The quadratic equantion(b-c)x^2+(c-a)x+(a-b)=0
has a repeated solution. Prove that b=(a+c)/2.
(Hint;We can rearrange b=(a+c)/2 as a+c-2b=0)
快d幫下我啦~~~~
2007-05-12 10:19 pm
回答 (2)
2007-05-12 10:33 pm
✔ 最佳答案
delta = (c-a)^2 - 4(b-c)(a-b) = 0 (for repeated solution)(c^2 - 2ac +a^2) - 4(ab - b^2 - ac +cb) = 0
c^2 - 2ac +a^2 - 4ab + 4 b^2 + 4ac - 4cb = 0
c^2 + 2ac +a^2 - 4(a + c)b + 4 b^2 = 0
(a + c)^2 - 4(a + c)b + 4 b^2 = 0
[(a + c) - 2b]^2 = 0
a + c - 2b=0
b=(a+c)/2
參考: me
2007-05-12 10:34 pm
due to the equation has two equal roots
therefore,
the discriminate = 0
(c-a)^2-4(b-c)(a-b) =0
c^2-2ac+a^2-4ab+4b^2+4ac-4bc=0
c^2+2ac+a^2-2(2b)(a+c)+4b^2=0
[(a+c)-2b]^2=0
a+c-2b=0
b=(a+c)/2
therefore,
the discriminate = 0
(c-a)^2-4(b-c)(a-b) =0
c^2-2ac+a^2-4ab+4b^2+4ac-4bc=0
c^2+2ac+a^2-2(2b)(a+c)+4b^2=0
[(a+c)-2b]^2=0
a+c-2b=0
b=(a+c)/2
收錄日期: 2021-04-25 17:17:58
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https://hk.answers.yahoo.com/question/index?qid=20070512000051KK02472