Tam G 發問於 科學及數學數學 · 6 年前

1) Given the quadratic equation ax^2+bx+c=0, where a, b and c are positive, does the

2) If the quadratic equation x^2+bx+c=0 has roots of opposite signs, is it possible that c is

### 4 個解答

• 最愛解答

知識+又多左個老師啦 =]

btw,

頭先我諗住打第1題，點解發表左之後會有1part內容會唔見左?....

2014-07-27 18:27:19 補充：

第1題 (睇黎都係要用連結...)

http://postimg.org/image/qy6nnzd29/

唔知我呢個寫法o唔ok??

2014-07-27 19:30:16 補充：

果 然 犀 利 ! !

2014-07-27 21:55:39 補充：

因為我嘅意見變左做

哈利波特2外傳 - 消失的意見

2014-07-27 22:09:13 補充：

各位知識友，

現家我想徵用各位的答案去回答發問者嘅問題，

唔知大家反唔反對？？

2014-07-28 20:11:24 補充：

Here are some methods to solve the questions.

1)

Method of 少年時 ( 專家 2 級 )：

ax²+bx+c=0

Let α and β be the roots of the equation,

as αβ = c/a, while a, b, c are all positive,

so α, β should both negative or both positive.

At the same time, α+β = -b/a, so they are both negative.

That is, the equation has no positive root.

Method of Masterijk ( 知識長 )：

If there exists a root x which is positive, with a, b, c being positive,

ax² + bx + c must be positive, it CANNOT BE zero.

Therefore, no positive root.

Method of Mr Kowk ( 博士級 1 級 )：

ax²+bx+c=0

Suppose the quadratic equation ax^2+bx+c=0 has a positive root,

that is x = [-b+√(b²-4ac)]/(2a) > 0

-b+√(b²-4ac) > 0

√(b²-4ac) > b

(b²-4ac) > b²

-4ac > 0

ac < 0

Since a,c are positive number, the multiplication of a and c MUST BE positive number.

∴The equation has no positive root.

2)

Method of Mr Kowk ( 博士級 1 級 )：

ax²+bx+c=0

αβ = c/a

when a = 1

αβ = c

∴α and β are opposite signs, αβ < 0, then c < 0

∴c must be negative.

2014-07-28 20:12:21 補充：

謝謝大家的支持 = ]

資料來源： 意見區
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• 6 年前

Too complicated, you can say,

Let α and β be the roots of the equation,

as αβ = c/a, while a, b, c are all positive, so α, β should both negative or both

positive. At the same time, α+β = -b/a, so they are both negative.

That is, the equation has no positive root.

答多一、兩題你就升級了，努力！

2014-07-28 07:49:59 補充：

當然不反對，既然你說明是徴用，那你連抄襲的意思都沒有，那有甚麼問題呢！

快點升級吧！

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• 6 年前

郭老師回來，大家很高興！

2014-07-27 17:47:05 補充：

哈哈哈～郭老幽默了～

HK~： 可能是又有字眼觸動了系統，小心。

你快去投票、評價賺點數，幾日內可以升級！！！

2014-07-27 21:02:40 補充：

我又想用另一個方法答呀！！！

If there exists a root x which is positive, with a, b, c being positive,

ax² + bx + c must be positive, it CANNOT BE zero.

Therefore, no positive root.

Done!

如何？

2014-07-27 21:13:52 補充：

聽講以前更強，後來有好多知識友被偉大的「系統」、「管理員」、人事鬥爭、惡意玩弄等等人性的醜惡一併欺壓，最後愔然離開了。

我地都唔可以確定下一個「愔然離開」的係咪我地．．．

2014-07-27 21:14:30 補充：

咦？

點解 HK~ 的 014 意見消失了！？！？

2014-07-27 22:01:42 補充：

你綜合大家的意見，去答題啦～

快Ｄ升級～

2014-07-27 22:18:58 補充：

我贊成。

［個人意見：大家應該不會介意，因為否則大家已經作答了。］

2014-07-28 08:20:34 補充：

對，大家都好想你快ｄ升級～！

加油！

╭∧---∧╮

│ .✪‿✪ │

╰/) ⋈ (\\╯

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• CK
Lv 7
6 年前

2 ax²+bx+c=0

αβ = c/a

when a = 1

αβ = c

α and β are opposite signs, αβ < 0; c < 0; c must be negative.

2014-07-27 17:22:52 補充：

多謝你們的 ( 惠康 )。

~~~~~~~~~~

2014-07-27 21:37:09 補充：

這個如何

x = [-b+√(b²-4ac)]/2a > 0

-b+√(b²-4ac) > 0

√(b²-4ac) > b

(b²-4ac) > b²

-4ac > 0

所以 假設 不成立。

2014-07-27 22:23:43 補充：

我唔反對。

~~~~~~~~~~~~~~~~~~~~

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