prove that the quadratic equation
-x^2+2(m+1)x-(2m^2+3)=0
has no real roots for any real values of m.
hint:use the fact that m^2-2m+1=(m-1)^2
f.5 mathematics help
2013-07-11 12:43 am
回答 (1)
2013-07-11 1:21 am
✔ 最佳答案
Δ=b^2 - 4ac=[2(m+1)]^2 - 4(-1)[-(2m^2+3)]
=4(m+1)^2 - 4(2m^2+3)
=4[(m+1)^2 - (2m^2+3)]
=4(m^2 + 2m + 1 - 2m^2 - 3)
=4(-m^2 + 2m - 2)
= -4(m^2 - 2m + 2)
= -4(m^2 - 2m + 1 + 1)
= -4[(m-1)^2 + 1] < 0 (as (m-1)^2 ≧ 0, then (m-1)^2 + 1 ≧ 1 > 0 )
as Δ < 0, -x^2+2(m+1)x-(2m^2+3)=0 has no real roots for any real values of m.
參考: myself
收錄日期: 2021-04-30 17:53:17
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