My Add Maths Modules Form 4 - Quadratic Equations - DOC

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					    My
 Additional
Mathematics
  Modules
  Form 4
            (Version 2012)

            Topic 2:
 Quadratic
 Equations
                      by

                   NgKL
M.Ed.,B.Sc.Hons.,Dip.Ed.,Dip.Edu.Mgt.,Cert.NPQH
IMPORTANT NOTES
2.1 QUADRATIC EQUATION AND ITS ROOTS (PERSAMAAN KUADRATIK DAN PUNCANYA)                                                          3.

          Exercise 2.1.1
      1. Write each of the following quadratic equation in general form.
           (Tuliskan setiap persamaan kuadratik berikut dalam bentuk am).


(a)   x( 2  x )  5                                                         (b)    x( x  4)  3x(2  5x)




                                                                                           x
(c)   2( x  3) 2  13                                                       (d)    x2      6
                                                                                           2




      1. Write whether the value given in each of the following quadratic equations is the root of the quadratic equation.
         (Tentukan sama ada nilai yang diberikan ialah punca bagi persamaan kuadratik berikut).

(a)   x 2  5 x  4  0; x  4                                                                                   1
                                                                             (b)   3x 2  7 x  2  0; x 
                                                                                                                 3




                                2                                                                                1
(c)   5x 2  17 x  6; x                                                    (d)    x(6 x  7)  1; x  
                                5                                                                                6




 2.2 SOLUTION of QUADRATIC EQUATIONS                                   (PENYELESAIAN PERSAMAAN KUADRATIK)

      To solve a quadratic equation means to find the roots of the quadratic equation.
       Menyelesaikan suatu persamaan kuadratik bererti mencari punca-punca bagi persamaan kuadratik itu.
           1.   Generally, there are threes methods to determine the roots of a quadratic equation ax 2  bx  c  0 ;
           Secara amnya terdapat tiga cara dalam menentukan punca suatu persamaan kuadratik ax2 + bx + c = 0.;
           (a) Factorisation, (Pemfaktoran)
           (b) Completing the square, (Penyempurnaan Kuasa Dua)
           (c) Quadratic Formula. (Rumus kuadratik)

      (A) Solution by Factorisation (Penyelesaian secara Pemfaktoran).
            Example:
                                                    2
            Solve each of the quadratics equation, x  3x  18 .
            Selesaikan setiap persamaan kuadratik yang berikut.

                       x2 – 3x = 18                                                               x                   +3   +3x
                       x2 − 3x – 18 = 0
                       (x + 3)(x – 6) = 0

                       Therefore, (Maka),                                                         x                   -6   -6x
                       x+3=0        or x – 6 = 0
                         x = − 3,          x=6                                                  x2                   -18   -3x
                                                                                                                               4.
             Exercise 2.1.2
        1. Solve each of the following quadratic equation by factorization.
              Selesaikan setiap persamaan kuadratik berikut dengan menggunakan keadah pemfaktora).


(a) x 2  x  6  0                                                              (c) 5x2 + x – 4 = 0




(b)         x 2  8x  15  0                                                    (d) 2 x 2  7 x  4  0




        (B) Solution by Completing the Square Method                          (Penyelesaian secara Penyempurnaan Kuasa Dua)

                 Example:
                 Solve the following quadratic equation by completing the square.
                 (Selesaikan persamaan kuadratik berikut secara penyempurnaan kuasa dua).


      (a)              x2 – 3x – 5 = 0                                             (b)               2x2 + 3x = 4
                            x2- 3x = 5                                                                       3    4
                                  2                  2                                                x2      x
                           3        3                                                                    2    2
              x2 − 3x +      5                                                                       2
                           2        2
                                                                                                                         2
                                                                                               3   3     3
                                    2                2                                      x + x+    2 
                                                                                            2

                             3        3                                                    2   4     4
                          x   = 5+   
                             2        2
                                                                                                              2          2
                                                                                                         3       3
                                    20  9                                                           x  =2 +  
                                  =                                                                      4       4
                                      4                                                                         32  9
                                    29                                                                        =
                                  =                                                                              16
                                    4                                                                           41
                                                                                                              =
                              3      29                                                                       16
                          x   
                              2       4                                                                 3      41
                                                                                                       x   
                                  x
                                        3
                                          
                                            29                                                            4      16
                                        2    4                                                                   3  41
                                                                                                              x 
                                       3     29                                                                  4  16
                                  x     
                                       2      4                                                                     3    41
                                 x  4.193 #                                                                  x 
                                                                                                                    4    16
                         or      x 
                                     3     29                                                                 x  0.8508 #
                                     2      4                                                                       3     41
                                 x  1.193 #                                                          or     x 
                                                                                                                    4    16
                                                                                                              x  2.351 #
                                                                                                             5.
       Exercise 2.1.3

    Solve the following quadratic equation by completing the square method.
    (Selesaikan persamaan kuadratik berikut dengan kaedah penyempurnaan kuasa dua).



