1.find the value(s) of k,if each of following equations has equal roots
a)kx^2+(2k+6)x+16=0
b)kx^2+2(k+4)x+25=0
急!!thank you!!!!!!!!!!!!!!!!!!!!!!!
maths急求!!
2008-09-24 7:01 am
回答 (2)
2008-09-24 9:08 am
✔ 最佳答案
A同b都係用delta黎計吧=.=delta啫係個三角形
DELTA=b^2-4ac
如果佢同你講係equal roots呢咁個delta啫係等如0
a.
(2k+6)^2-4k(16)=0
4k^2+24k+36-64k=0
K^2-10k+9=0
(k-9)(k-1)=0
k=9or k=1
b.
[2(k+4)]^2-4k(25)=0
4[k^2+8k+16]-100k=0
4k^2+32k+64-100k=0
4k^2-68k+64=0
K^2-17k+16=0
(k-16)(k-1)=0
k=16or k=1
有咩唔明再問我啦^^
參考: 自己未遺棄既f5數學
2008-09-24 8:43 am
For the eqtn has equal roots => discriminant=0
a) (2k+6)^2-4(16)(k)=0
4k^2-40k+36=0
k^2-10k+9=0
(k-9)(k-1)=0
k=1 or 9
b) [2(k+4)]^2-4(25)(k)=0
k^2-16k+16=0
k=8+4sqrt(3) or 8-4sqrt(3)
a) (2k+6)^2-4(16)(k)=0
4k^2-40k+36=0
k^2-10k+9=0
(k-9)(k-1)=0
k=1 or 9
b) [2(k+4)]^2-4(25)(k)=0
k^2-16k+16=0
k=8+4sqrt(3) or 8-4sqrt(3)
收錄日期: 2021-04-13 21:57:42
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