Question

Dik a dan b akar-akar persamaan 2x^2 + 6x - 3 = 0 . Tentukan persamaan yang akar-akarnya (a + 3Beta) dan (3a + beta )!

Answer 1

2x²+6x-3=0
akar-akarnya = α dan β

x1+x2 = -b/a
x1+x2 = -6/2 = -3

x1*x2 = c/a
x1*x2 = -3/2

x1+x2 = α+3β + 3α+β
x1+x2 = 4α + 4β
x1+x2 = 4(α+β)
x1+x2 = 4(-3)
x1+x2 = -12

x1*x2 = (α+3β) * (3α+β)
x1*x2 = 3α² + 10αβ + 3b²
x1*x2 = 3α² + 3β² + 10(-3/2)
x1*x2 = 3(α+β)²-2αβ -15
x1*x2 = 3(-3)²+3 -15
x1*x2 = 15

x²-(x1+x2)x+x1*x2=0 (membuat persamaan kuadrat baru)
x² -(-12)x+15=0
x²+12x+15=0