(a) x 2  6 x  4  0                                                    (b) x 2  10 x  3  0




(c) 3x 2  5  4 x                                                       (d) 2 x 2  3( x  1)




(e) 2x 2  4x  1  0                                                      (f) 2( x  2 )  4  3 x2  3 x
    (C) Solution by Quadratic Formula                                 (Penyelesaian secara rumus kuadratik)
                                                                                                                                                         6.

          The quadratic formula is obtained by completing the square method as shown below.
         ( Rumus kuadratik diperolehi dengan kaedah penyempurnaan kuasa dua persamaan kuadratik seperti ditunjukkan di bawah).


                   ax 2  bx  c  0                                                         Example:
                                 b   c
                        x   2
                                 x   0                                                    Solve the quadratic equation 4x2 – 8x + 1 = 0
                                 a   a
                                                                                             using quadratic formula.
                                                                                             (Selesaikan persamaan kuadratik 4 x  8x  1  0 secara rumus
                                                                                                                                2
                                       b       c
                                x2      x                                                 kuadratik).
                                       a       a

                                         2                                2
                                                                                             Solution: (Penyelesaian).
                   2    b   b                     c  b 
               x        x                          
                        a   2a                    a  2a 
                                                                                              4 x 2  8x  1  0
                                         2
                                                                                              a  4,      b  8,            c 1
                                b                  b2
                                                     c
                            x                                                                                                   b  b 2  4ac
                                2a           4a2   a                                       Using quadratic formula, x 
                                                                                                                                          2a

                                             
                                                 b 2  4 ac                                                           ( 8 )  ( 8 )2  4( 4 )( 1 )
                                                              2                                                 x
                                                         4a                                                                      2( 4 )
                                                                                                                   8  64  16
                                    b                     b 2  4 ac                                             
                                 x                                                                                   8
                                    2a                       2a
                                                                                                                   8  48
                                                                                                                 
                                                                                                                      8
                                                 b              b 2  4 ac
                                       x                                                                         8  48 8  6.928
                                                                  2a                         Therefore,         x        
                                                                                                                      8        8
                                                                                                                    8  6.928
                                                                                                                x
                                    b  b  4ac                  2                                                     8
          Quadratic formula x                                                                                     14.928
                                         2a                                                                      
                                                                                                                      8
          can be use to solve any quadratic equation
                                                                                                                x  1.866
          even though the equation can be solve by
                                                                                                           or
          either factorisation or completing the square                                                              8  6.928
          methods.                                                                                              x
                                                                                                                         8
                                                     2
                                       b       b        4 ac
          (Rumus Kuadratik x                        boleh digunakan                                                 1.072
                                         2a                                                                      
           untuk menyelesaikan sebarang persamaan kuadratik tanpa
                                                                                                                       8
           mengira sama ada persamaan itu boleh diselesaikan dengan
           menggunakan kaedah pemfaktoran dan penyempurnaan                                                     x  0.134
           kuasa dua atau tidak).




      Exercise 2.1.4

    Solve each of the following quadratic equations by using the quadratic formula
    (Selesaikan setiap persamaan kuadratik berikut dengan menggunakan kaedah rumu)


(a) 2 x 2  5x  4                                                                                2
                                                                                         (b) − 3 x  4 x  11  0
                                                                                                                           7.
(c) x 2  8x  3                                                               (d) 3( x 2  1)  7 x




2.2 FORMING QUADRATIC EQUATIONS FROM ROOTS
    PEMBENTUKAN PERSAMAAN KUADRATIK DARIPADA PUNCA-PUNCANYA


   1. If ( x   )( x   )  0 , then x    0 or x    0 and the roots are  and  .
        Jika ( x   )( x   )  0 , maka x    0 atau x    0 dan punca-puncanya ialah  dan  .

