f.4quadratic equations
2010-09-30 4:01 am
1.Prove that the quadratic equation (x-k)(x-(k-1))=1 has two distinct real roots for any real values of k .
回答 (1)
2010-09-30 4:09 am
✔ 最佳答案
(x-k)(x-(k-1))=1(x - k)(x - k + 1) - 1 = 0(x - k)^2 + (x - k) - 1 = 0Let x - k be y ,y^2 + y - 1 = 0△ = 1^2 - 4(1)(-1) = 5 > 0So y = x - k have two distinct real values , therefore x have two distinct real values 。
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