TNDOT Friday 11: Algebra

October 18, 2024

4:46 pm

This is a continuation of last week’s TNDOT, the next four questions from the 1911 pass Algebra Matric exam (one of which, the either-or Q6, has already been posted and solved as part of TNDOT 5). This post seems to contain, by the way, a very early Victorian exam error.

I’ve also posted solutions to Q1-Q3 of last week’s TNDOT. I’m leaving Q4 for now, since readers have not yet offered much on it. Which means you guys are behind on your homework. (And we’re not going outside to play rounders until you have finished your homework.)

Have fun.

5. State and prove the Remainder Theorem; and shew that $\color{blue}\boldsymbol{a-x}$ is a factor of

$\color{blue}\boldsymbol{\aligned &bc(bc-ax)(x-a)^2 + ca(ca-bx)(x-b)^2\\+ ab(ab&-cx)(x-c)^2 +(ax-bc)(bx-ca)(cx-ab).\endaligned}$

Factorize this expression completely.

6(α) if a men or b boys can dig m acres in n days, find the number of boys whose assistance will be required to enable a – p men to dig m + p acres in n – p days.

(β) To complete a certain work A requires a times as long as B and C would take if they worked conjointly ; B requires b times as long as C and A and would take if they worked conjointly ; and C requires c times as long as A and B would take if they worked conjointly. Prove that

$\color{blue}\boldsymbol{\qquad \dfrac1{1+a} +\dfrac1{1+b}+\dfrac1{1+c}=1 \,; }$

and find the time which each would take separately to do the work, assuming that if they all worked together they would require h hours.

7. Find the condition that*

$\color{blue}\boldsymbol{ax^2 + 2hxy+by^2 +2gx+2fy + \color{OliveGreen}c}$

may be the product of two factors of the first degree in $\color{blue}\boldsymbol{x}$ and $\color{blue}\boldsymbol{y}$ .

Prove that this condition is satisfied for the expression

$\color{blue}\boldsymbol{6\left(x^2-y^2\right) +5xy-10x+11y -4;}$

and find its factors.

*) In Red Five’s photocopy, the first quadratic is missing a constant term, the expression ending with a hanging plus sign. So, above I’ve added in the c. This could be RF’s bad photocopying but it doesn’t appear to be. I have no idea if the error appeared on the actual exam papers or if the error was created later during reproduction.

8. Employ the method of graphs to find four sets of solutions to the simultaneous equations

$\color{blue}\boldsymbol{\aligned &x^2-3y=13\\&y^2+3x=13.\endaligned}$

UPDATE (25/10/24)

Thanks, as always, to commenters for their hints and thoughts and solutions. As usual, I’ll update the post with solutions question by question.

Q5. Here is the polynomial again:

$\color{Plum}\boldsymbol{\aligned P =\ & bc(bc-ax)(x-a)^2 + ca(ca-bx)(x-b)^2\\[3\jot] \qquad &+ ab(ab-cx)(x-c)^2 +(ax-bc)(bx-ca)(cx-ab).\endaligned}$

It is worth comparing to TNDOT 3 (which also includes a discussion of proving the remainder theorem). As tom and Alasdair noted, the key is that the polynomial is symmetric: swapping around the a, b and c keeps the polynomial identical.

So, once we know (x – a) is a factor, we also know (x – b) and (x – c) are as well. Then, since the polynomial is (at most) cubic, we just have to figure out the coefficient out the front. But that is easy to do, just by looking for the x³ terms in the cubic: these sum to a coefficient of -3abc + abc = -2abc. So, our polynomial is,

$\color{Plum}\boldsymbol{P = -2abc(x-a)(x-b)(x-c)\,.}$

Finally, to prove (x – a) is factor, of course we apply the remainder theorem, plugging in x = a and crossing our fingers that everything cancels. Which it does, but not obviously so. Plugging in x = a, we get,

$\color{Plum}\boldsymbol{\aligned P(a) = \ & 0 + ca(ca-ba)(a-b)^2 + ab(ab-ca)(a-c)^2 +\left(a^2-bc\right)(ba-ca)(ca-ab)\\[3\jot] =\ &a^2(c-b)\left[c(a-b)^2 -b(a-c)^2 +\left(a^2-bc\right)(b-c) \right]\\[3\jot] =\ &a^2(c-b)\left[ca^2 -2abc +cb^2 -ba^2 +2abc -bc^2 +a^2b - a^2c -b^2c + bc^2 \right]\\[3\jot] =\ &0\,.\endaligned}$

UPDATE (25/10/24)

The solution to Q6 is reproduced from TNDOT 5.

