{"id":3888,"date":"2023-10-06T06:19:05","date_gmt":"2023-10-06T06:19:05","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3888"},"modified":"2023-10-06T06:20:05","modified_gmt":"2023-10-06T06:20:05","slug":"week-5-ss3-first-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-5-ss3-first-term-mathematics-notes\/","title":{"rendered":"Week 5 – SS3 First Term Mathematics Notes"},"content":{"rendered":"

WEEK 5
\n<\/strong>Factorization of Quadratic Expressions.
\nFactorize the following:<\/p>\n

    \n
  1. 6a2<\/sup> + 15a + 9\n<\/li>\n
  2. 6a2<\/sup> \u2013 19ax – 36x2<\/sup>\n\t\t<\/li>\n
  3. (5x \u2013 1) ( x \u2013 3) – ( x \u2013 5) ( x \u2013 3)\n<\/li>\n
  4. 35 \u2013 2b – b2<\/sup>\n\t\t<\/li>\n
  5. x2<\/sup> \u2013 y2<\/sup> + ( x + y ) 2<\/sup>\n\t\t<\/li>\n
  6. 25a2<\/sup> – 4 ( a \u2013 2b ) 2<\/sup>.\n<\/li>\n<\/ol>\n

    Solutions.
    \n6a2 + 15x + 9.
    \nSince 3 is a common factor to all the terms, first take out 3 as the common factor:
    \n6a2<\/sup> + 15a + 9 = 3 (2a2+ 5a + 3)
    \n = 3 (2a2<\/sup> + 2a + 3a + 3)
    \n = 3 (2a (a+1) + 3 (a+1)
    \n= 3 ( (a+1) (2a+ 3)).
    \nHence.
    \n\"\"\/6a2<\/sup> + 15a + 9 = 3 (a+ 1) (2a + 3)
    \n = 3 (a+1) ( 2a + 3)<\/p>\n

    \u00a0<\/p>\n

      \n
    1. 6a2<\/sup> \u2013 19ax \u2013 36x2<\/sup>\n\t\t<\/li>\n<\/ol>\n

      1st<\/sup> step: Find the product of the first and last terms.
      \n\u00a0\u00a0\u00a0\u00a06a2<\/sup> x -36x2<\/sup> = -216a2<\/sup>x2<\/sup>
      \n\t2nd<\/sup> step: Find two terms such that their products is \u2013 216a2x and their sum is -19ax ( the middle term).
      \nFactors of -216a2<\/sup>x2<\/sup> sum of factors .<\/p>\n

      \u00a0a. +27ax and -8ax + 19ax
      \nb. -27ax and +8ax – 19ax
      \nc. +9ac and -24ax – 15ax
      \nd. -9x and 24ax + 15ax.
      \nOf these only b gives the required result.
      \n3rd<\/sup> step: Replace -19ax in the given expression by -27ax and 8ax then factorize by grouping:
      \n6a2<\/sup> \u2013 19ax \u2013 36x2<\/sup>
      \n\t= 6a2<\/sup> \u2013 27ax + 8ax \u2013 36x2<\/sup>
      \n\t= 3a (2a \u2013 9x) + 4x (2a-9x)
      \n=(2a \u2013 9x ) ( 3a + 4x)
      \nhence,
      \n6a2<\/sup> \u2013 19ax \u2013 36x2<\/sup> = (2a -9x) (3a + 4x)<\/p>\n

        \n
      1. (5x \u2013 1 (x -3) – (x -5)(x \u2013 3)\n<\/li>\n<\/ol>\n

        \"\"\/= (x \u2013 3) ( 5x -1) – (x \u2013 5)
        \n= ( x -3) ( 5x \u2013 1 – x + 5 )
        \n= ( x \u2013 3) ( 5x – x + 5 \u2013 1)
        \n= ( x \u2013 3) ( 4x + 4)
        \n= ( x \u2013 3) + ( x + 1)
        \n= ( x -3) 4(x+1)
        \n= 4(x-3) (x+1)
        \n 4. 35 \u2013 2b \u2013b2<\/sup>
        \n\t or – b2<\/sup> \u2013 2b + 35
        \n = -b2 \u2013 7b + 5b + 35
        \n = -b (b+7) + 5 (b + 7)
        \n = (b+7) ( -b + 5)
        \n = (b+7) ( 5-b)
        \nor (7+b) ( 5 \u2013 b)<\/p>\n

        \u00a05. x2<\/sup> \u2013 y2<\/sup> + (x + y)2<\/sup>
        \n\tSince
        \nX2<\/sup> \u2013 y2<\/sup> = (x)2<\/sup> \u2013 (y)2<\/sup>
        \n\t= ( x + y) ( x \u2013y)
        \nthen x2<\/sup> \u2013 y2<\/sup> + (x + y)2<\/sup> = (x + y)(x-y) + (x+ y)2<\/sup>
        \n\t= (x +y)( x-y + (x + y)
        \n= (x + y ( x \u2013y + x + y)
        \n= ( x + y ) ( x + x
        \n= (x + y) (2x)
        \n(x +y ) (2x) = 2x ( x +y)<\/p>\n

        \u00a0EVALUATION.
        \nFactorize the following <\/p>\n

          \n
        1. m2<\/sup> \u2013 15mm \u2013 54n2<\/sup>\n\t\t<\/li>\n
        2. 8a2<\/sup> \u2013 18b2<\/sup>\n\t\t<\/li>\n
        3. If 17x = 37 52<\/sup> – 3562<\/sup>\n\t\t<\/li>\n<\/ol>\n

