Matematika

Pertanyaan

Diketahui f(x) = x^5 + ax^2 + 4x - 10 dan f(2) = 26 nilai a adalah
A. - 3
B. - 2
C. -1
D. 1
E. 2

2 Jawaban

  • f(x) = x^5 + ax^2 + 4x - 10
    26 = 2^5 + a(2)^2 + 4(2) - 10
    26 = 32 + 4a + 8 - 10
    26 = 30 + 4a
    26 - 30 = 4a
    4a = - 4
    a = -4/4
    a = -1
  • f(x)= x^5 + ax^2 + 4x-10 = 0
    f(2)= (2)^5 + a(2)^2 + 4(2) -10 =26
    f(2)= 32 + 4a + 8 - 10 = 26
    f(2)=30 + 4a = 26
    f(2)= 4a = 26 - 30
    f(2)= 4a = -4
    f(2)= a = -4 / 4
    a = -1

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