Question

Mohon dibantu yah !!

Answer 1

3x²+2x+5 = 0  => akar2 nya x1 dan x2

ax²+bx+c = 0 => akar2 nya α = x1+1 , β = x2+1

o 3x²+2x+5 = 0

x1 + x2 = -2/3

x1.x2 = 5/3

o ax²+bx+c = 0

α+β = x1+1 + x2+1

      = x1+x2+2  <= x1+x2 = -2/3

     = -2/3 + 2

    = 4/3

α.β = (x1+1)(x2+1)

     = x1.x2 + 2(x1+x2) + 1  <= x1.x2 = 5/3 , x1+x2 = -2/3

    = 5/3 + 2.(-2/3) + 1

    = 5/3 - 4/3 + 1

    = 1/3 + 1

αβ = 4/3

rumus PKB :

x² - (α+β)x + (α.β) = 0  <= α+β = αβ = 4/3

x² - 4/3 x + 4/3 = 0 <= kali 3 semua

3x² - 4x + 4 = 0 <= a = 3 , b = -4 , c = 4

2a+b+c = 2(3) - 4 + 4

            = 6