          Find the value of x
          \nASSIGNMENT
          \nFactorize the following :<\/p>\n

            \n
          1. a2<\/sup> + 5ab \u2013 36b2<\/sup>\n\t\t<\/li>\n
          2. r2<\/sup> \u2013 25\n<\/li>\n
          3. 10p2<\/sup> \u2013 41p – 45\n<\/li>\n
          4. 12m2<\/sup> \u2013 4mn -5n2\n<\/li>\n
          5. 12a2<\/sup>b2<\/sup> + 11ab \u2013 5.\n<\/li>\n<\/ol>\n

            Theory.
            \nFactorize the following <\/p>\n

              \n
            1. 8c2<\/sup> \u2013 14c \u2013 9\n<\/li>\n
            2. x2<\/sup>y2<\/sup> \u2013 xy \u2013 30.\n<\/li>\n<\/ol>\n

              \u00a0If I is a root of the equation 5x2<\/sup> + kx \u2013 3 = 0
              \nFind the other root.
              \n3. Given that \u00bd and -3 are the roots of the equation 0 = ax2 <\/sup>+ bx + C, find a, b,c, where a, b and c are the least possible integers.<\/p>\n

              \u00a0Solutions<\/p>\n

                \n
              1. Since 2 \u00bd is a root of the equation, then (x \u2013 2 \u00bd ) is a factor, similarly, if -1 is a root of the equation then (x \u2013 (1-1) is a factor. i.e (x + 1) is a factor.\n<\/li>\n<\/ol>\n

                Thus, the required equation is : (x \u2013 2 \u00bd ) ( x + 1) = 0
                \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x (x + 1) \u2013 2 \u00bd ( x + 1) = 0
                \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> + x – 2 \u00bd x – 2 \u00bd = 0
                \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> \u2013 1 \u00bd x – 2 \u00bd = 0.
                \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0X2<\/sup> \u2013 3<\/sup>\/2 x – 5<\/sup>\/2 =0
                \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x2<\/sup> \u2013 3x \u2013 5 =0
                \n2. Since I is a root of the given equation
                \n\u00a0\u00a0\u00a0\u00a05x2<\/sup> + kx \u2013 3 = 0
                \nthen it must satisfy the given equation.
                \nSubstitute I for x,
                \n5(1)2<\/sup> + k(1) – 3 = 0
                \n5 x 1 + k \u2013 3 = 0
                \n5 + k \u2013 3 = 0
                \n\\5-3 + k = 0
                \n2 + k = 0
                \nk = 0-2
                \nk = -2.
                \nHence the given equation becomes
                \n5x2<\/sup> \u2013 2x \u2013 3 = 0
                \nby factorization method, this quadratic equations can be solved as follows :
                \n5x2<\/sup> \u2013 2x \u2013 3 = 0
                \n5x2<\/sup> \u2013 5x + 3x \u2013 3 = 0
                \n5x (x -1) + 3 ( x -1) = 0
                \n(x-1) ( 5x + 3) = 0
                \n(x -1) = 0 or 5x + 3 = 0
                \ni.e. x = 0+ 1 or 5x = 0-3
                \nx = 1 or 5x = -3
                \nx = 1 or x = –3<\/sup>\/5.
                \nHence the other roots of the given quadratic equation is x = 3<\/sup>\/5<\/sub>.<\/p>\n

                  \n
                1. Since the given roots of the equation 0 = ax2<\/sup> + bx + C are \u00bd and -3, then the factors are (x \u2013 \u00bd ) and (x \u2013 c \u2013 3)\n<\/li>\n<\/ol>\n

                  i.e. ( x \u2013 \u00bd ) and ( x + 3)
                  \nThus, the required equation is
                  \n(x \u2013 \u00bd ) ( x + 3) = 0
                  \nx(x+ 3) – \u00bd (x + 3) = 0
                  \nx2<\/sup> + 3x \u2013 \u00bd x – 3\/2 = 0
                  \n2x2<\/sup> + 6x \u2013 x – 3 = 0
                  \n2x2<\/sup> + 5x \u2013 3 = 0
                  \nComparing this with the given equation ax2<\/sup> + bx + C = 0
                  \nThen
                  \nA= 2, b= 5, c = -3.
                  \nEVALUATION
                  \nI. 7 and -3 are the roots of the quadratic equation x2<\/sup> +kx -21 = 0,what is the value of k?
                  \n2. -5 is a root of the
                  \n ASSIGNMENT
                  \n1.Find the quadratic equation whose roots are 2 and -3.
                  \n (a) (x2<\/sup> + x -6 (b) x2<\/sup> \u2013 x -6 (c ) x2<\/sup> + 5x \u2013 6 (d) x2<\/sup> \u2013 5x + 6 (e) x2<\/sup> + 3x \u2013 6.
                  \n2. (2x + 3) is a factor of 6x2<\/sup> + – 12. The other factor is \u2026\u2026\u2026\u2026.
                  \n (a) (x + 6) (b) (2x \u2013 3) (c) (3x + 4) (d) (3x \u2013 4) (e) (4x \u2013 9).
                  \n3. Find the roots of the equation x2 + 12x \u2013 28 = 0 the <\/p>\n

