1. Diketahui persamaan kuadrat x^2-3x-4=0 mempunyai akar akar x1 dan x2 persamaan kuadrat akar akarnya (x1/x2 -1) dan (x2/x1 -1) adalah

x1 + x2 = b/a
x1.x2 = -c/a

x1² + x2²
= (x1 + x2)² - 2.x1.x2
= 3² - 2(-4)
= 17

(x1/x2 - 1) + (x2/x1 - 1)
= (x1² + x2²)/(x1.x2) - 2
= -17/4 - 2
= -25/4
(x1/x2 - 1)(x2/x1 - 1)
= 2 - (x1² + x2²)/(x1.x2) 
= 2 - (-17/4)
= 25/4

x²- (-{(x1/x2 - 1) + (x2/x1 - 1)} + {(x1/x2 - 1)(x2/x1 - 1)}
x² + 25x/4 + 25/4 = 0 ==> dikali 4
4x² + 25x + 25 = 0