1) it is given that -3 is a root of the quadratic equation (x+k)^2-x-1=0, where k is a constant. find all possible values of k.
2) if the function y=2x^2-4x(2)^1/2-m touches the x-axis, find the value of m.
remark:唔識打開方,開方2=2^1/2
f.4 maths
2011-01-05 1:45 am
回答 (2)
2011-01-05 3:48 am
✔ 最佳答案
Q1 有冇打錯題目Q2 y=2x^2-4x+[(2)^1/2-m]?
2011-01-04 19:48:00 補充：
1)
Sub. x=-3 into the eq.
(-3+k)^2-(-3)-1=0
k^2-6k+9+2=0
k^2-6k+11=0...(*)
Discriminant of (*)=6^2-4(11)<0
So, there is no real values for k.
2)
If y=2x^2-(4√2)x-m touches the x-axis, discriminant=0
(-4√2)^2-4(2)(-m)=0
32+8m=0
m=-4
2011-01-05 3:10 am
第1冇錯
第2係 y=2x^2+4開方2x-m
( x冇開方)
第2係 y=2x^2+4開方2x-m
( x冇開方)
收錄日期: 2021-04-22 00:52:07
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https://hk.answers.yahoo.com/question/index?qid=20110104000051KK00775