Q1 (4k^2 - 2k + 7)x^2 + (2k - 7)x^2 + 3 => 0
find the range of k for all real values of x
Q2 (-3k^2 + 4k - 9)x^2 - (2k^2 - 3k + 31) = 0
find the range of x for all real values of k
Intequality and Quadratic equation
2007-05-28 10:16 pm
回答 (3)
2007-05-28 11:25 pm
✔ 最佳答案
As follows:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/May07/Crazyquadratic6.jpg
參考: My Maths knowledge
2007-06-01 1:13 am
1)
(4k^2 - 2k + 7)x^2 + (2k - 7)x^2 + 3 => 0
delta <= 0 and 4k^2 - 2k + 7 > 0
(2k - 7)^2 - 4(4k^2 - 2k + 7)(3) <= 0
4k^2 - 28k + 49 - 48k^2 + 24k - 84 <= 0
-44k^2 - 4k - 35 <= 0
44k^2 + 4k + 35 => 0
hence,all real values of k.
2)
(-3k^2 + 4k - 9)x^2 - (2k^2 - 3k + 31) = 0
since -3k^2 + 4k - 9 < 0 and
2k^2 - 3k + 31 > 0 for all real values of k
hence,no solution.
參考: eason mensa
2007-05-28 11:05 pm
Q1
(4k^2 - 2k + 7)x^2 + (2k - 7)x^2 + 3
= (4k^2)x^2 + 3
if (4k^2 - 2k + 7)x^2 + (2k - 7)x^2 + 3 => 0 is true for all real x
then , (4k^2)x^2 + 3 => 0 is true for all real x
since 4k^2 => 0 for all real value of k
hence, for all real value of k , (4k^2)x^2 + 3 => 0 for all value of x
Q2
(-3k^2 + 4k - 9)x^2 - (2k^2 - 3k + 31) = 0
x^2 = (2k^2 - 3k + 31)/(-3k^2 + 4k - 9)
x = √((2k^2 - 3k + 31)/(-3k^2 + 4k - 9))
or
x = - √((2k^2 - 3k + 31)/(-3k^2 + 4k - 9))
(4k^2 - 2k + 7)x^2 + (2k - 7)x^2 + 3
= (4k^2)x^2 + 3
if (4k^2 - 2k + 7)x^2 + (2k - 7)x^2 + 3 => 0 is true for all real x
then , (4k^2)x^2 + 3 => 0 is true for all real x
since 4k^2 => 0 for all real value of k
hence, for all real value of k , (4k^2)x^2 + 3 => 0 for all value of x
Q2
(-3k^2 + 4k - 9)x^2 - (2k^2 - 3k + 31) = 0
x^2 = (2k^2 - 3k + 31)/(-3k^2 + 4k - 9)
x = √((2k^2 - 3k + 31)/(-3k^2 + 4k - 9))
or
x = - √((2k^2 - 3k + 31)/(-3k^2 + 4k - 9))
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