Matematika

Pertanyaan

bila (ax-3) (x+b) = (p+1) x2 + 7x - 6 , maka p =

2 Jawaban

  • (ax - 3)(x + b) = (p+1)x^2 +7x - 6
    ax^2 +(ab-3)x - 3b = (p+1)x^2 + 7x - 6
    ** -3b = -6
         b = 2
    ** (ab - 3) = 7
        (a(2) - 3) = 7
        (2a - 3) = 7
               2a = 10
                 a = 5
    ** a = p + 1
        p + 1 = a
        p + 1 = 5
              p = 4
  • [tex](ax - 3)(x + b) = (p + 1)x^2 + 7x - 6 \\ ax^2 + (ab - 3)x - 3b = (p + 1)x^2 + 7x - 6 \\ - 3b = - 6 \\ b = 2 \\ 2a - 3 = 7 \\ a = 5 \\ maka: \\ a = p + 1 \\ 5 = p + 1 \\ p = 4[/tex]

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