the equation x^2+(m+n)x+(m^2+n^2)=0,where m and n are real numbers,has equal roots.
Find m in term of n.
x^2 是x的2次方
m^2是m的2次方
n^2是n的2次方
a.maths的問題
2007-08-17 3:05 am
回答 (1)
2007-08-17 3:40 am
✔ 最佳答案
For x^2+(m+n)x+(m^2+n^2)=0 to have equal real root,Discriminant = 0
(m+n)^2 - 4(1)(m^2 + n^2) = 0
3m^2 - 2nm + 3n^2 = 0........................(*)
Discriminant of (*) = (-2n)^2 - 4(3)(3n^2)
...........................= 4n^2 - 36n^2
...........................= -32n^2
Since n is a real number, n^2 >or = 0
i.e. Discriminant < or = 0
since n is real, Discriminant must be = 0
0 = -32n^2
n = 0
similarly, m = 0
2007-08-16 19:41:31 補充：
Over, m = n
參考: Keith
收錄日期: 2021-04-13 12:53:43
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