                  \u00a0Examples<\/p>\n

                  \u00a0<\/p>\n

                    \n
                  1. \n
                    The electrical resistance R ohms of a wire varies directly as the length cm and inversely as the square roots of the diameter d cm3.\n<\/div>\n
                      \n
                    1. Express d in terms of l,R and the constant of variation k.\n<\/li>\n
                    2. Find the value of d, correct to 2 decimal places, when 1 = 15cm, R = 0.13ohms and k = 1.25 x 10-3.\n<\/li>\n<\/ol>\n<\/li>\n
                    3. V varies jointly as the square of x and inversely as y, if V =18 when x = 3 and y =4, find V when x = 6 and y =9.\n<\/li>\n<\/ol>\n

                      Solutions.<\/p>\n

                      \u00a0The greater of the two roots is \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026
                      \n(a) -14\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) -2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c ) 2 (d) 14
                      \n4. What is the product of the roots of the equation x2<\/sup> -3x + 2 =0?
                      \n\u00a0\u00a0\u00a0\u00a0(a) -3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) -1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d ) 3 \u00a0\u00a0\u00a0\u00a0 (e)5
                      \n5. If x2<\/sup> \u2013 10x -24 =0, the sum of the roots of the equation is \u2026\u2026\u2026\u2026.
                      \n\u00a0\u00a0\u00a0\u00a0(a) \u2013 10 (b) 10 (c ) -24 (d) -2
                      \nTHEORY<\/p>\n

                        \n
                      1. find the quadratic equation whose roots are (3 + \u221a5).\n<\/li>\n
                      2. If 1 is root of the equation 5×2 + kx \u2013 3 =0, find the other root.\n<\/li>\n<\/ol>\n

                        \u00a0
                        \n\u00a0Quadratic equation n2<\/sup> \u2013 8n \u2013 65 =0, what is the other root of the equation?
                        \nTopic: Solution of Quadratic Equation by Factorization Method
                        \n<\/strong>Examples:
                        \nSolve the following quadratic equations by the factorization methods.<\/p>\n

                          \n
                        1. m2<\/sup> = 11m + 42\n<\/li>\n
                        2. 2x2<\/sup> \u2013 x \u2013 21 =0\n<\/li>\n
                        3. 2p2<\/sup> + 11p = 30.\n<\/li>\n<\/ol>\n

                          Solutions.
                          \n(1 ) m2<\/sup> = 11m + 42
                          \nthen m2<\/sup> \u2013 11m \u2013 42 =0
                          \n1st<\/sup> step: Find the product of the first and last terms :
                          \nm2<\/sup> x \u2013 42<\/sup> =-42m2<\/sup>
                          \n\t2nd<\/sup> step:find two terms such that their product is -42m2 and their sum is -11m ( the middle term).
                          \nFactors of -42m2<\/sup> sum of factor s
                          \n(a) + 3m and \u2013 14m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0– 11m
                          \n(b) – 6m and +7m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0+ 11m
                          \n( c) -6m and + 7m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0+m
                          \n(d) +6m and -7m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-m
                          \nOut of these, only (a) gives the required result.
                          \n3rd<\/sup> step: Replace -11 in the given expression by +3m and -14m. then factorise by grouping:
                          \nm2<\/sup> -11m -42 = 0.
                          \nM2<\/sup> + 3m – 14m \u2013 42 =0
                          \nM(m+ 3) (m \u2013 14) = 0
                          \nM + 3 = 0 or m \u2013 14 =0
                          \ni.e m = 0 -3, or m = 0 + 14.
                          \nM = -3 or m = 14.
                          \n2. 2x2<\/sup> \u2013x -21 =0
                          \n1st<\/sup> Step: Product of first and last terms = 2x2<\/sup> x -21
                          \n\u00a0\u00a0\u00a0\u00a0= – 42x2<\/sup>
                          \n\t2nd<\/sup> step: Find two terms such that their product is 42x2<\/sup> and their sum is \u2013x (the middle term).
                          \nFactors of-42x2<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0sum of factors
                          \n(a) + 2x and -21x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-19x
                          \n(b) -2x and 21x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0+19x
                          \n( c)-3x and + 14x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0+ 11x
                          \n(d) + 3x and -14x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-11x
                          \n( e) +6x and -7x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-x
                          \n(f) -6x and + 7x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0+x
                          \nout of these, only (e)gives the required result.
                          \n3rd<\/sup> step: Replace \u2013x in the given expression by +6x and -7x. then factorize by grouping
                          \n2x2 <\/sup> – x -21 =0
                          \n2x2<\/sup> +6x -7x \u2013 21 = 0
                          \nx ( x + 3) -7 ( x + 3 ) = 0
                          \n( x + 3) ( x-7) =0
                          \neither x + 3 = 0 or 2x -7 =0
                          \ni.e x = 0-3 or 2x = 7
                          \nx = -3 or x =7<\/sup>\/2
                          \nx = -3 or x = 3 \u00bd .<\/p>\n

                          \u00a03. 2p2<\/sup> + 11p = 30.
                          \n 2p2<\/sup> + 11p \u2013 30 = 0
                          \n1st<\/sup> step: Product of the first and last terms
                          \n 2p2<\/sup> x -30 =- 60p2<\/sup>
                          \n\t2nd<\/sup> step: find two terms such that their products is -60p2<\/sup> and their sum is +11p (the middle term )
                          \nFactors of +11\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0sum of factors
                          \n(a) + 6p and \u2013 10p\u00a0\u00a0\u00a0\u00a0-4p
                          \n(b) -15p and +4p\u00a0\u00a0\u00a0\u00a0+4p
                          \n( c) -15p and +4p\u00a0\u00a0\u00a0\u00a0-11p
                          \n(d) +15p and -4p\u00a0\u00a0\u00a0\u00a0+11p.
                          \nOut of these, only (d) gives the more required solution. Replace + 11p by +15p and -4p in the given expression .
                          \n2p2<\/sup> + 15p -4p -30 = 0
                          \np ( 2p + 15) \u2013 2 (2p + 15) = 0
                          \n(2p + 15) (p \u2013 2) = 0
                          \neither 2p+ 15 = 0 or p -2 = 0
                          \n2p = 0 \u2013 15 or p = 0 + 2
                          \n2p = -15 or p = 2
                          \np = -15 or p = 2
                          \np = –15<\/sup>\/2 or p = 2
                          \np = -7 \u00bd or p = 2.
                          \nEVALUATION.
                          \nSolve the following quadratic equations by factorization method.<\/p>\n

