Prove the equation has real roots.

2006-11-04 8:16 am
(b-c)x²+(c-a)x+(a-b)=0
a. Show that if a , b , c are real , then the roots of the equation are real.

回答 (1)

2006-11-04 8:32 am
✔ 最佳答案
(a) Discriminant
= (c - a)^2 - 4(b - c)(a - b)
= c^2 - 2ac + a^2 - 4(ab - ac - b^2 + bc)
= c^2 - 2ac + a^2 - 4ab + 4ac + 4b^2 - 4bc
= c^2 + 2ac + a^2 - 4ab - 4bc + 4b^2
= c^2 + ac + ac - 2bc + a^2 - 2ab - 2ab - 2bc - 2bc + 4b^2
= c^2 + ac - 2bc + ac + a^2 - 2ab - 2bc - 2ab + 4b^2
= c(c + a - 2b) + a(c + a - 2b) - 2b(c + a - 2b)
= (c + a - 2b)(c + a - 2b)
= (c + a - 2b)^2, which is non-negative.
Thus the roots are real.


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