TNDOT Friday 7: Algebra

September 20, 2024

9:39 am

algebra, exams, polynomials, rational functions, TNDOT

Yes, it’s TNDOT Week, here in CrapLand.

We’ve decided to try to make TNDOT a regular Friday thing. So, the plan is to post an old exam problem sometime each Friday day (Melbourne time), and then readers can ponder and discuss over the week-end. Once readers have provided enough in the way of solutions, we’ll update the post, presumably early each following week.

There will be no hard and fast rule about when readers’ full solutions should be postable, but maybe Sunday evening Melbourne time makes sense? Of course before then readers are permitted, and encouraged, to share approaches and provide hints and the like, as is natural for the discussion at hand. General principle: this is supposed to be fun and educational for all, not a competition. So be looking to help each other without too much limelight hogging.

This one is in a similar vein to the previous, and is from the 1916 Victorian Pass Algebra Matric exam. Enjoy.

3. Find the relation between the coefficients if the equations

$\color{Blue}\boldsymbol{ax^2 + bx + c = 0}$

and $\color{Blue}\boldsymbol{a'x^2 + b'x + c' = 0}$

have a common root.

Prove that if the equation $\color{Blue}\boldsymbol{ax^2 + bx + c = 0}$ has a root in common with $\color{Blue}\boldsymbol{bx^2 + cx + a = 0}$ , then its other root satisfies

$\color{Blue}\boldsymbol{a^2bx^2 + ax\left(2b^2-ac\right) + a^3 + b^3 -abc = 0\,.}$

UPDATE (24/09/24)

Thanks to Franz and hjm and Rusty Polly and Red Five for their solutions and suggestions. Here is the solution to the first part, guided by their ideas. I’m still pondering what is most natural and simplest for the second part.

We’ll switch notation and use capitals, just because it seems a little clearer. So, we have $\boldsymbol{P}$ the common solution to two equations:

$\color{Plum} \boldsymbol{(1) \qquad ap^2 + bp + c = 0\,,}$

$\color{Plum} \boldsymbol{(2)\qquad Ap^2 + Bp + C = 0\,.}$

We now want to find a relationship between the coefficients. (As noted by RF, this was possibly a standard question of the time, expecting a formulaic response.)

RF does this by a direct approach. Here, we’ll use Viéta’s formulas (see the previous TNDOT), to illustrate their use and since it seems a little cleaner. So, let $\boldsymbol{q}$ and $\boldsymbol{Q}$ be the second solutions to the given quadratic equations. Then Viéta gives us,

$\color{Plum} \boldsymbol{(3) \qquad p + q = -\dfrac{b}{a}\qquad p + Q = -\dfrac{B}{A}\,.}$

$\color{Plum}\boldsymbol{(4) \qquad pq = \dfrac{c}{a}\qquad pQ = \dfrac{C}{A}\,.}$

We then have to do something simultaneous-ish, but it’s not too hard to see a way in. Subtracting the equations in (3) and then the equations in (4), we obtain two expressions for $\boldsymbol{Q-q}$ :

$\color{Plum} \boldsymbol{(5) \qquad Q-q = \dfrac{b}{a}-\dfrac{B}{A}= \dfrac{Ab-aB}{aA}\,,}$

$\color{Plum} \boldsymbol{(6) \qquad p(Q-q) = \dfrac{C}{A}-\dfrac{c}{a}= \dfrac{aC-Ac}{aA}\,.}$

Dividing (6) by (5), we then have,

$\color{Plum} \boldsymbol{(7) \qquad p = \dfrac{aC-Ac}{Ab-aB}\,.}$

From here, there are a few ways to go, but neatest seems to take $\boldsymbol{B(1)-b(2)}$ . (Why do this? Because it spits out the denominator in our expression for $\boldsymbol{p}$ , and because the $\boldsymbol{x}$ coefficient disappears.) So, we obtain,

$\color{Plum} \boldsymbol{(8) \qquad (aB-Ab)p^2 + (cB-Cb)=0\,.}$

Plugging in the expression for $\boldsymbol{p}$ from (7), and watching the minus signs, we obtain,

$\color{Plum} \boldsymbol{\phantom{(8)} \qquad \dfrac{\left(aC-Ac\right)^2}{Ab-aB} = (cB-Cb)\,,}$

which simplifies to,

$\color{Plum} \boldsymbol{\phantom{(9)} \qquad {\left(aC-Ac\right)^2} = ({Ab-aB})(cB-Cb)\,.}$

This is the desired relation for the coefficients.

