1. Given that x^2+kx+12=0 has roots α and 2α , find the value of α,
then find the value of k.
2. Determune the range of k for which 3x^2+8x+2k=0 has 2 different negaive real solutions.
F4 A.maths Root-Coefficient7
2009-08-26 9:26 am
回答 (1)
2009-08-26 1:30 pm
✔ 最佳答案
Q1. a + 2a = 3a = -k........(1)(a)(2a) = 2a^2 = 12 ...(2)
From (2), a = +/- sqrt 6.
Sub. into (1) k = -/+ 3 sqrt 6.
Q2.
Let the 2 roots be a and b and both are negative.
ab = 2k/3, so k must be positive.
for real and unequal solution, delta > 0, that is
(8)^2 - 4(3)(2k) > 0
64 - 24k > 0
64 > 24k
8/3 > k
so range of k is 8/3 > k > 0.
收錄日期: 2021-04-25 22:36:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090826000051KK00174