Find the range of values of k such that each of the following quadratic equations has real roots.
k(2x+1)= 2- kx^2
2. A man bought some oranges for $240.Unfortunately,five were rotten. To cover the loss,he had to sell each of the remaing oranges at aprice of $0.2 more than it's cost. How many oranges did he buy?
中四數2條
2007-10-07 11:44 pm
回答 (2)
2007-10-08 12:56 am
✔ 最佳答案
1. k(2x+1)= 2 - kx^2
2kx + k = 2 - kx^2
kx^2 + 2kx + (k - 2) = 0 (k > 0) ....(*)
'.' (*) has real roots.
.'. discriminant >= 0
(2k)^2 - 4(k)(k - 2) >= 0
4k^2 - 4k^2 + 8k >= 0
8k >= 0
k > = 0
'.' k > 0 and k >= 0
.'. the solution is k > 0
2.
Let x be the number of oranges he bought,
then (x - 5) is the number of remaining oranges.
Cost of one orange = $(240 / x)
Price of one of the remaining oranges = 240 / x + 0.2
(240 / x + 0.2)(x - 5) = 240
(240 + 0.2x)(x - 5) = 240x
240x - 1200 + 0.2x^2 - x = 240x
0.2x^2 - x - 1200 = 0
x^2 - 5x - 6000 = 0
(x - 80)(x + 75) = 0
'.' x - 80 = 0 or x + 75 =0
.'. x = 80 or x = -75 (rejected)
.'. He bought 80 oranges.
參考: me .^ ^
2007-10-08 12:13 am
1)
k(2x+1)=2-kx²
2kx+k=2-kx²
kx²+2kx+(k-2)=0
∵ equations has real roots,
∴∆≥0
(2k)²-4(k)(k-2)≥0
4k²-4k²+8k≥0
8k≥0
k≥0
k(2x+1)=2-kx²
2kx+k=2-kx²
kx²+2kx+(k-2)=0
∵ equations has real roots,
∴∆≥0
(2k)²-4(k)(k-2)≥0
4k²-4k²+8k≥0
8k≥0
k≥0
收錄日期: 2021-04-29 00:15:08
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