UPDATE (26/09/24)

(27/09/24. See minor update below)

For the second part, we have $\boldsymbol{p}$ the common root of

$\color{Plum} \boldsymbol{(1) \qquad ap^2 + bp + c = 0\,,}$

and

$\color{Plum} \boldsymbol{(10) \qquad bp^2 + cp + a = 0\,,}$

Now, let $\boldsymbol{q}$ be the other root of (1). Then we want to show

$\color{Plum}\boldsymbol{a^2bq^2 + aq\left(2b^2-ac\right) + a^3 + b^3 -abc = 0\,.}$

We’ll do this by simply hammering the LHS until it disappears. In order to do so, and to give the clue to doing this, we use the first Viéta formula, (3), which gives

$\color{Plum} \boldsymbol{(11) \qquad p + q = -\dfrac{b}{a}\,.}$

Rearranging, we then have

$\color{Plum} \boldsymbol{(12) \qquad aq + b = -ap \,.}$

How doe this help? Well, rearranging our mess, we have

$\color{Plum}\boldsymbol{\aligned \mbox{\bf MESS} &= a^2bq^2 + aq\left(2b^2-ac\right) + a^3 + b^3 -abc \\[3\jot] &= a^2bq^2 + 2aqb^2 + b^3 - a^2cq + a^3 -abc \\[3\jot] &= b\left(a^2q^2 + 2aqb + b^2\right) - aq(ac) + a^3 -abc\\[3\jot] &= b\left(aq+b\right)^2 - aq(ac) + a^3 -abc\,.\endaligned}$

We now use the Viéta equation (12) on both the quadratic term and the linear term:

$\color{Plum}\boldsymbol{\aligned \mbox{\bf MESS} &= ba^2p^2 + ac(ap+b) + a^3 -abc\\[3\jot] & = ba^2p^2 + a^2cp + abc + a^3 -abc\\[3\jot] &= a^2(bp^2 + cp + a)\\[3\jot] &= 0\,,\endaligned}$

where the last line follows from (10). Done.

There’s a moral there, but I don’t know what it is.

UPDATE (27/09/24)

Having posted this, it occurred to me that probably what was expected was to plug (11) for $\boldsymbol{p}$ directly into (10), and then you get what you get. So, going from (10) to MESS, rather than from MESS to 0.

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46 Replies to “TNDOT Friday 7: Algebra”

Red Five says:

September 20, 2024 at 5:15 pm

This is not going to be an attempt to answer either problem but more of an attempt to start a discussion about possible approaches to the question.

Suppose equations $ax^2+bx+c=0$ and $a'x^2+b'x+c'=0$ have a common root, this can mean a few different things.

If we let $ax^2+bx+c=a(x-p)(x-q)$ and $a'x^2+b'x+c'=a'(x-r)(x-s)$ and we assume for the sake of argument that $p\leq q$ and $r\leq s$ then the most obvious conclusion is that $p=r$ and/or $q=s$ .

But it is also possible that neither of these are true and instead the common root is because $p=s$ or $q=r$ .

Each of these leads to a different equation to be solved/simplified and so there may be multiple answers here.

Can anyone see an issue with this approach or suggest why it may not be an efficient path to solution?

Reply
1. Franz Lemmermeyer says:
  
  September 21, 2024 at 5:16 am
  
  Let $p$ denote the common root . . . .
  
  Reply
Sentinel says:

September 21, 2024 at 12:38 am

One thing I would like to clarify – do we need to consider complex numbers/coefficients/roots?