6(α) seems to require care and perseverance but no real cleverness. Each man can dig m/na acres per day, and each boy can dig m/nb acres per day. So, if we have a – p men and, say, B boys, then in n – p days they will dig together

$\color{Plum}\boldsymbol{\frac{m(a-p)(n-p)}{na}+ \frac{mB(n-p)}{nb}}\,.$

We require m + p acres to be dug. Equating with the above, and doing some obvious simplification, this gives,

$\color{Plum}\boldsymbol{{(a-p)b + Ba}= \frac{nab(m+p)}{m(n-p)}}\,.$

We can then solve for B, although we can’t see any particular way to make the answer pretty:

$\color{Plum}\boldsymbol{{ B}= \frac{nb(m+p)}{m(n-p)}-\frac{(a-p)b}{a}}\,.$

6(β) seems trickier, although not bad when you get the right angle on it. Let’s have A also stand for the rate (work-units/hour) at which A can work, and similarly for B and C. Then B and C working together work at the rate B + C, and so on. With this terminology, we are then told

$\color{Plum}\boldsymbol{B +C = aA\qquad A + C = bB\qquad A + B = cC }\,.$

Solving for a, b and c respectively in these equations, and then adding 1, we have,

$\color{Plum}\boldsymbol{1+ a= \frac{A + B+C}{A}\qquad 1+ b= \frac{A + B+C}{B} \qquad 1+ c= \frac{A + B+C}{C} }\,.$

Taking the reciprocals and summing gives the required equation.

For the second part, the rate at which A and B and C work together is A + B + C. From the above equations (or thinking about the initial information), this is 1 + a times the rate at which A works alone. So, if all of them working together takes h hours, then A alone will take (1 + a) times as long, and thus A will take (1 + a)h hours. Similarly, B and C will take (1 + b)h and (1 + c)h hours, respectively.

UPDATE (25/10/24)

Q7. As James and tom have noted, this is naturally interpreted as a conic sections question. It can then be tackled with determinants and/or rotation of coordinates. I think, however, James is right, that straight-forward algebraic crunching was expected, and we’ll basically reproduce James’s solution. It also feels like a canned question, that students knew they were expected to learn this derivation (however it was intended to be done).

We begin with

$\color{Plum}\boldsymbol{ax^2 + 2hxy+by^2 +2gx+2fy + c\,,}$

and we think of it as a quadratic in x:

$\color{Plum}\boldsymbol{ax^2 + (2hy+2g)x + (by^2 + 2fy + c)\,.}$

The quadratic will factor (as an x thing), if the discriminant is a perfect square (so that square root in the quadratic formula doesn’t give us rooty things). The discriminant is,

$\color{Plum}{\boldsymbol{\aligned \Delta &= (2hy+2g)^2 - 4a\left(by^2 + 2fy + c\right)\\[3\jot] &= 4h^2y^2+8ghy+4g^2 - 4aby^2 - 8afy - 4ac\,.\endaligned}$

This is a quadratic in y:

$\color{Plum}\boldsymbol{\Delta = 4\left(h^2-ab\right)y^2 + 8(gh-af)y + 4\left(g^2-ac\right).}$

Now recall that we want $\boldsymbol{\Delta}$ to be a perfect square. So, we calculate a second discriminant $\boldsymbol{\delta}$ , and the question is when $\boldsymbol{\delta}$ is zero. So,

$\color{Plum}\boldsymbol{\delta = 64(gh-af)^2 - 64\left(h^2-ab\right)\left(g^2-ac\right)\,.}$

This can be simplified a little, but there seems to be no particular value in doing so.

We are then given,

$\color{Plum}\boldsymbol{Q = 6\left(x^2-y^2\right) +5xy-10x+11y -4\,.}$

So,

$\color{Plum}\boldsymbol{a = 6,\ h = \dfrac52,\ b = -6,\ g = -5,\ f =\dfrac{11}2,\ c = -4\,.}$

Then,

$\color{Plum}\boldsymbol{\dfrac{\delta}{64} = \left(-\dfrac{25}{2} -\dfrac{66}2\right)^2 - \left(\dfrac{25}{4}+36\right)\left(25+24\right) = \left(-\dfrac{91}{2}\right)^2 -\left(\dfrac{169}{4}\right)\left(49\right)=0\,.}$

Now that we know Q factorises, it is not difficult to guess how, but we can also use the calculations here to note $\boldsymbol{\Delta = (13y-14)^2}$ . It then follows that we can factorise,

$\color{Plum}\boldsymbol{\aligned Q &= 6\left[x +\dfrac{5(y-2)+(13y-14)}{12}\right]\left[x +\dfrac{5(y-2)-(13y-14)}{12}\right]\\[3\jot] &= 6\left(x +\dfrac{18y-24}{12}\right)\left(x +\dfrac{-8y+4}{12}\right)\\[3\jot] &= 6\left(x +\dfrac{3y-4}{2}\right)\left(x +\dfrac{-2y+1}{3}\right)\\[3\jot] &= \left(2x + 3y-4\right)\left(3x -2y+1\right)\,. \endaligned}$

UPDATE (25/10/24)

Q8. We have to solve the system

$\color{Plum}\boldsymbol{\left\{\aligned &x^2-3y=13\\&y^2+3x=13.\endaligned\right.}$

As tom notes, this is easy to do by first subtracting the two equations. Presumably, as Alasdair suggests, “the method of graphs” means that the students are intended to graph the things and note the symmetry about the line y = -x. See below. But, that doesn’t seem to help much with the actual solving.

