一元二次方程

👨🏻‍🏫 學術台數學

kawan

2008-10-05 10:54 pm

已知一元二次方程x((2k-1)x+3)(3k+1)+9=0有兩個相等實根.求

a)k的值

b)上述方程的根

已知一元二次方程ax^2+(a+b)x+b=0 .a和b是實數

證明該方程必有實根

實心圓錐斜高25cm　總表面積224兀　cm^2 求圓錐底半徑

詳細解釋

更新1:

ABCD是邊長1CM的正方形.ABFE and CDFG都是梯形.若CDFG的面積是 ABFE面積的2倍求AE的長度一個長方形BDEF內接於一直角三角解ABC 其中角B是直角已知AB=14 CM BC=28CM DE=X CM 以x表示長度以x表示長方形BDEF面積 Y= -2X^2+18圖像與x軸相交於a b 兩點而與y軸相交於c點求三角形abc面積

回答 (2)

用戶7639

2008-10-06 1:17 am

✔ 最佳答案

As follow AS~~~

圖片參考：http://www.photo-host.org/img/539961screenhunter_03_oct._05_17.13.gif

2008-10-05 17:17:26 補充：
r = 7cm

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momo

2008-10-06 1:20 am

x((2k-1)x+3)(3k+1)+9=0
(2k-1)(3k+1)x^2 + 3(3k+1)x + 9 = 0
(6k^2 -k - 1)x^2 + (9k+3)x + 9 = 0
有兩個相等實根
b^2 - 4ac = 0
(9k+3)^2 -4(6k^2 - k -1)(9) = 0
(3k+1)^2 - 4(6k^2 -k-1) = 0
9k^2+6k+1-24k^2+4k+4 = 0
15k^2-10k-5 = 0
3k^2 - 2k - 1 = 0
(3k +1)(k-1) = 0
k = -1/3 or 1
(b)
when k = -1/3
[6(-1/3)^2 -(-1/3) - 1)x^2 + [9(-1/3)+3]x + 9 = 0
9 = 0
k = -1/3 doesn't work
when k = 1
[6(1)^2 -1 - 1)x^2 + [9(1)+3]x + 9 = 0
4x^2 + 12x + 9 = 0
(2x+3)^2 = 0
x = -3/2
(2)
ax^2 +(a+b)x+b = 0
discriminant = (a+b)^2 - 4(a)(b)
= (a^2 +2ab +b^2) -4ab
= (a^2 - 2ab+b^2)
= (a-b)^2 >= 0
當discriminant >或 = 0, 該方程必有實根
(3)
設圓錐底半徑 = r, 圓錐斜高 = l = 25
πr^2 + πrl = 224π
224π = πr^2 + πr(25)
r^2 +25r - 224 = 0
(r + 32)(r - 7) = 0
r = 7cm or -32(rejected)

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