Reply
1. Red Five says:
  
  September 21, 2024 at 8:38 am
  
  In 1916 complex roots would have been examined in the Algebra Honours paper. This was the Pass exam, so no.
  
  Also, after some further playing around with the idea I floated before, I believe there IS a simpler approach.
  
  Whether it is that much simpler I am yet to determine.
  
  In short: suppose $ax^2+bx+c=0$ and $a'x^2+b'x+c'=0$ have a common root, as Franz suggested, let it be $x=p$ .
  
  Then it follows that $x=p$ must also be a root of $(a-a')x^2+(b-b')x+(c-c')=0$ .
  
  Whether this makes for any meaningful simplification of the problem I’m not sure.
  
  Reply
  1. Rusty Polly says:
    
    September 22, 2024 at 8:46 pm
    
    If a, b, c are real and a common root is complex non-real, would not both other roots be complex conjugate(s) of the same, so both roots are then matched?
    
    Reply
    1. Red Five says:
      
      September 22, 2024 at 8:53 pm
      
      Fair point. Yes, I think that is valid.
      
      Reply
2. marty says:
  
  September 21, 2024 at 9:21 am
  
  Hi, Sentinel. As RF suggests, complex roots would probably not be expected here. But I’d also be surprised if whether or not the roots are complex or not will play a role in the solution.
  
  Reply
marty says:

September 21, 2024 at 12:34 pm

In case it matters, I should point out that I don’t know how to do this question. Whenever choosing a TNDOT, I choose a question that makes me react, “Huh. I wonder how you do that”. I tend to then think no more about it until readers have begun to make suggestions, and then not much until the solutions come in.

With this question, I have a couple thoughts, different from RF’s and Franz’s, but I’ve spent no more than a minute or two coming to those thoughts.

Reply
Red Five says:

September 21, 2024 at 2:00 pm

Looking at the second part of the problem for a moment:

If $ax^2+bx+c=0$ and $bx^2+cx+a=0$ have a common root, it can be shown that $a+b+c=0$ .

I will leave it to others for now to decide if (and how) this is helpful.

Reply
1. Rusty Polly says:
  
  September 22, 2024 at 9:14 pm
  
  subtracting both equations gets:
  (a-b)Xsq +(b-c)X + (c-a) =0 (3)
  For X=1
  a-b +b-c +c-a =0
  so X = 1 is a root of (3)
  We kind of know it is this common root, but this needs to be shown….
  
  If a+b+c = 0 is provided, then
  aXsq + bX +c |X=1 = a+b+c =0, so 1 is a root of (1)
  Similarly, bXsq +cX +a | X=1 = b+c+a = 0 and so X=1 is this common root of (1) and (2).
  The other root of (1) would then be, say q
  p+q =-b/a
  q= -b/a -1 = – (b+a)/a = c/a
  Verifying that other root satisfies (4), Y9 work now.
  
  How many problems in this 1916 exam? How much time? How many of them solved required to pass?
  
  Reply
  1. marty says:
    
    September 22, 2024 at 9:20 pm
    
    Thanks, RP. I’ll look at your, and others’, solution tomorrow.
    
    I’m not sure how long the exam was. I’d guess three hours but that’s only a guess. The students had to answer eight questions, from a choice of ten. It’s not clear what constituted a pass, or how that was determined. My sense from reading some of the history is that plenty of students failed.
    
    Reply
2. marty says:
  
  September 24, 2024 at 10:07 pm
  
  RF, what is your argument that a + b + c = 0?
  
  Reply
  1. Red Five says:
    
    September 25, 2024 at 8:25 am
    
    I was not entirely correct with that statement upon reflection, because it is also possible that $a=b=c$ .
    
    The relation came up a few times in similar questions.
    
    Reply
    1. Rusty Polly says:
      
      September 25, 2024 at 5:40 pm
      
      If a=b=c equations are identical, simplify to Xsq +x +1 = 0
      solutions are complex (conjugates) and are the 2 cube roots of unity (non real ones).
      Are they not? Just had a go at this scenario.
      Whether p2 satisfies the equation….
      
