Matematika

Pertanyaan

Jika α dan β akar-akar persamaan kuadrat 4x² - 2x - 3 = 0, maka persamaan kuadrat yang akar-akarnya α + 1 dan β +1 adalah

2 Jawaban

  • X1 + X2 = -b/a = 2/4 = 1/2
    X1.X2 = c/a = -3/4
    α = α + 1
     β = β +1
    α+β = α + 1 + β +1
           = α+ β + 2
           = 1/2 + 2 = 5/2
    α.β = (α + 1)(β +1)
          = α.β  + 2(α+β) + 2
          = -3/4 + 2(1/2) + 2
          = 9/4
    PKB ; x" - (α+β)x + α.β
            = x" - 5/2x + 9/4 (x4)
            = 4x" - 10x + 9
  • α + 1 = X1
    β +1  = X2

    α + β = -b/2a = -(-2)/2(4) = ¼
    α
    β = c/a = -3/4

    X
    ² - (XI+X2)X + (X1.X2) = 0
    X - ( α + 1+ β +1 )X + (( α + 1 )(β +1 )) = 0

    X
    ² - ( α + β +2)X + (α β + α + β +1 ) = 0
    X
    ² - (1/4 + 2)X + (-3/4 + ¼  + 1 ) = 0
    X
    ² – ( 9/4)X + ( ½ ) = 0    <DIKALIKAN 4 SEMUA>
    4X
    ² -9X + 2 =0


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