jika akar persamaan kuadrat 3xkuadrat +6x+12=0 adalah αdan β, maka nilai 1perαkuadrat + 1 per βkuadrat =....

• a+b = -6/3
a+b = -2

• ab = 12/3
ab = 4

• 1/a^2 + 1/b^2
= (a^2 + b^2) / a^2 . b^2
= [(a+b)^2 - 2ab] / (ab)^2
= [(-2)^2 - 2(4)] / (4)^2
= [4-8] / 16
= -4/16
= -1/4

Solusi
3x² + 6x + 12 = 0
a = 3, b = 6 dan c = 12

α + β = -6/3 = -2
αβ = 12/3 = 4
α² + β² = (α + β) - 2αβ = (-2)² - 2(4) = -4

Dengan demikian:
= 1/α² + 1/β²
= (α² + β²)/(αβ)²
= (-4)/(4)²
= -4/16
= -1/4