      Reply
      1. marty says:
        
        September 25, 2024 at 7:33 pm
        
        Thanks, Rusty Polly. I’ll look at this more tomorrow.
        
        ps It’s preferable if you LaTeX (or html) your maths. It’s easy to do for simple things like polynomials and fractions.
        
        Reply
hjm says:

September 21, 2024 at 9:18 pm

Partial solution with Mathematica:

Let $p_1=b/a$ , $q_1=c/a$ , $p_2=b'/a'$ , $q_2=c'/a'$ .

Then the two equations have a common root iff

$p_1^2 q_2 + p_2^2 q_1 + (q_1 - q_2)^2 = p_1 p_2 ( q_1 + q_2 )$ .

For the second pair of equations, we have $p_1=c/b$ and $q_2=a/b$ . The condition then becomes

$a^3 + b^3 + c^3 = 3 a b c$ .

Reply
1. marty says:
  
  September 21, 2024 at 9:29 pm
  
  “Mathematica”.
  
  Reply
2. Red Five says:
  
  September 21, 2024 at 10:12 pm
  
  The question wants an equation that contains $a,b,c,a',b',c'$ only. There is quite a nice solution to the first part requiring algebra skills that are well within the grasp of a good Year 9 student but it does require not thinking in the way modern textbooks and their exercises seem to want people to think.
  
  Your (partial) solution to the second part has merit but needs some work.
  
  Reply
Red Five says:

September 21, 2024 at 10:16 pm

OK. I’ll post one more hint before Monday when I had planned to post a more complete solution:

There is a special case to be aware of – namely what happens if $ab'=a'b$ but beyond this, there is no need to consider complex roots nor complex coefficients.

Reply
Red Five says:

September 22, 2024 at 8:49 pm

Having read a few examination papers from around the time of this paper, I have formed the belief that students are likely to have been taught the conditions for two quadratic equations to have a common root and this exam question may not have been asking for the derivation of the rule, but rather evidence that students were aware of the relation.

Pretending that is not the case (or a student had forgotten the rule), here is one way to arrive at the result:

Let $ax^2+bx+c=0$ and $a'x^2+b'x+c'=0$ have a common root and let that common root be $x=r$ .

Then we can write $ar^2+br+c=0$ and $a'r^2+b'r+c'=0$

Rearranging, these equations become $ar^2+br=-c$ and $a'r^2+b'r=-c'$

If we consider this a system of two equations where the unknowns are $r^2$ and $r$ we can solve simultaneously to obtain $r^2=\frac{bc'-b'c}{b'a-ba'}$ and $r=\frac{ac'-a'c}{a'b-ab'}$

Squaring the second equation and equating the two expressions for $r^2$ gives $\frac{(ac'-a'c)^2}{(a'b-ab')^2}=\frac{bc'-b'c}{b'a-ba'}$

And therefore $(ac'-a'c)^2=(bc'-b'c)(b'a-ba')$

It may or may not have been expected to say that if $b'a-ba'=0$ then the equations have $two$ common roots.

I will allow other commenters to give a solution to the second part (although hjm has covered the crux of it in a previous comment)

Reply
1. marty says:
  
  September 22, 2024 at 9:16 pm
  
  Thanks, RF. That’s an interesting question, of what would have been expected to be proved, or simply recited. I noticed another exam from that time had the same preamble “relation” question.
  
  Reply
  1. Red Five says:
    
    September 22, 2024 at 9:51 pm
    
    This was the December 1916 examination. Look at Q2 from the February 1916 paper (presumably the February paper was for students who did not do so well in the December 1915 paper?) if you have it. The first parts are identical.
    
    What that means, I am not sure.
    
    Reply
    1. marty says:
      
      September 22, 2024 at 9:54 pm
      
      Thanks, RF. Yes that was the other question I spotted. I have no understanding of the purpose of the February paper, but I guess it may be something like you suggest.
      
      Reply
marty says:

September 24, 2024 at 11:02 pm

I’ve updated the post with a solution to the first part of the problem. I’m still pondering the most natural approach to the second part.

