1.solve
25^(x^(1/2)-1/2)+5^(x^(1/2)+1)>250
2.a,b,c are real
a+b+c=8 and a^2+b^2+c^2=24
(i)express ab in terms of c
(ii)show that a and b are roots of the quadratic equation
t^2-(8-c)t+(20-8c+c^2)=0
(iii)deduce from these results that
4/3<=u<=4 where u =a,b or c
<= denote less than or equal
equation
2009-11-22 4:53 pm
回答 (1)
2009-11-22 5:11 pm
✔ 最佳答案
(1)圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Nov09/Crazyeqn1.jpg
(2i) a + b + c = 8
(a + b + c)2 = 64
a2 + b2 + c2 + 2ab + 2bc + 2ca = 64
24 + 2ab + 2c(a + b) = 64
ab + c(8 - c) = 20
ab = c2 - 8c + 20
(ii) Sum of roots = a + b = 8 - c
Product of roots = ab = c2 - 8c + 20
Hence a and b are roots of equation:
t2 - (8 - c)t + (c2 - 8c + 20) = 0
(iii) Since a, b, c are real, the discriminant of the equation t2 - (8 - c)t + (c2 - 8c + 20) = 0
should be non-negative, i.e.
(8 - c)2 - 4(c2 - 8c + 20) >= 0
64 - 16c + c2 - 4c2 + 32c - 80 >= 0
3c2 - 16c + 16 <= 0
(3c - 4)(c - 4) <= 0
1/4 <= c <= 4
Using similar approach, we can obtain 1/4 <= a <= 4 and 1/4 <= b <= 4
參考: Myself
收錄日期: 2021-04-13 16:57:05
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