   2. On the other hand, if given  dan  as the roots of a quadratic equation, then,
                                                                     3.
      Sebaliknya, jika diberi  dan  ialah punca-punca persamaan kuadratik, maka,

       Method 1: Steps to form a quadratic equation are, (Langkah-langkah membentuk persamaan kuadratik ialah)
                 ( x   )( x   )  0
                      x 2  (   ) x    0
          where,         is the product of roots (POR) (ialah Hasil Tambah Punca [(HTP)]
                       is the sum of roots(SOR) (ialah Hasil Darab Punca [(HDP)]
       Method 2: Steps to form a quadratic equation are; (Langkah-langkah membentuk persamaan kuadratik daripada ialah;)
                       (i) Determine the POR, (Hitungkan HTP),
                       (ii) Determine the SOR, and (Hitungkan HDP, dan),
                       (iii) Form the quadratic equation by, (Bentukkan persamaan kuadratik yang dikehendaki iaitu)
                        x2 – (SOR)x + POR = 0
                                                                            2
                                                                        x        ( HTP) x  HDP  0

       Example:
       Form the quadratic equation in the form of ax 2  bx  c  0 which has the roots of 4 and –9.
       (Bentukkan persamaan kuadratik yang mempunyai punca-punca 4 dan - 9 dalam bentuk ax2 + bx + c = 0).


       Method 1:                                                                   Method 2:

           (x – 4) (x + 9) = 0                                                          SOR = 4 + (−9) = −5
                                                                                       POR = 4(−9) = −36
           x 2 + 9x – 4x – 36 = 0                                                  Then, the quadratic equation;
                                                                                       x 2  ( SOR ) x  POR  0
           x2 + 5x – 36 = 0
                                                                                        x 2  ( 5 ) x  ( 36 )  0
                                                                                            x 2  5 x  36  0
      Note:                                                                                                              8.
      If given the equation as ax 2  bx  c  0 , then the x coefficient need to be expressed into the value of 1.
      ( Nota: Jika diberi ax  bx  c  0 , perlu diungkapkan dahulu pekali x2 supaya menjadi satu, iaitu)
                            2



                                              b     c
                                                x 0       x2 
                                              a     a
                                                  b                                                             b
      Then, (maka), sum of roots, SOR           (Hasil tambah Punca,                 HTP                )
                                                  a                                                             a
                                                 c                                                      c
                    product of roots, POR         (Hasil darab Punca,                  HDP          )
                                                 a                                                      a
      Example:
      If  and  are the roots of the quadratic equation 3x 2  2x  5  0 , form the quadratic equation which has the
      roots  2 and  2
      Solution:
      Given the quadratic equation, 3x 2  2x  5  0
      Then, a  3, b  2 and c  5
      The roots of the quadratic equation are  dan 
                                              b        2 2
      Then, SOR                                
                                              a        3 3
                                         c        5
              POR                        
                                         a        3
      The new roots are  2 dan  2

              SOR   2   2  (   ) 2  2                             POR  ( 2 )(  2 )
                        2                                                        (  ) 2
                    2       5  4 10
                      2                                                           2
                    3       3 9 3                                                5           25
                                                                                   -        
                               34                                                    3           9
                       
                               9
       The new quadratic equation formed;
                  2
              x        ( SOR ) x  POR  0

                           2        34        25
                       x                x        0
                                    9         9
                           9x2 – 34x + 25 = 0


  Exercise 2.2.1

1. Form the quadratic equation from the given roots as shown in the table:

                                                                             Quadratic Equation
             Roots
                                                      Factorisation Method                              SOR-POR Method




(a)         2 and -3




(b)       −2 and 5
                                                                                                                                 9.
                   Root                      Factorisation Method                                    SOR-POR Method




                   3
  (c)               and 6
                   4




 (d)          2       1
                and -
              3       5




 (e).         2k and −4




2. Find the value of m and k for each of the following quadratic equations with the roots given.
   (Cari nilai m dan nilai k bagi setiap persamaan kuadratik dengan puncanya diberi).


                                                     1                                                                     1
(a) 3x 2  mx  k  0 with roots −5 dan                .                       (b) 2 x 2  mx  k  0 with roots 1 dan      .
                                                     3                                                                     2
                                                           m
(c) 2 x 2  9  (k  1) x with roots −3 and                 .               (d) x 2  3  4 x with roots m and k.                                10.
                                                           2




3. Find the value of p for each of the following quadratic equations.
     (Cari nilai p bagi setiap persamaan kuadratik berikut ).