Reply
1. Red Five says:
  
  September 25, 2024 at 8:28 am
  
  My first thought was to name the other root $q$ and then say $q=\frac{c}{a} \times \frac{1}{p}$ where $p$ is the common root of the two equations.
  
  Whether this is neater than trying $q=\frac{-b}{a}-p$ I cannot yet decide and I suspect there is another, much nicer solution hiding somewhere that for now eludes me.
  
  Reply
  1. marty says:
    
    September 25, 2024 at 10:19 am
    
    Thanks, RF. The common root p is a cube root of unity, and I *think* the problem can be hammered through similar to the manner outlined by RP. But I’m not sure I get the second part of this problem, what was intended by the examiners. I’ll look more later.
    
    Reply
    1. Red Five says:
      
      September 25, 2024 at 12:45 pm
      
      It is odd, I’ll give it that. I’ve seen quite a few similar questions on JSE (an examination in India) papers but none that use this specific form, which is a bit surprising as the common root comes up rather frequently.
      
      Reply
      1. marty says:
        
        September 25, 2024 at 1:12 pm
        
        Yes, it’s a weird question. I think I’ve figured out what they want, except it didn’t work. Will check my work in a bit.
        
        Reply
        
        Red Five says:
        
        September 25, 2024 at 2:32 pm
        
        OK. I tried briefly but couldn’t see how the sum was zero, so assume I made an algebraic error.
        
        I might do hjm’s trick and see what Mathematica gives me later.
        
        The main concern I have is that if the common root is $p$ , then the second root can be found either as $\frac{-b}{a}-p$ or as $\frac{c}{a} \times \frac{1}{p}$ and these two approaches keep giving me (seemingly) different results.
        
        Reply
        
        marty says:
        
        September 25, 2024 at 2:33 pm
        
        I think the trick is to use both, along with p being a root of unity.
        
        Reply
Red Five says:

September 25, 2024 at 8:17 pm

I have given the second part of this question much thought over a few days (school holidays are meant for just this type of activity, right…?) and have come to some intermediate conclusions:

1. If we can show that the common root is $x=1$ then the problem becomes much easier and writing the other root as $x=\frac{-b}{a}-1$ and then substituting pretty much solves the whole thing.

2. This was the Pass level exam, so I wonder if students would have instinctively tested $x=0$ and $x=1$ to see if they worked.

3. The command term “Prove” has me worried that point 1 may be irrelevant.

Reply
1. Rusty Polly says:
  
  September 25, 2024 at 11:18 pm
  
  Is there some room for considering cx^2 +bx +a = 0 ? For a full circle.
  Solution to all 3 simult. is…. x=1.
  And a+b+c =0
  If we could prove either ( p=1 or a+b+c = 0 ) then the rest is smooth sailing, correct.
  
  For a moment I thought there was a close connection between Part A and Part B.
  If the condition found by you and Marty implies there is a common root, the OTHER root of Eq 1 and the Eq 3 have a common root, which could be shown by putting
  A = a^2b, B = a(2b^2-ac) and C= a^3+b^3 +abc into it… did not work for me very well.
  Substituting x= -(a+b)/a , c=-(a+b) works .
  
  Reply
  1. marty says:
    
    September 25, 2024 at 11:58 pm
    
    RP, again please do proper typesetting. It ain’t that hard.
    
    I’ll indicate below where I am, which might fit in with what you’re doing. I’ll look again tomorrow.
    
    Reply
Rusty Polly says:

September 25, 2024 at 11:27 pm

Is the relation found “iff” or not?

Reply
1. marty says:
  
  September 26, 2024 at 12:00 am
  
  What relation? The relation I gave in the update? I don’t know, but you could work ti back to see (modulo zero denominators). But I don’t see that it would help with the second question.
  
  Reply
marty says:

September 26, 2024 at 12:13 am

OK, here are my (unchecked) thoughts earlier today on the second part of the question. We have for the shared root $\boldsymbol{p}$ ,

$\boldsymbol{ap^2 + bp + c = 0}$

and

$\boldsymbol{bp^2 + cp+ a = 0}$

If we multiply the first equation by $\boldsymbol{p}$ and then subtract the two equations, we get

$\boldsymbol{a p^3- a = 0}$ .

