? 發問於 科學及數學數學 · 1 十年 前

2 quadratic equations questions...

1. Suppose a and b are two distinct roots of the equation x^2 +3x -2=0

(a) Find the values of a^2 +3a and b^2 +3b

*(b) Hence fin the value of a+b Ans:-3

2. Solve x^2+2ax+b=0

Ans: -a+開方(a^2-b) 或 -a-開方(a^2-b)

2 個解答

評分
  • cipker
    Lv 5
    1 十年 前
    最佳解答

    1(a)

    As a and b are the roots of x²+3x−2=0,

    we have

     a²+3a−2=0

     b²+3b−2=0

    which leads to

     a²+3a=2

     b²+3b=2

    (b)

    Using the RESULT of(a),

      a²+3a    = 2

    −) b²+3b    = 2

    -------------------------------

      a² +3a−(b²+3b) =2−2

      a²−b² +3a−3b=0

    (a−b)(a+b)+3(a−b)=0

      (a−b) [(a+b)+3]=0

             a−b=0     or (a+b)+3=0

            a=b(rejected)  or   a+b=−3

    2

        x²+2ax +b=0

    x²+2ax+a²+b−a²=0

      (x+a)² +b−a²=0

          (x+a)² = a²−b

           x+a = ±√(a²−b)

          x=−a ±√(a²−b)

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  • Danny
    Lv 6
    1 十年 前

    2) x^2 + 2ax + b = 0

    x^2 + 2ax + a^2 -a^2 + b = 0

    ( x + a )^2 - a^2 + b = 0

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

    x + a = (a^2 - b)^(0.5) or x + a = - (a^2 - b)^(0.5)

    x = - a + (a^2 - b)^(0.5) or - a - (a^2 - b)^(0.5)

    1a) a is root of x^2 +3x -2=0: a^2 + 3a - 2 = 0 => a^2 + 3a = 2

    b is root of x^2 +3x -2=0: b^2 + 3b - 2 = 0 => b^2 + 3b = 2

    1b) By formula, sum of roots = a+b = -3/1 = - 3

    Alternatively, using the result of Question 2,

    sum of roots = [ - a + (a^2 - b)^(0.5) ] + [- a - (a^2 - b)^(0.5)]

    = - 2 a

    Compare x^2+2ax+b=0 and x^2 +3x -2=0

    we get, a = 1.5

    So sum of roots = -2a = -2*1.5 = -3

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