(a) One of the roots of the quadratic equation                               (b) One of the roots of the quadratic equation
    3x2 – px + 54 = 0 is twice of the other root.                                x2 − px + 12 = 0 is thrice of the other root.
                                          2                                                                         2
     (Satu daripada punca persamaan 3 x        px  54  0 ialah dua kali       (Satu daripada punca persamaan x        px  12  0 ialah tiga
     punc yang satu lagi).                                                       kali punc yang satu lagi).
(c) One of the roots of the quadratic equation                                       (d) One of the roots of the quadratic equation                       11.
    x2 − px+ 8 = 0 is square root of the other root.                                     x2 − 6x = 2px – 27 is square of the other root.
      (Satu daripada punca persamaan x2 − px + 8 = 0 ialah punca kuasa                    (Satu daripada punca persamaan x2 − 6x = 2px – 27 ialah
      dua punca yang satu lagi).                                                          kuasa dua punca yang satu lagi).




 2.3 CONDITION FOR TYPES OF ROOTS OF QUADRATIC EQUATIONS
       (SYARAT UNTUK JENIS PUNCA PERSAMAAN KUADRATIK)

      1.    Types of roots of quadratic equations ax2+ bx + c = 0 depend to the value of b2- 4ac which derived from
                      b  b 2  4ac
             x
                           2a
      2.    (Jenis    punca     persamaan       kuadratik      ax 2  bx  c  0     bergantung   kepada      nilai   b 2  4ac yang   wujud   daripada   rumus
                                         2
                              b    b        4 ac
            kuadratik, x                             ).
                                    2a
            The foolowing table shows the types of roots of quadratic equations.
             (Jadual di bawah menunjukkan sifat punca persamaan kuadratik).


             b 2  4ac  0                                               b 2  4ac  0                                         b 2  4ac  0
           Two different roots                                          Two equal roots                                           No root
             (Dua punca berbeza)                                          (Dua punca sama)                                   (Tiada punca nyata)




                                                                         tangen
Straight line intersects the curve at Straight line meet the curve at
two different points.                 point.
(Garis lurus menyilang garis lengkung pada dua              (Garis lurus menyentuh lengkung pada satu titik     Straight line does not intersect, touch
titik yang berlainan).                                      sahaja).
                                                                                                                or meet the curve.
                                                                                                                (Garis lurus tidak menyilang atau menyentuh
                                               b  4ac  0
                                                 2                                                              lengkung).



                              Two distinctive roots (Dua punca nyata)
     Example:                                                                                                                            12.

     Determine the type of roots of each of the following quadratic equations without solving the equation.
      (Tentukan jenis punca bagi setiap persamaan kuadratik yang berikut tanpa menyelesaikan persamaan itu).


     (a) 2 x 2  7 x  4  0                                                   (b) x 2  3x  5  0
     (a) Given 2 x 2  7 x  4  0                                             (b) Given x 2  3x  5  0
         Then, a  2, b  7, c  4                                                Then, a  1, b  3, c  5
          Then, b 2  4ac  (7) 2  4(2)(4)                                          Then, b 2  4ac  (3) 2  4(1)(5)
                           49  32                                                                    9  20
                           17                                                                         11
          Since b  4ac  0 ,
                 2
                                                                                      Since b  4ac  0 ,
                                                                                             2

          The equation has two different roots.                                       The equation does not have any root.
          (Persamaan itu mempunyai dua punya nyata yang                               (Persamaan itu tidak mempunyai punya nyata).
          berbeza).


     Example of Applicatio:

     Find the possible values of m if a straight line y = mx – 1 is the tangent to a curve y = x2 – 7x + 7m.
     (Carikan nilai-nilai yang mungkin bagi m jika garis lurus y  mx  1 ialah tangen kepada lengkung y  x  7 x  7 m. )
                                                                                                            2


     Solution:
                                                                            y = x2 – 7x + 7m

                                                                               y = mx - 1
                                                             tangent line
                Given,                      y  mx  1 ……………..(1)
                                     y  x 2  7 x  7m. ………(2)
                Subtitute (2) dalam (1)
                Then,                            x 2  7 x  7m  mx  1
                                      x 2  7 x  mx  7m  1  0
                                     x 2  (7  m) x  7m  1  0
                And then,             a  1, b  7  m, c  7m  1
                Since the straight line is a tangent which meet the curve only at a point, therefore the condition of the
                solution is b2 – 4ac = 0.
                                                                                                                2
                (Oleh kerana tangen garis lurus hanya menyentuh lengkung pada satu titik, maka penyelesaian b        4 ac  0 ).
                                                            b  4ac  0
                                                                   2


                                            (7  m)  4(1)(7m  1)  0
                                                       2

                                            49  14m  m 2  28m  4  0
                                                     m 3  14m  45  0
                                                     (m  5)(m  9)  0
                                                      m = 5 or m = 9


    Exercise 2.3.1

   1. State the condition of the roots for each of the following quadratic equations
        (Tentukan keadaan punca bagi setiap persamaan kuadratik berikut).


