Prove that kx^2+2x-(k-2)=0 has real roots for any value of k.?
回答 (3)
2020-05-12 9:02 pm
✔ 最佳答案
kx² + 2x - (k - 2) = 0Discriminant, Δ
= b² - 4ac
= 2² - 4k[-(k - 2)]
= 4 + 4k(k - 2)
= 4 + 4k² - 8k
= 4(k² - 2k + 1)
= 4(k - 1)²
For any value of k, Discriminant, Δ = 4(k - 1)² ≥ 0
Hence, kx² + 2x - (k - 2) = 0 has real roots for any value of k.
2020-05-12 9:20 pm
A quadratic equation of the form ax² + bx + c = 0 has roots given by:
x = [-b ± √(b² - 4ac)]
For roots to exist, b² - 4ac ≥ 0
i.e. 2² - 4(k)(2 - k) ≥ 0
so, 4 - 8k + 4k² ≥ 0
or, k² - 2k + 1 ≥ 0
i.e. (k - 1)² ≥ 0
This is true for any value of k.
:)>
x = [-b ± √(b² - 4ac)]
For roots to exist, b² - 4ac ≥ 0
i.e. 2² - 4(k)(2 - k) ≥ 0
so, 4 - 8k + 4k² ≥ 0
or, k² - 2k + 1 ≥ 0
i.e. (k - 1)² ≥ 0
This is true for any value of k.
:)>
2020-05-13 12:04 am
kx² + 2x - (k - 2) = 0
real b² - 4ac = 0 an b² - 4ac > 0
for b² - 4ac = 0
2² - 4k(-(k - 2)) = 0
4 + 4k(k - 2) = 0
4 + 4k² - 8k = 0
4(k - 1)² = 0
real → k = 1
for b² - 4ac > 0
4(k - 1)² > 0
real → {k < 1 or k > 1}
real b² - 4ac = 0 an b² - 4ac > 0
for b² - 4ac = 0
2² - 4k(-(k - 2)) = 0
4 + 4k(k - 2) = 0
4 + 4k² - 8k = 0
4(k - 1)² = 0
real → k = 1
for b² - 4ac > 0
4(k - 1)² > 0
real → {k < 1 or k > 1}
收錄日期: 2021-04-24 09:51:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200512125052AASFaO5