                            \n
                          1. 4e2 <\/sup> – 20 e + 25 = 0\n<\/li>\n
                          2. \n
                            4a2<\/sup> \u2013 11a = 3\n<\/div>\n

                            \u00a0<\/li>\n<\/ol>\n

                            ASSIGNMENT.
                            \n1. If x2<\/sup> – 10 x \u2013 24 = 0, then x = 12 or \u2026\u2026\u2026\u2026\u2026
                            \n\u00a0\u00a0\u00a0\u00a0(a) -3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) -2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) -1 (d) 1 (e) 2
                            \n2. if x2<\/sup> + Kx + 16\/9 is a perfect square, then K = \u2026\u2026\u2026\u2026\u2026\u2026..
                            \n\u00a0\u00a0\u00a0\u00a0(a) 1<\/sup>\/3 (b) 2<\/sup>\/3 (c) 4<\/sup>\/2 (d) 8<\/sup>\/3 (e) 16<\/sup>\/3
                            \n3. What is the sum of the roots of the equation
                            \n\u00a0\u00a0\u00a0\u00a0x2<\/sup> -3x +2 =0?
                            \n\u00a0\u00a0\u00a0\u00a0(a) \u2013 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) -1 ( c) 2 (d )3 (e) 5
                            \n4. Find the roots of the equation
                            \n\u00a0\u00a0\u00a0\u00a0x2<\/sup> + 12x \u2013 28 = 0
                            \nThe greater of the two roots is \u2026\u2026\u2026\u2026.
                            \n(a) -14 (b) -2 (c ) 2(d)7 (e ) 14.
                            \n5. What are the factors of 6×2 + x \u2013 12?
                            \nA.(2x+ 3)( x + 6) B. (2x +3) ( 2x \u2013 3) C. (2x +3)(3x + 4) D. (2x + 3) (3x -4)
                            \nE. (2x + 3 ) ( 4x -9).<\/p>\n

                            \u00a0THEORY.
                            \nSolve the following equations”
                            \n1. 4y2<\/sup> – 28y + 49 =0
                            \n2. 2x2<\/sup> + 11x + 5 = 0.<\/p>\n

                            \u00a0
                            \n\u00a0Completing the square method
                            \n<\/strong>Given a quadratic equation,
                            \n<\/strong>a +bx=0 (where a, b and c are constants)
                            \n\t\t<\/strong>a +bx =-c (subtract c from both sides)
                            \n<\/strong> + (divide both sides by a)
                            \n<\/strong>The coefficient of x=
                            \n<\/strong>Divide by 2 =
                            \n\t\t\t<\/strong>Square and add to both sides i.e
                            \n\t\t<\/strong> + = i.e.
                            \n<\/strong>(X + = or
                            \n<\/strong>X +
                            \n<\/strong>X +
                            \n<\/strong>X =
                            \n<\/strong>X = . Q.E.D.
                            \n<\/strong>
                            \n\u00a04Y + 4 = 0
                            \nSOLUTION
                            \n4Y = 4 (Re-arrange the equation)
                            \n (Divide through by coefficient of )
                            \n4Y+ ( = . Half Y, square it and add it to both sides.
                            \n =
                            \n(Y = 0 Y2 =
                            \nY = 0 +2 Y = +2 or 2
                            \nOR using method of difference of two squares
                            \n (Y2)(Y+2) = 0
                            \nY 2 = 0 or Y + 2 = 0
                            \nY = 2 or Y = 2.
                            \nExample – Solve the equation using completing the square method.<\/strong>
                            \n\t(1)\u00a0\u00a0\u00a0\u00a0X<\/sub>2<\/sup> \u2013 8x + 3 = 0
                            \n(2)\u00a0\u00a0\u00a0\u00a03x2<\/sup> \u2013 5x \u2013 7 = 0
                            \nSolution
                            \n1.\u00a0\u00a0\u00a0\u00a0x2<\/sup> \u2013 8x + 3 = 0
                            \n\u00a0\u00a0\u00a0\u00a0Step 1:<\/strong> Take C to the other side x2<\/sup> \u2013 8x = -3
                            \n\u00a0\u00a0\u00a0\u00a0
                            \nStep 2:<\/strong> Divide through by the coefficient of x2<\/sup>
                            \n\t\u00a0\u00a0\u00a0\u00a0x2<\/sup> \u2013 8x = -3
                            \n\t\u00a0\u00a0\u00a0\u00a0 1 1 1<\/p>\n

