F.4 a. maths

2006-11-30 7:37 pm
Show that the graph y= 3x^2+12x+7+k(x^2-1) cuts the x-axis for all real values of k.

回答 (1)

2006-11-30 7:45 pm
✔ 最佳答案
y = 3x² + 12x + 7 + k(x²-1)
y = 3x² + 12x + 7 + kx² - k
y = (3+k)x² + 12x + (7-k)
Determinant △
= (12)² - 4(3+k)(7-k)
= 144 - 4(21 - 3k + 7k - k²)
= 144 - 4(21 + 4k - k²)
= 144 - 84 - 16k + 4k²
= 4k² - 16k + 60
= 4(k² - 4k + 15)
= 4(k² - 4k + 4 + 11)
= 4[(k-2)² + 11]
> 0 for all real values of k as (k-2)² >= 0 for all real values of k.
So y = 3x² + 12x + 7 + k(x²-1) has real roots for all real values of k.
So y = 3x² + 12x + 7 + k(x²-1) cuts x-axis for all real values of k.

2006-11-30 13:29:10 補充:
小小補充:尾二行應該是:「So 3x² + 12x + 7 + k(x²-1) = 0 has real roots for all real values of k.」


收錄日期: 2021-04-12 17:46:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061130000051KK00957

檢視 Wayback Machine 備份