Smart Solutions for Some Equations (Quadratic, Polynomial and Rational)


Let’s look at a few problems on solving equations of different forms.

Find the smaller root for the following.

Problem (a).  (2-x) (3-x) = 2 / 3

Problem (b).   x (x-1) (7-2x) = (3-x) (4-x) (4-2x)

Problem (c).   [x (x-1) ⁄ (3-x)] + [x (x-1) ⁄ (4-x)] = 4 – 2x

Relating to equations above, we plot several curves using Geogebra (an app which can produce the graph for given functions – of course, this is only one feature of the many).

What’s interesting is all the three equations as in (a)(b)(c) share a common solution, and it can be found through algebraic manipulation. Take a forward look (click) here – you will see amazing graphs; the solution (roots) shall be some intersection points.


Of these questions, the challenging levels are in the order of ascending: from average level like (a), to somewhat challenging like (b), to very challenging in (c). For better understanding, we suggest the reader to try solve questions (a) (b) (c) first; and then compare with the solution provided in this file (pdf),
where we solve algebraically only for (a), but applying a combination of algebraic and graphic solution to tackle problems (b) and (c).  It’s called smart solution — we are applying a variety of tools at hand, not just following a fixed procedure.


Combining graphing with algebra facilitate us to guess where the root(s) of the equation are, or (sometimes) make observation of the roots easier. It’s also fun!

Try work out these equations first, then you can check out the solutions (pdf) here.

Now you can take a second look at the graphic solution. Have you seen and understood more, about the question, and about the how algebraic manipulation plays out in function graphs?

Amazing Detection of Parallel-o-gram: Unusual But Fairly Reasonable

Talking about shapes, one of the most familiar is the parallelogram. And about it we have a problem as follows.

Given quadrilateral ABCD, E, F is the midpoint of AD, resp. DC. Lines BE and AF intersect at point G. (And the figure is as below.)

Suppose it happens that GE : GA : GF : GB = 1 : 2 : 3 : 4. Attempt to prove ABCD is a parallelogram.

How can the division ratio GE : GA : GF : GB as given, be used as the hints to deduct the parallelogram?

Well, to put the known condition in proper perspective is important. If first attention is directed to GA : GF = 2 : 3, you might wonder how that can be utilized. Suppose you notice that GE : GA = GA : GB = 1 : 2 : 3 : 4, and suppose AF and BE are perpendicular to each other, then we conclude that the two triangles EGA and AGB are similar right triangles .. But how that can be used to help recognize a parallelogram?

By observing AF = AG + GF, and BE = BG + GE, and we find that AF = BE (by applying properties of the proportion, AF : BE = (2 + 3) : (4 + 1) = 1 : 1). So we have two hanging line segments AF and BE that are congruent (equal in length), and both segments are drawn from one vertex (of ABCD) to midpoint of a non-adjacent side. The information in its current form seems much helpful.

Now passing through D let’s draw a parallel line of BE; suppose the line intersects BC at point J: DJ // BE. Suppose that DJ intersects AF at point Q (Q should be between G and F). We work out that DQ = 2 EG (notice E is the midpoint of AD), as well as GQ = AG. So if it held that DQ : QJ = 2 : 3, we would prove that DJ = BE.

(Why doing that? If BE = DJ were proved, then JDEB would be a parallelogram. Parallelogram JDEB implies that AD // BC. Can you see that? Just note ED // BJ, and  segment ED is on line AD, resp. BJ is on BC; so we have, finally, AD // BC). So we’ve proved that AD and BC, as one pair of the opposite sides of a quadrilateral, is parallel. We can analogously attempt for AB // CD. Now by definition, ABCD is a parallelogram.

However to work out DQ : QJ = 2 : 3 some effort is needed, and the good news is we are on the track.

A rather interesting way (of course challenging but a pleasant experience), isn’t it? For anyone who wants to do more systematically, or automatically, is there another way?

Yes, there is. That way is to use coordinate system.