                            \u00a0Step 3:<\/strong> Divide the coefficient of x by 2, then square and add it to both sides.
                            \nCo efficient of x = 8
                            \n\u00a0\u00a0\u00a0\u00a0:\u00a0\u00a0\u00a0\u00a08 = 4
                            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02
                            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a042<\/sup> = 16
                            \n=x2<\/sup> \u2013 8x + (-8)2<\/sup> = -3 + (- 8)2<\/sup>
                            \n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02
                            \nx2<\/sup> \u2013 8x + (-3)2<\/sup> = -3 + (-4)2<\/sup>
                            \n\t(x \u2013 4)2<\/sup> = -3 + 16
                            \n(x \u2013 4)2<\/sup> = 13
                            \nx \u2013 4 = + 13
                            \nx = +4 + 13
                            \nx = 4 + 3.61
                            \nx = 4 \u2013 3.61\u00a0\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\u00a04 + 3.61
                            \n\u00a0\u00a0\u00a0\u00a00.39\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\u00a07.61<\/p>\n

                            \u00a02.\u00a0\u00a0\u00a0\u00a03x2<\/sup> \u2013 5x \u2013 7 = 0
                            \n\u00a0\u00a0\u00a0\u00a03x2<\/sup> \u2013 5x = +7
                            \n\u00a0\u00a0\u00a0\u00a03x2<\/sup> \u2013 5x = 7
                            \n\t\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a03 3\u00a0\u00a0\u00a0\u00a0
                            \n\u00a0\u00a0\u00a0\u00a0x2<\/sup> \u2013 5x = 7
                            \n\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0 3
                            \n\u00a0\u00a0\u00a0\u00a0x2<\/sup> \u2013 5x + (-5)2<\/sup> = 7 + (-5)2<\/sup>
                            \n\t\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0 6
                            \n\u00a0\u00a0\u00a0\u00a0(x \u2013 5)2<\/sup> = 7 + 25
                            \n\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0 3 36
                            \n(x \u2013 5)2<\/sup> = 84 + 25
                            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 36<\/p>\n

                            \u00a0\"\"\/\u00a0\u00a0\u00a0\u00a0(x -5)2<\/sup>\u00a0\u00a0\u00a0\u00a0 = +109
                            \n\t\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0 36
                            \n\"\"\/\u00a0\u00a0\u00a0\u00a0
                            \nx = 5 + +109
                            \n\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0 6
                            \n\u00a0\u00a0\u00a0\u00a0x \u2013 5 = +10.44
                            \n\u00a0\u00a0\u00a0\u00a0 6 6
                            \nx \u2013 5 = +10.44
                            \n\u00a0\u00a0\u00a0\u00a0 6 6
                            \n\u00a0\u00a0\u00a0\u00a0x = 5 + 10.44 or\u00a0\u00a0\u00a0\u00a05 \u2013 10.44
                            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6
                            \n\u00a0\u00a0\u00a0\u00a0x = 15.44\u00a0\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\u00a0-5.44
                            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a06\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6
                            \n\u00a0\u00a0\u00a0\u00a0x = 2.5.573\u00a0\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\u00a0– 0.906
                            \n\u00a0\u00a0\u00a0\u00a0 = 2.57\u00a0\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\u00a0– 0.91
                            \n\u00a0\u00a0\u00a0\u00a0m = 7 + 2.24
                            \n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2
                            \n\u00a0\u00a0\u00a0\u00a0m = 7 + 2.24 or\u00a0\u00a0\u00a0\u00a07 \u2013 2.24
                            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2
                            \n\u00a0\u00a0\u00a0\u00a0m = 9.24\u00a0\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\u00a05.24
                            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2
                            \n\u00a0\u00a0\u00a0\u00a0m = 4.62 or 2.62<\/p>\n

                            \u00a02.\u00a0\u00a0\u00a0\u00a03x2<\/sup> + 7x -3 = 0
                            \n\u00a0\u00a0\u00a0\u00a03x2<\/sup> + 7x = 3
                            \n\u00a0\u00a0\u00a0\u00a03x2<\/sup> + 7x = 3
                            \n\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a03 3
                            \n\u00a0\u00a0\u00a0\u00a0x2<\/sup> + 7x = 1
                            \n\u00a0\u00a0\u00a0\u00a0 3
                            \n\u00a0\u00a0\u00a0\u00a0x2<\/sup> + 7x + (7)2<\/sup> = 1 + (7)2<\/sup>
                            \n\t\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6
                            \n\u00a0\u00a0\u00a0\u00a0(x + 7)2<\/sup> = 1 + 49
                            \n\t\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 36
                            \n\u00a0\u00a0\u00a0\u00a0(x + 7)2<\/sup> = 1 + 36 + 49
                            \n\t\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 36
                            \n\u00a0\u00a0\u00a0\u00a0(x + 7)2<\/sup> = 85
                            \n\t\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0 36
                            \n\"\"\/\u00a0\u00a0\u00a0\u00a0x + 7 = +85
                            \n\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0 36
                            \n\"\"\/\u00a0\u00a0\u00a0\u00a0
                            \nx + 7 = +85
                            \n\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0 6
                            \n\u00a0\u00a0\u00a0\u00a0x + 7 = +9.22
                            \n\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0 6
                            \n\u00a0\u00a0\u00a0\u00a0x = -7 + 9.22 or -7 \u2013 9.22
                            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6
                            \n\"\"\/\"\"\/\"\"\/\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n

                            \u00a0Examples
                            \nSolve the following equations by completing the square method.<\/p>\n

                              \n
                            1. n2<\/sup> – 12n + 1 = 0\n<\/li>\n
                            2. y2<\/sup> +7y – 30 =0\n<\/li>\n
                            3. m2<\/sup> -7m + 11 = 0\n<\/li>\n<\/ol>\n

