[SNMPTN 2009] Jika kedua akar persamaan saling berlawanan tanda, tetapi mempunyai nilai mutlak yang sama. Maka nilai m sama dengan ?

[tex] \frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1} \\ (m + 1)x^2 - b(m + 1)x = a(m - 1)x - c(m - 1) \\ (m + 1)x^2 - b(m + 1)x - a(m - 1)x + c(m - 1) = 0 \\ (m + 1)x^2 - (b(m + 1) + a(m - 1))x + c(m - 1) = 0 [/tex]

"Akar berlawanan tanda"
"Nilai mutlak keduanya sama"

x1 = -x2
x1 + x2 = 0
[tex] \frac{-[-(b(m + 1) + a(m - 1))]}{m + 1} = 0 \\ \frac{b(m + 1) + a(m - 1)}{m+ 1} = 0 \\ b + a.\frac{m - 1}{m + 1} = 0 \\ \frac{m - 1}{m + 1} = \frac{-b}{a} \\ am - a = -bm - b \\ am + bm = a - b \\ m(a + b) = a - b \\ m = \frac{a - b}{a + b} [/tex]