Let r = (x+1)/(x^2+x+1). Find the range of values of r for all real values of x.
Why the answer is -1/3 smaller or equal than r greater than or equal 1
A. maths(F.4)
2006-11-01 5:41 am
回答 (2)
2006-11-01 5:56 am
✔ 最佳答案
r = (x + 1) / (x^2 + x + 1)r(x^2 + x + 1) = x + 1
rx^2 + rx + r - x - 1 = 0
rx^2 + (r - 1)x + (r - 1) = 0.................(*)
As x is a real number, for suitable values of r, (*) must have real roots.
Thus discriminant ≥ 0
(r - 1)^2 - 4r(r - 1) ≥0
(r - 1)[(r - 1 - 4r)] ≥ 0
(r - 1)(-3r - 1) ≥ 0
(r - 1)(3r + 1) ≤ 0
-1 / 3 ≤ r ≤ 1
2006-11-02 2:58 am
r = (x + 1) / (x^2 + x + 1)
r(x^2 + x + 1) = x + 1
rx^2 + rx + r - x - 1 = 0
rx^2 + (r - 1)x + (r - 1) = 0.................(*)
As x is a real number, for suitable values of r, (*) must have real roots.
Thus discriminant ≥ 0
(r - 1)^2 - 4r(r - 1) ≥0
(r - 1)[(r - 1 - 4r)] ≥ 0
(r - 1)(-3r - 1) ≥ 0
(r - 1)(3r + 1) ≤ 0
-1 / 3 ≤ r ≤ 1
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r(x^2 + x + 1) = x + 1
rx^2 + rx + r - x - 1 = 0
rx^2 + (r - 1)x + (r - 1) = 0.................(*)
As x is a real number, for suitable values of r, (*) must have real roots.
Thus discriminant ≥ 0
(r - 1)^2 - 4r(r - 1) ≥0
(r - 1)[(r - 1 - 4r)] ≥ 0
(r - 1)(-3r - 1) ≥ 0
(r - 1)(3r + 1) ≤ 0
-1 / 3 ≤ r ≤ 1
希望幫到你 求你選我最佳答案
求下你
唔該!!!!!!
唔該!!!!!!
參考: me
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