   Quadratic Equation                                Value of b 2  4ac                                         Condition of the Roots


(a) x 2  2 x  5  0
      Quadratic Equation                              Value of b 2  4ac                                  Condition of the Roots   13.



(b) x 2  6 x  9  0




(c) x 2  3x  6




(d) 3x 2  8x  3




(e) x(1  4x)  3



2. The following quadratic equations have two equal roots, determine the possible value of p.
       ( Persamaan kuadratik berikut mempunyai dua punca yang sama, cari nilai yang mungkin bagi p.)


(a)     px 2  2 x  p  0                                                    (b)    px 2  8 x  2  0




3. Determine the range of p values if the following quadratic equations that have two distinctive different roots.
   (Cari julat nilai p jika persamaan kuadratik berikut mempunyai dua punca yang berbeza).
           (a) x 2  4 x  1  p  0                                          (b) x 2  2 x  p  2  0
4. Find the range of k values if the following quadratic equations do not have any distinctive root.                                 14.
     (Cari julat nilai k jika persamaan kuadratik berikut tiada punca).


(a) x 2  3x  k  0                                                         (b) 2 x 2  4 x  3  k  0




5. Express a relationship between p and q if the following quadratic equations have two equal roots.
    (Terbitkan suatu perkaitan antara p dengan q jika persamaan kuadratik berikut mempunyai dua punca yang sama).


(a) px 2  9qx  4 p  0                                                     (b)    px 2  5qx  9 p  0




 6. Express h in terms of k if the quadratic equation                        7. Express p in terms of q if the quadratic equations
    5kx2  hx  5k  0 have two distinctive different                           x2  6 x  p  q  0 do not have any root.
    roots.
   Past Year SPM Questions
                                                                                                                     15.

1. Solve the quadratic equation                             2. The quadratic equation x(x + 1) = px – 4 has two
   2x(x – 4) = (1 –x)(x+2).                                    distinct roots. Find the range of values of p.
   Give your answer correct to four significant figures.                                                      [3 marks]
                                                [3 marks]                                               SPM2003/Paper 1
                                         SPM2003/Paper 1




3. Form the quadratic equation which has the roots          4. The straight line y = 5x – 1 does not intersect the
   −3 and ½ , in the form ax2 + bx + c = 0, where a,           curve y = 2x2 + x + p.
   b and c are constant.                                       Find the range of values of p.
                                                [2 marks]                                                     [3 marks]
                                          SPM2004/Paper 1                                               SPM2005/Paper 1
                                                                                                                     16.


5. Solve the quadratic equation x(2x – 5) = 2x – 1.            6. A quadratic equation x2 + px + 4 = 2x has two equal
   Give your answer correct to three decimal places.              roots. Find the possible values of p.
                                                   [3 marks]                                                    [3 marks]
                                             SPM2005/Paper 1                                              SPM2006/Paper 1




7. (a) Solve the following quadratic equation;                 8. It is given that – 1 is one of the roots of the
       3x2 + 5x – 2 = 0.                                          quadratic equation x2 – 4x – p = 0.
                                                                  Find the value of p.
  (b) The quadratic equation hx2 + kx + 3 = 0, where                                                            [3 marks]
      h and k are constants, has two equal roots.                                                         SPM2008/Paper 1
      Express h in terms of k.                    [4 marks]
                                            SPM2007/Paper 1
                                                                                                                    17.


9. The quadratic equation x2 + x = 2px – p2, where p is       10. The quadratic equation (1 – p)2 – 6x + 10 = 0,
   a constant, has two different roots. Find the range            where p is a constant, has two different roots.
    of values of p.                             [3 marks]         Find the range of values of p.
                                           SPM2009/Paper 1                                                   [3 marks]
                                                                                                       SPM2010/Paper 1




11. The quadratic equation x2 – 5x + 6 = 0 has roots          12. The quadratic equation mx2 + (1 + 2m)x + m – 1= 0
    h and k, where h > k.                                         has two equal roots. Find the value of m.
                                                                                                               [3marks]
    (a) Find the value of h and of k.             [2 marks]                                             SPM2011/Paper 1

    (b) Using the values of h and k from 11(a), form
        the quadratic equation which has roots h + 2
        and 3k – 2.                              [2 marks]
                                            SPM2009/Paper 2

				
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