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Tam G 發問於 科學及數學數學 · 6 年前

F.4 maths Quadratic Equations

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

equation have a positive root? Explain your answer.

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

positive?Explain your answer.

求教! 請詳細解釋和列出解題過程,感謝萬分!!

4 個解答

評分
  • 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.

    ∴It is a contradiction.

    ∴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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