                              Solutions
                              \n1. n2<\/sup> – 12n + 1 = 0
                              \n n2<\/sup> \u2013 12n = 0 \u2013 1
                              \n n2<\/sup> \u2013 12n = -1.
                              \nAdd to both sides the square of –12<\/sup>\/2<\/sub> ( -6).\u00a0\u00a0\u00a0\u00a0
                              \nn2<\/sup> \u2013 12n ( -6) 2<\/sup> = -1 + (-6)2<\/sup>
                              \n\t(n \u2013 6) 2 <\/sup> = -1 + (-6) 2<\/sup>
                              \n\t(n \u2013 6) 2<\/sup> =+ 35.
                              \nTake square root of both sides.
                              \n n-6 = \u00b1 \u221a35.
                              \ni.e n = \u00b1 \u221a35 + 6
                              \n n = + \u221a35 + 6 or n = -\u221a35 + 6
                              \n n = 6 + \u221a35 or n = 6 – \u221a35
                              \n n = 6 + 5.916 or n = 6 \u2013 5.916
                              \n n = 11.916 or n = 0.084.
                              \n i. e . n = 11.92 or n = 0.08 to 2 decimal places.<\/p>\n

                              \u00a02.y2<\/sup> +7y \u2013 30 = 0
                              \n y2<\/sup> + 7y = 0 + 30
                              \n y2<\/sup> + 7y = 30
                              \nAdd to both side the square of 7<\/sup>\/2
                              \ny2<\/sup> +7y + ( 7<\/sup>\/2)2<\/sup> = 30 + (7<\/sup>\/2) 2<\/sup>
                              \n\ti.e. ( y +7<\/sup>\/2) 2 <\/sup> = 30+ 49
                              \n 1 4.
                              \n( y + 7<\/sup>\/2<\/sub>) 2<\/sup> = 120 + 49
                              \n 4
                              \n( y + 7<\/sup>\/2<\/sub>)2<\/sup> = 169
                              \n 4.
                              \nTake square root of both sides :
                              \n:. Y + 7<\/sup>\/2<\/sub> = \u00b1 \u221a169\/4\/
                              \ny + 7<\/sup>\/2<\/sub> = \u00b1 13<\/sup>\/2<\/sub>
                              \n\ti.e y = \u00b1 13<\/sup>\/2<\/sub> –7<\/sup>\/2<\/sub>
                              \n\t.e y = +13<\/sup>\/2 – 7<\/sup>\/2 or y = – 13- 7
                              \n 2.
                              \ny = +6 or y = -20
                              \n 2 2
                              \ny = +3 or y = -10.<\/p>\n

                              \u00a03.m2<\/sup> \u2013 7m + 11 = 0
                              \nm2<\/sup> \u2013 7m = 0 \u2013 11
                              \nm2<\/sup> -7m = -11
                              \nAdd to both sides the square of –7<\/sup>\/2<\/sub>
                              \n\tM2<\/sup> \u2013 7m ( –7<\/sup>\/2<\/sub>)2<\/sup> = -11 + (-7<\/sup>\/2<\/sub>)2<\/sup>
                              \n\t( m \u2013 7<\/sup>\/2<\/sub>)2<\/sup> = -11 + 49
                              \n\t 4
                              \n( m \u2013 7<\/sup>\/2<\/sub>)2<\/sup> = -44 + 49
                              \n 4.
                              \n( m \u2013 7<\/sup>\/2<\/sub>)2<\/sup> = 5<\/sup>\/4<\/sub>
                              \n\tTake square root of both sides:
                              \nm –7<\/sup>\/2<\/sub> = \u00b1 \u221a5<\/sup>\/4<\/sub>
                              \n\tm= \u00b1\u221a5<\/sup>\/4<\/sub> + 7<\/sup>\/2<\/sub>
                              \n\tm=\u00b1 \u221a+ 7<\/sup>\/2<\/sub>
                              \n\ti.e. m = + \u221a5 + 7 or m = – \u221a5 + 7
                              \n 2 2 2 2.
                              \nm = + 2.236 +7 or m = – 2.236 + 7
                              \n 2 2 2 2
                              \nm=+1.118 + 3.5 or m = -1.118 + 3. 5
                              \nm = + 4.618 or m = 2.418
                              \ni.e m = 4.62orm = 2.42.
                              \nto 2 decimal places
                              \nEVALUATION.
                              \nSolve the following method of completing the squares:<\/p>\n

                                \n
                              1. x2<\/sup> \u2013 8x \u2013 1 = 0\n<\/li>\n
                              2. p2<\/sup> -10p + 15 = 0\n<\/li>\n<\/ol>\n

                                WEEKEND ASSIGNMENT
                                \n1. The greater of the two roots of the equations (2x -5) (3x +10 ) = 0 is
                                \n\u00a0\u00a0\u00a0\u00a0(a) \u2013 50 (b) -10 ( c) -3 (d) 2 \u00bd ( e) 5.
                                \n2. If one of the roots of the equation ( n-13)2 = 9 is 10, what is the other root?
                                \n (a) 4\u00a0\u00a0\u00a0\u00a0(b) 16 ( c) 22 (d) 68 ( e ) 94.
                                \n3. Find the value of k such that x2<\/sup> + 5x +k is a perfect square.
                                \n\u00a0\u00a0\u00a0\u00a0(a) 2 \u00bd (b) 4 \u00bc ( c ) 6 \u00bc ( d) 25 ( e) 100.
                                \n4. If x2<\/sup> \u2013 10x = 24, what is the values of x ?
                                \n\u00a0\u00a0\u00a0\u00a0(a) 12,-3) (b) ( -2, 12) ( c) -1, 12) (d) ( 1, 12) (e) (2,12).
                                \n5. What must be added to a2<\/sup> + 3<\/sup>\/5a to make it a perfect square?
                                \n\u00a0\u00a0\u00a0\u00a0(a) 3\/10 (b) 6\/5 ( c) 9\/100 (d) 100\/9 ( e) 36\/5
                                \nTHEORY
                                \nSolve the following equations, giving answers correct to 1 decimal place.<\/p>\n

