Quod Erat Demonstrandum

2011/10/16

無聊改卷後2

Filed under: NSS — johnmayhk @ 2:40 下午

上次談同事擬的中五 Core Mathematics 測驗,見:

https://johnmayhk.wordpress.com/2011/10/12/boring-chatting-after-marking-paper/

還有後續?是,因為要改正嘛。從學生的改正,可以看出他們是有心學習的:

擬題同事建議答案中的一小步,即 3x^2+2x-2\sqrt{5}-15=0 \Rightarrow (x-\sqrt{5})(3x+2+3\sqrt{5})=0,我把它化為一大步,回答學生,見:

另外,果然青出於藍,以下是羅同學的改正,非常好:

容我再改寫一下:

Solve 3x^2+2x-2\sqrt{5}-15=0 and express the answers in the form of a+b\sqrt{c} where a,b,c are rational numbers and \sqrt{c} is irrational.

Solution

3x^2+2x-2\sqrt{5}-15=0 ………. (*)

3x^2+2x=2\sqrt{5}+15

3x^2+2x=3(\sqrt{5})^2+2(\sqrt{5})

Hence, \sqrt{5} is a root of (*).

Let \alpha be another root, then

\alpha+\sqrt{5}=-\frac{2}{3}

\alpha=-\frac{2}{3}-\sqrt{5}

Thus, the roots of (*) are

\sqrt{5} and -\frac{2}{3}-\sqrt{5}

感謝學生!

2 則迴響 »

  1. \[3(x^2-5)+2(x-\sqrt{5})=0\]
    \[3(x+\sqrt[5])(x-\sqrt{5})+2(x-\sqrt{5})=0\]
    \[(x-\sqrt{5})(3x+3\sqrt{5}+2)=0\]

    迴響 由 abc — 2011/10/16 @ 11:25 下午 | 回覆


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