if the roots of the equation x 二 次 + k x+ ( k + 1 5 ) = 0 are real number,then
k的範圍?
一條maths mc 值20分
2007-04-22 2:45 am
回答 (6)
2007-04-22 11:47 pm
✔ 最佳答案
k^2 -4(1)(k+15) >=0k^2 - 4k - 60 >=0
k <= -6 or k>= 10
參考: ME
2007-05-03 10:12 pm
Which number is bellow -6 and higher than 10?
2007-04-22 3:51 am
二次方程, 找根o既公式為
x=0.5(-b+-sqrt(b^2-4ac))
當sqrt(b^2-4ac)係real number 時, 根就唔會係complex number, 即係real number
而要sqrt(b^2-4ac)為real, 即b^2-4ac要>=0
所以, 在此需要
k^2 -4(1)(k+15) >=0
k^2 - 4k - 60 >=0
(k - 10)(k +6)>=0
k <= -6 or k>= 10
x=0.5(-b+-sqrt(b^2-4ac))
當sqrt(b^2-4ac)係real number 時, 根就唔會係complex number, 即係real number
而要sqrt(b^2-4ac)為real, 即b^2-4ac要>=0
所以, 在此需要
k^2 -4(1)(k+15) >=0
k^2 - 4k - 60 >=0
(k - 10)(k +6)>=0
k <= -6 or k>= 10
參考: Personal work
2007-04-22 2:57 am
x^2 + kx + (k+15) = 0
if the roots are real no.
then the discriminant , D ≧ 0
k^2 - 4(1)(k+15) ≧ 0
k^2 - 4k - 60 ≧ 0
(k-10)(k+6) ≧0
k ≧ 10 or k </= -6 (smaller than or equal to -6)
p.s ≧ - larger than or equal to
if the roots are real no.
then the discriminant , D ≧ 0
k^2 - 4(1)(k+15) ≧ 0
k^2 - 4k - 60 ≧ 0
(k-10)(k+6) ≧0
k ≧ 10 or k </= -6 (smaller than or equal to -6)
p.s ≧ - larger than or equal to
參考: my maths knowledge
2007-04-22 2:52 am
If the roots are real, then discriminant should be greater than or equal to 0
k^2 -4(1)(k+15) >=0
k^2 - 4k - 60 >=0
k <= -6 or k>= 10
k^2 -4(1)(k+15) >=0
k^2 - 4k - 60 >=0
k <= -6 or k>= 10
參考: me
2007-04-22 2:51 am
b^2+4ac>0
即係k^2+4(k+15)>0
即係k^2+4(k+15)>0
收錄日期: 2021-04-24 08:39:02
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