maths 問題 equations and graphs

2011-10-24 7:36 am
1.Which of the following intervals must contain a root of 2x^3 - x^2 -x -3 = 0?

1. -1<x<1 2. 0<x<2 3. 1<x<3

A. 1 only B. 2 only C. 3 only D. 1 and 2 only E. 2 and 3 only

(要步驟)

2.The difference of the roots of the equation 2x^2 -5x +k =0 is 7/2. Find k.



3. Suppose the graph of y =x^2 -2x -3 is given. In order to solve the quadratic equation 2x^2 -6x -3=0, Find the straight lines should be added to the given graph.



4. Let k be a constant. If X and B are the roots of the equation x^2 -3x +k = 0, then X^2 +3B =?


各位識既麻煩幫小弟解答一下><
最好詳細D,因為我怕睇唔明=3=THX
更新1:

第3題唔係好明x² - 2x - 3 = x + 3/2 點樣整出來=.= 同埋答案是 y=x- 3/2 應該唔知邊度正負號調轉左-.-

更新2:

同埋第1題點解2個答案乘埋細過0就係contain a root既=.=?

更新3:

其實唔係好明=.= 我照你講既求其整條式: y=x^2 -3x -18 個root既左邊定係右邊Y都係>0 既=.= 同埋如果1正1負乘埋<0 咁點解<0就姐係contain a root既=.= >0點解唔得既=0=?

回答 (1)

2011-10-24 8:16 am
✔ 最佳答案
1) Let f(x) = 2x³ - x² - x - 3f(-1) = - 2 - 1 + 1 - 3 = - 5 < 0
f(1) = 2 - 1 - 1 - 3 = - 3 < 0∴ f(-1) * f(1) > 0 So -1 < x < 1 are not contain a root of f(x).f(0) = - 3 < 0
f(2) = 16 - 4 - 2 - 3 = 7 > 0∴ f(0) * f(2) < 0 So 0 < x < 2 must contain a root of f(x).f(1) < 0
f(3) = 54 - 9 - 3 - 3 > 0∴ f(1) * f(3) < 0 So 1 < x < 3 must contain a root of f(x).Answer : E 2 and 3 only
2) (α - β)² = (7/2)²α² - 2αβ + β² = 49/4α² + 2αβ + β² - 4αβ = 49/4(α + β)² - 4αβ = 49/4(- (-5)/2)² - 4(k/2) = 49/42k = 25/4 - 49/4k = - 3
3)2x² - 6x - 3 = 0x² - 3x - 3/2 = 0x² - 2x - 3 = x + 3/2The required straight lines is y = x + 3/2i.e. 2x - 2y + 3 = 0
4) Since x² - 3x + k = 0So x² = 3x - k ;x² + 3B= (3x - k) + 3B= 3(x + B) - kBy the sum of roots , = 3(- (-3)/1) - k= 9 - k

2011-10-25 23:55:47 補充:
謝謝樓主提醒 , 修正如下 :


3)

2x² - 6x - 3 = 0

x² - 3x - 3/2 = 0

x² - 3x - 3/2 + x - 3/2 = x - 3/2

x² - 2x - 3 = x - 3/2

The required straight lines is y = x - 3/2

2011-10-26 00:22:40 補充:
第1題點解2個答案乘埋細過0就係contain a root既=.=?

好間單 , 你想像下條開口向上的拋物線 , 個 root 是 x軸 和 拋物線的交點 ,

個 root 左邊少少的數會令 y > 0 , 個 root 會令 y = 0 , 個 root 右邊少少的數
會令 y < 0 ......

所以一正一負相乘會 < 0 。

OK ?

2011-10-27 17:10:05 補充:
y=x^2 -3x -18

y = 0 when x = - 3 or 6

f(- 4) > 0
f(- 3) = 0
f(- 2) < 0
所以 - 4 < x < - 2 有 root - 3。

f(5) < 0
f(6) = 0
f(7) > 0
所以 5 < x < 7 有 root 6。

不是看 - 4 和 7

2011-10-27 17:14:01 補充:
>0點解唔得既=0=?

只是說過 < 0 肯定有 root 。

並沒有話過 > 0 無 root !!

如 f(- 4) > 0 , f(7) > 0
事實上 - 4 < x < 7 有兩個 roots !!

但有些情況也可能無。


收錄日期: 2021-04-21 22:24:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111023000051KK01137

檢視 Wayback Machine 備份