Suppose the two hanging segments AF and BE are perpendicular to each other, and let their intersection (G) be the origin: so G(0,0). With the given GE : GA = GA : GB = 1 : 2 : 3 : 4 (we’ll call it1-2-3-4 condition), we draw the following diagram that involves G, E, A, B, F to help us understand, and to figure out coordinates to be E(0,1), A (2,0), F (-3,0) and B (0,-4).

Coordinates of the other two vertices can be worked out from the conditions that E is the midpoint of AD, as well as F is the midpoint of DC. (Noting the positive direction of x-axis points from G to A.) We worked out D (-2,2) and C (-4,-2). If it’s indeed that ABCD is one parallelogram, then the two diagonals AC and BD bisect each other, which means that the midpoint of the two diagonals are the same, as verified straightforwardly:

Midpoint of AC: ((2-4)/2, (0-2)/2) ) = (-1, -1)

Midpoint of BD: ((0-2)/2, (-4+2)/2 ) = (-1,-1)

Since AC and BD bisect each other, ABCD is a parallelogram. But what if the AB and BE are not perpendicular? Well ABCD is still a parallelogram in case – but explanation is needed – for which we can pursue later. It deserves to point out, by changing the x- and y- axis from being orthogonal to arbitrary, whether the shape is parallelogram or not will have the same answer.

Let me ask if you would like this problem. If like, then share it with your friends!

Continuing from the problem discussed above, suppose that apart from AE, BF, we draw CJ and DK, where J is midpoint of AB and K is the midpoint of BC. Here is your turn to take a ramp-up challenge:
(If you truly understand what’s going on in the above discussion, then it’s just a piece of cake!)

(i) Can you show that there are exactly four intersections among the four lines AE, BF, CJ and DK; and the four intersection points connect to make a parallelogram?

(ii) What type of parallelogram is the one mentioned in question (i)? Is it a rectangle or rhombus?

*(iii) Is it possible for the original parallelogram to be a rectangle ABCD? In that case, if GE : GA : GF : GB = 1 : 2 : 3 : 4, what would be the aspect ratio AB : BC (aspect ratio is the ratio of width-to-length in a rectangle shape).

不一样的数学探索 —有趣,直观又有用

可以吗?在平面上任给5个点 (where no three points are on the same line),可以选3个点用直线连成三角形,让另外2点都在这三角形的内部吗?有些情况下这能做到,但是总能作到吗?当这个仿佛天边飘过来的问题(既简单又似无从下手)提给 15左右的学生时,不少就表现出跃跃欲试的兴趣。有一个立即发表意见:"总是可以,只要找到最外面的三个点,余下的就在里面。"

Yes Interior -2同伴反驳了,他举了一个例子:取呈正方形的四个点,无论取哪3个,都不能把另一点包含在其中。第五点呢,比方说在正方形内部。还是没办法取出含另两点在内部的三角形。

Not-Interior - 1教育中从简单开放的问题出发,易于参与。比方说 什么叫“最外面的点”?反驳的孩子的思路也好棒,他是举例说明有不行的情况。提醒注意,要鼓励的是探索,动脑筋思考,发表意见要重事实和逻辑,不嘲笑攻击别人。这才是安全探索的环境。和同行讨论时有共识,启发思考,东西方都有好的做法,有共通的东西。简单贴上东方,西方教育的标签,捧一个打一个,是我们一直不赞成的。


数学家,数学教育家 Polya 讲过(大意),好老师善于提问题引导学生深入。通过上面讨论,文章开头的问题不是总能做得到。有时能,有时不能。那么什么时候能,什么时候不能呢?可以想一下,答案在文中说明能和不能的两幅插图中找。先注重观察特征,然后再想怎么表达清楚。有兴趣的切磋一下,问问你的孩子。





(English version is being prepared and corrected – to be posted soon)

你听到过正整数的关联Murray数吗?其实是一个蛮有趣的问题,基础在于完全平方数的简单推理–简单归简单,可是要严密哦!今年的CIMC(加拿大中级数学竞赛 – 910年级组)就出了这样的问题。


For each positive integer n, the Murray number of n is the smallest positive integer M, with M >n, for which there exists one or more distinct integers greater than n and less than or equal to M whose product times n is a perfect square.