                                  \n
                                1. x2<\/sup> + 14x + 8 = 0\n<\/li>\n
                                2. 2p2<\/sup> +9p – 30 =0\n<\/li>\n<\/ol>\n

                                  \u00a0
                                  \n\u00a0Quadratic Formula
                                  \n<\/strong>Quadratic formula is derived by using the method of completing the square. Considering the general form of quadratic equation:
                                  \n\"\"\/\u00a0\u00a0\u00a0\u00a0ax2<\/sup> + bx + c = 0\u00a0\u00a0\u00a0\u00a0<\/em>X = The required quadratic formula.
                                  \n\t\t\t\t<\/em><\/strong>Example – Solve x2<\/sup> + 3x \u2013 2 = 0 using formula method
                                  \n\"\"\/a = 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a06 = 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0C = – 2
                                  \n\"\"\/x = -b b2<\/sup> \u2013 4ac
                                  \n\t\"\"\/\"\"\/\u00a0\u00a0\u00a0\u00a0 2a\u00a0\u00a0\u00a0\u00a0
                                  \nx = -3 32<\/sup> \u2013 4(i) (-2)
                                  \n\"\"\/\u00a0\u00a0\u00a0\u00a0 2(i)
                                  \n\"\"\/x = -3 9 + 8
                                  \n\u00a0\u00a0\u00a0\u00a0 2
                                  \n\"\"\/x = -3 17
                                  \n\"\"\/\u00a0\u00a0\u00a0\u00a0 2
                                  \nx = – 3 4.12
                                  \n 2
                                  \nx = -3 + 4.12\u00a0\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\u00a0-3 \u2013 4.12
                                  \n\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2
                                  \nx = 0.56\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\u00a0– 3.56<\/p>\n

                                  \u00a0
                                  \n\u00a02.\u00a0\u00a0\u00a0\u00a0m2<\/sup> \u2013 7m + 11 = 0
                                  \n\u00a0\u00a0\u00a0\u00a0Solve using formula method.
                                  \n\u00a0\u00a0\u00a0\u00a0m2<\/sup> \u2013 7m + 11 = 0
                                  \n\"\"\/\u00a0\u00a0\u00a0\u00a0a = 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0b = – 7\u00a0\u00a0\u00a0\u00a0c = 11
                                  \n\"\"\/\u00a0\u00a0\u00a0\u00a0m = – b b2<\/sup> \u2013 4ac
                                  \n\"\"\/\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2a
                                  \n\u00a0\u00a0\u00a0\u00a0m = – (-7) 72<\/sup> \u2013 4(i) (ii)
                                  \n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02(i)
                                  \n\"\"\/\u00a0\u00a0\u00a0\u00a0m = 7 49 \u2013 44
                                  \n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2
                                  \n\"\"\/\u00a0\u00a0\u00a0\u00a0m = 7 5
                                  \n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02
                                  \n\u00a0\u00a0\u00a0\u00a0m = 7 2.23
                                  \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2<\/p>\n

                                  \u00a0\u00a0\u00a0\u00a0\u00a0m = 7 + 2.23\u00a0\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\u00a07 \u2013 2.23
                                  \n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0 2
                                  \n\u00a0\u00a0\u00a0\u00a0m = 4.62\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0or \u00a0\u00a0\u00a0\u00a02.62
                                  \nEVALUATION
                                  \n<\/strong>1.\u00a0\u00a0\u00a0\u00a03x2<\/sup> + 7x \u2013 3 = 0\u00a0\u00a0\u00a0\u00a0solve using formula method
                                  \n2.\u00a0\u00a0\u00a0\u00a0Using completing the square and formula method solve 3x2<\/sup> \u2013 12x + 10 = 0
                                  \nSum & Product of Roots of a Quadratic Equation
                                  \n<\/strong>The expression for sum & product of roots of quadratic equation is gotten from the general expression of quadratic equation. If the distinct roots are \u03b1 and \u03b2, then
                                  \n\u03b1 + \u03b2 = -b\/a (sum of roots)
                                  \n\u03b1\u03b2 =c\/a (product of roots)
                                  \nExample 1 \u2013 find the sum and products of 2x2<\/sup> + 3x – 1 = 0
                                  \nSolution
                                  \n2x2<\/sup> + 3x \u2013 1 = 0
                                  \na =2, b = 3, c = -1
                                  \nLet \u03b1 and \u03b2 be the roots of the equation, then
                                  \n\u03b1+\u03b2= -b\/a= -3\/2
                                  \n\u03b1\u03b2 = c\/a = -1\/2
                                  \nExample 2 – find the sum and products of 3x2<\/sup> \u2013 5x -2 = 0
                                  \nSolution
                                  \n3x2<\/sup> -5x -2 =0
                                  \na= 3, b= -5, c= -2
                                  \nlet \u03b1 and \u03b2 be the roots of the equation, then
                                  \n\u03b1+\u03b2= -b\/a = 5\/3
                                  \n\u03b1\u03b2= c\/a= -2\/3
                                  \n\t\t<\/strong>NB: <\/strong>The quadratic equation whose root are \u03b1 and \u03b2 is
                                  \n (X – \u03b1 )(X – \u03b2) = 0
                                  \n<\/strong>X2<\/sup> \u2013 (\u03b1 +\u03b2)X + \u03b1\u03b2 = 0
                                  \n<\/strong>Example \u2013 Find the quadratic equation whose roots are 3 & -2
                                  \nSolution
                                  \n\u03b1=3 and \u03b2=-2
                                  \n\u03b1+\u03b2 = 3 + (-2) = 1
                                  \n\u03b1\u03b2 = 3 x -2 = -6
                                  \nX2<\/sup> \u2013 (\u03b1 +\u03b2)x + \u03b1\u03b2 = 0
                                  \nX2 <\/sup>\u2013 (1)X + (-6) = 0
                                  \nX2<\/sup> \u2013 X -6 = 0
                                  \nSymmetric Properties of Roots
                                  \n<\/strong>Certain relations involving \u03b1 and \u03b2 can also be determined from \u03b1+\u03b2 and\u03b1\u03b2 even when we do not know \u03b1 and \u03b2 distinctly. Such relations are usually said to be symmetric. They are symmetric in the sense that if \u03b1 and \u03b1 are interchanged, either the relation remains the same or is multiplied by -1
                                  \nExample \u2013 If \u03b1 \u2260 \u03b2, determine whether or not each of the following is symmetric
                                  \n(a) \u03b1+\u03b2 (b) \u03b1\u03b2 (c) \u03b12<\/sup> + \u03b22<\/sup> (d) \u03b12<\/sup> – \u03b22<\/sup> (e) 3\u03b1 +2\u03b2 (f) \u03b13<\/sup> + \u03b23<\/sup>
                                  \n\tSolution<\/p>\n

