✔ 最佳答案
1)α - β = 4(α - β)^2 = 16α^2 + β^2 - 2αβ = 16(α + β)^2 - 4αβ = 16[- 3(p-1) / -2]^2 - 4(-10/-2) = 16[9(p-1)^2]/4 - 20 = 16(p - 1)^2 = 16p - 1 = 4 or - 4p = 5 or - 3
2)2(x+2)^2 - (x+2) - 12 = 02(x^2 + 4x + 4) - x - 2 - 12 = 02x^2 + 7x - 18 = 0
3)(x - k)[x - (k+1)] = 1x^2 - kx - (k+1)x + k(k+1) = 1x^2 - (2k+1)x + (k^2 + k - 1) = 0△ = (2k+1)^2 - 4(k^2 + k - 1) = 4k^2 + 4k + 1 - 4k^2 - 4k + 4= 5 > 0所以對任意實數k,二次方程(x-k)[x-(k+1)]=1都有兩個相異實根.
2010-10-10 16:15:32 補充：
2)答案應是
2x^2 + 7x - 6 = 0
(不是 2x^2 + 7x - 18 = 0)
2010-10-10 19:13:29 補充：
4)方程px^2+qx+r=0有兩個相異實根,
所以 △ = q^2 - 4pr > 0 ,
px^2+2qx+r=0
△ = (2q)^2 - 4pr
= 4q^2 - 4pr
= 3q^2 + (q^2 - 4pr)
3q^2 >= 0 , (q^2 - 4pr) > 0 ,
所以△ = 3q^2 + (q^2 - 4pr) > 0 ,
即px^2+2qx+r=0 有兩個相異實根。