对于正整数n, 指定n 的(关联)Murray数是有如下性质的正整数中最小的那个:存在一或多个不同的大于 n 但不超过M的整数, 这些数的乘积 再乘上 n 得到一个完全平方数。

本文注:(关联)Murray数可就叫Murray 数。“关联”一词主要强调 Murray M和给定整数n 间是一个数学关系。完全平方数就是整数的平方,如1, 4, 9, 16, 25, .. 一直下去。

例子:3 × 6 × 8 = 122, 且找不到一或多个大于3 但是都比8更小的整数使其积再乘3是完全平方。

这个定义要多读两遍,确保理解。定义是写在题目开头的,所以无需事先知道,但是做题前一定要弄清意义。然后看下面的–就是CIMC 的问题 。

(甲)6 Murray 数被发现是 12。说明为什么。

(乙)决定 8 Murray 数。


比如这样的说明:因为 6 × 8 × 12 = 576 = 242 而且 6 < 8 < 12, 所以 12 合乎Murray数的定义。注意了!这个说明有点问题。因为按定义,任何数的关联Murray数只有一个,要尽量的小。所以验证 12 符合条件还不够,必得说明在6 12之间其他的数(7, 8, 9, 10, 11)都不行,从而排除。

现在说明 7 (或 11)不能成为 6 Murray数。他们不能在定义的完全平方式中出现。小于7的数字不能含素数因子7,所以7再次出现要等以后(至少等到14 吧)。11 是同样道理。9 是完全平方,假使在完全平方式中现身,那末去掉因子9以后还是完全平方数。8 10 也不行(理由从略)。

现在来看问题(乙)。答卷中有认为 8 Murray 数是18 的,理由是:

8 x 18 = 12 2
或 8 x 9 x 18 = 36 2

9 18 都大于8,且不超过18

这解不正确!有些细节可能被忽略了。第一是Murray 数只有一个,是满足如此如此性质中最小的那个。以上验证不足说明 8 Murray 数是 18,只是说明Murray数肯定小于 18 (依定义也知大于8)。此外在 完全平方的表达式中应有几个因子?定义里没说,唯一要求是这些因子相异。(不能因为例子中给出3 个因子,就把特例当一般。)看下面的式子:
8 x 10 x 12 x 15 = 1202
或者 8 x 9 x 10 x 12 x 15 = 3602

按定义,8 Murray 数必得不超过15. 且找不到更小M <15可用n, M 间的因子做积再乘上8是完全平方数。故 15 是解。

这问题真让人痒痒 – 答题时或者说明不严谨,或者理解有点偏差;于是与正确的解失之交臂。


**********  ***********  **********  **********  **********


(丙)证明有无限多个这样的正整数n 满足:n 不是完全平方数,n Murray 数小于 2n.

这题的证明很容易跑题。要求既不是说明 n Murray 数可以小于 2n,也不是要说明有无限多个正整数n。也非 n Murray 数一定小于2n。这样说吧(也许从反面理解更容易)假如有人声称他已经找到了所有关联Murray数小于2n — 即原数2 倍的情形;那你一定可以对他讲“且慢,你一定漏掉了一些”。

(丙)其实不难,有兴趣的可钻研。我们给两个提示(窗户纸已捅破 – 再往前走一步便是证明了)。

假如有人声称他已经找到了所有关联Murray数小于2n — 即原数2 倍的情形; 那么一定可以告诉你最大的一组。现在你对他讲:我还有更大的一组,你一定漏掉了。这样开启了如下对话:(假定你是 B

A. 我已经找到了所有 n关联Murray数小于2n — 即原数2 倍的情形。一共有有限个。

B. 你找到所有这种情形的数了,并且有限个。那其中一定有最大的。能告诉我最大的n 吗?

A. 当然。n = xxxx.

B. 哦,4n Murray 数也一定会小于4n 2 ; 所以你把 4n 漏掉了。

A. 那就再加上 4n, 其中 n 是我刚讲的数。

B. 你一定还漏掉了4 (4n), 也就是 16n.

看出来了吧…… 这样下去一定没完没了。