Persamaan kuadrat x^2-3x-4=0 mempunyai akar x1 dan x2 .(x1/x2-1) dan (x2/x1-1) adalah.. A. 4x^2+9x-9=0 B. 4x^2+17x-17=0 C. 4x^2+-17x+17=0 D. 4x^2+25x+25=0 E. 4x

x^2 - 3x - 4 = 0
x1 + x2 = -b/a = -(-3)/1 = 3
x1.x2 = c/a = -4/1 = -4
(x1)^2 + (x2)^2 = (x1 + x2)^2 - 2x1.x2 = (3)^2 - 2(-4) = 9 + 8 = 17
x1/x2 + x2/x1 = (x1^2 + x2^2)/(x1.x2) = 17/-4

Misal p = x1/x2 - 1 dan q = x2/x1 - 1
p + q = (x1/x2 - 1) + (x2/x1 - 1) = x1/x2 + x2/x1 - 2 = -17/4 - 2 = -25/4
p.q = (x1/x2 - 1)(x2/x1 - 1) = 1 - (x1/x2 + x2/x1) + 1 = 1 - (-17/4) + 1 = 25/4
Persamaan kuadrat baru
x^2 - (p + q)x + pq = 0
x^2 - (-25/4)x + 25/4 = 0 ======> kali 4
4x^2 + 25x + 25 = 0