                                    \n
                                  1. \n
                                    \u03b1+\u03b2 = \u03b2 +\u03b1 , therefore \u03b1+\u03b2 is symmetric\n<\/div>\n<\/li>\n
                                  2. \n
                                    \u03b1\u03b2 =\u03b2\u03b1, therefore \u03b1\u03b2 is symmetric\n<\/div>\n<\/li>\n
                                  3. \n
                                    \u03b12<\/sup> + \u03b22<\/sup> = \u03b22<\/sup> + \u03b12<\/sup>, therefore \u03b12<\/sup> + \u03b22<\/sup> is symmetric\n<\/div>\n<\/li>\n
                                  4. \n
                                    \u03b12<\/sup> – \u03b22<\/sup> = – (\u03b22<\/sup> – \u03b12<\/sup>), therefore\u03b12<\/sup> – \u03b22<\/sup> is symmetric\n<\/div>\n<\/li>\n
                                  5. \n
                                    3\u03b1 +2\u03b2 \u2260 3\u03b2 + 2\u03b1 since \u03b1 \u2260 \u03b2, therefore 3\u03b1 +2\u03b2 is not symmetric\n<\/div>\n<\/li>\n
                                  6. \n
                                    \u03b13<\/sup> + \u03b23<\/sup> = \u03b23<\/sup> + \u03b13, <\/sup> therefore \u03b13<\/sup> + \u03b23<\/sup> is symmetric\n<\/div>\n<\/li>\n<\/ol>\n

                                    EVALUATION
                                    \n<\/strong><\/p>\n

                                      \n
                                    1. \n
                                      Find the quadratic equation whose roots are \u00bd and 5\n<\/div>\n<\/li>\n
                                    2. \n
                                      Find the sum & product of roots of the equation 3x2<\/sup> \u2013 5x \u2013 2 = 0\n<\/div>\n<\/li>\n<\/ol>\n

                                      \u00a0 ASSIGNMENT
                                      \n<\/strong><\/p>\n

                                        \n
                                      1. Solve for x ; x2<\/sup> \u2013 9 = 0 (a) 3, 3 (b) 3, -3 (c) 9, 3 (d) -3, -3\n<\/li>\n
                                      2. Find the sum and products of 2x2<\/sup> + 3x – 1 = 0 (a) -3\/2, \u00bd (b) 3\/2 , -1\/2 (c) -3\/2, -1\/2 (d) 3, 2\n<\/li>\n
                                      3. Solve for x : 6x2<\/sup> \u2013 13x + 5 = 0 (a) 5\/3, \u00bd (b) -5\/3,1\/3 (c) 5\/3, -1\/3 (d) 3\/5, 1\/3\n<\/li>\n
                                      4. Factorize 6x2<\/sup> + 5x -6 (a) (3x – 2) (2x + 3) (b) (3x + 2) (2x + 3) (c)(3x – 2) (2x – 3) (d) (2x -3) (2x+ 2)\n<\/li>\n
                                      5. Find the sum and products of x2<\/sup> \u2013 4x \u2013 3 = 0\n<\/li>\n<\/ol>\n

                                        Theory
                                        \n<\/strong><\/p>\n

                                          \n
                                        1. \n
                                          Solve S = ut + \u00bd at2<\/sup> using the completing the square method\n<\/div>\n<\/li>\n
                                        2. \n
                                          Find the quadratic equation whose roots are \u00be and \u00bd\n<\/div>\n<\/li>\n<\/ol>\n

                                          \u00a0
                                          \n\t\t\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

                                          WEEK 5 Factorization of Quadratic Expressions. Factorize the following: 6a2 + 15a + 9 6a2…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,302],"tags":[],"class_list":["post-3888","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss3-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3888","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3888"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3888\/revisions"}],"predecessor-version":[{"id":3889,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3888\/revisions\/3889"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3888"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3888"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3888"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}