So either $\boldsymbol{a = 0}$ , giving a non-quadratic case, or we have

$\boldsymbol{p^3=1\,.}$

Now, maybe for this being a pass exam, we are supposed to ignore complex possibilities and conclude $\boldsymbol{p = 1}$ . If not, or anyway, let $\boldsymbol{q}$ be the other root of the original equation. We are then looking for a totally different quadratic equation satisfied by $\boldsymbol{q}$ , the mess given in the statement of the question.

I know how to get such a quadratic, but not the given mess.

By Viéta, we have

$\boldsymbol{p = -q -\dfrac{b}{a}}$

and

$\boldsymbol{pq=\dfrac{c}{a}\,.}$

It’s now easier to multiply through both equations by $\boldsymbol{a}$ .

In any case, cubing the first equation gives a cubic for $\boldsymbol{q}$ . But we get an expression for $\boldsymbol{q^3}$ by cubing the second equation. Plugging that in, we then obtain a quadratic for $\boldsymbol{q}$ . But unless I goofed somewhere, we don’t get the desired quadratic, the mess given in the statement of the question. The quadratic we get can be rewritten, by using the relationship established in the first part of the problem, but it doesn’t quite seem to work.

Reply
Red Five says:

September 26, 2024 at 9:00 am

Assume that we are only looking for real roots. Then $x=1$ is the common root.

The other root is then $x=\frac{-b}{a}-1=\frac{-a-b}{a}$

This means $x^2=\frac{a^2+2ab+b^2}{a^2}$

From there we can show that $a^2bx^2+ax(2b^2-ac)=a^2b-b^3+a^2c+abc$

Adding the constant term of $a^3+b^3-abc$ makes the strange expression $a^2b+a^2c+a^3$

Which factorises to $a^2(a+b+c)$

And, as Rusty Polly showed, $a+b+c=0$ and we are done.

(I actually think we can use $x=\frac{-b}{a}-p$ and then use the fact that $x=p$ is a common root without need to show $p=1$ but I haven’t managed to properly crunch the algebra on that one as yet. May or may not happen.)

Reply
1. marty says:
  
  September 26, 2024 at 10:00 am
  
  Thanks, but even if this works, it feels unnatural. One would expect that the given quadratic, or something close to it, arises from a natural procedure, not that it can simply be crunched out.
  
  Reply
  1. Red Five says:
    
    September 26, 2024 at 11:00 am
    
    Absolutely agree. I just cannot find said “natural” procedure.
    
    Reply
marty says:

September 26, 2024 at 9:56 pm

OK, I figured it out, although I still don’t get it. Will update the post in a little while.

Reply
1. Red Five says:
  
  September 27, 2024 at 2:37 pm
  
  Looking at a few papers from the time, I think the point of the question was to be an algebra work out more than anything else. If that was the case, I think it succeeds.
  
  Reply
  1. marty says:
    
    September 27, 2024 at 2:38 pm
    
    Yes, and the algebra wasn’t that hard. It is interesting how looking for something too clever can distract from the straight-forward. Which is the moral, I guess.
    
    Reply
marty says:

September 26, 2024 at 10:27 pm

The post is updated. A short rest and then TNDOT 8 …

Reply
1. Red Five says:
  
  September 27, 2024 at 8:56 am
  
  You have used (11) twice. I think the final Vieta equation should be numbered (12). Otherwise, very nicely done.
  
  Reply
  1. marty says:
    
    September 27, 2024 at 9:28 am
    
    Ah thanks. Will fix.
    
    Reply
  2. marty says:
    
    September 27, 2024 at 10:51 am
    
    Fixed, plus a small update.
    
    Reply

UPDATE (24/09/24)

UPDATE (26/09/24)

UPDATE (27/09/24)

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46 Replies to “TNDOT Friday 7: Algebra”

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