If john starts 2 sec later, he will take 8 secs to catch up with mary.
                      If john starts to run only when mary is 8m ahea 5 sec to catch up with her.
                      Form two eqn in x an y.

                      1 Reply Last reply Reply Quote 0
                      • G Offline
                        greenflora
                        last edited by

                        Hi, can anyone help with this question?


                        The scale of map X is 1 : x and the scale of map Y is 4 : y.
                        If the actual area of a park is represented by 10cm^2 and 2.5cm^2 on maps X and Y respectively, find x : y in its simplest form.

                        1 Reply Last reply Reply Quote 0

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