Anyway, to solve the system, subtract the first equation from the second, giving,

$\color{Plum}\boldsymbol{y^2 +3x - x^2 + 3y = 0\,.}$

That easily factorises to give,

$\color{Plum}\boldsymbol{(y-x+3)(y+x) = 0\,.}$

Then y = x – 3 lead to the quadratic equation $\boldsymbol{x^2 - 3x - 4=0}$ . So, x = 4 or x = -1, giving solutions (4, 1) and (-1, -4).

Alternatively, y = -x leads to the quadratic equation $\boldsymbol{x^2 + 3x - 13=0}$ , So, $\boldsymbol{x=\dfrac{-3\pm \sqrt{61}}{2}}$ . That gives solutions $\boldsymbol{\left(\dfrac{-3\pm \sqrt{61}}{2}, \dfrac{3\mp \sqrt{61}}{2}\right)}$ .

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22 Replies to “TNDOT Friday 11: Algebra”

Alasdair says:

October 18, 2024 at 5:58 pm

I may or may not give these a go later on, but the use of A, B and C in 6(beta) reminds me of this excellent piece by the Canadian humorist Stephen Leacock: https://www.online-literature.com/stephen-leacock/literary-lapses/40/ I may have posted this before, in which case tell me to stop repeating myself, and remove this post.

Reply
1. marty says:
  
  October 18, 2024 at 5:59 pm
  
  “Stop repeating yourself”, but your comment can stay. The piece is hilarious.
  
  Reply
2. Red Five says:
  
  October 18, 2024 at 6:34 pm
  
  I think this is one of the funniest things I have read all year.
  
  Many thanks for sharing.
  
  Reply
tom says:

October 18, 2024 at 7:44 pm

Do we post answers here?

Reply
1. marty says:
  
  October 18, 2024 at 7:45 pm
  
  Yes. The general idea is to give people a go. So ideas and broad thoughts for a day or two, and then full solutions are welcome and encouraged.
  
  Reply
Ed Barbeau says:

October 19, 2024 at 2:50 am

If you give the problem about A, B and C and the “certain work” to modern kids, you might ask them for a particular situation (i.e. numbers for which it works) and whether it is necessarily the case that each individually requires the same amount of time. I think there is a lot to be gained by asking students to look for examples.

I wonder what sort of preparation the students would have had in 1911 to tackle questions like this. One can imagine that they might have been drilled in a lot of the “tricks of the trade”.

Reply
tom says:

October 19, 2024 at 9:05 am

Q5 symmetry
Q6 Problems like this were still current when I was at school. Enough already.
Q7 Taking a lazy birds-eye view, one could perform a translation to remove the linear terms, then a rotation to eliminate the $xy$ term. Then we have $pX^2+qY^2+r$ . If $p$ and $q$ are different signs then we could factor as the difference of squares? But I’m too lazy to actually do all this. And a direct approach is probably what was intended anyway?
Q8 What’s the difference?

Reply
marty says:

October 19, 2024 at 9:34 am

Thanks, tom. Any sense of what “the method of graphs” means?

Reply
1. Red Five says:
  
  October 19, 2024 at 2:23 pm
  
  One would assume it meant to plot or sketch graphs of both relations on the same axes but exactly what this means for the level of accuracy expected of answers is then up for debate.
  
  I assume students in schools in 1911 would have used graph paper.
  
  Reply
  1. marty says:
    
    October 19, 2024 at 2:24 pm
    
    Really? I would not assume graph paper.
    
    Reply
2. tom says:
  
  October 20, 2024 at 12:57 pm
  
  No, I guessed they meant draw the graphs, but my graphs were too rough to help much. So I just used algebra.
  
  Reply
  1. marty says:
    
    October 20, 2024 at 1:00 pm
    
    Thanks, tom. I’m guessing the point was to graph/sketch enough to appreciate the symmetry, as indicated by Aladasdair below. But clearly “method of graphs” was an expression of the time, and I wonder what the method was supposed to encompass.
    
    Reply
Alasdair says:

October 19, 2024 at 5:15 pm

Tom beat me to it for (5); symmetry plus the expression being cubic in x. 6(β) – Stephen Leacock aside – can be managed in terms of the fractions of work each of A, B, C can do in say, one hour. It’s a neat result though. For (8), once you observe that the graphs of the equations are symmetric about the line y = -x, you see that two solutions are on that line: (a, -a) and (b, -b), and the other two are symmetric about it: (c,d) and (-d,-c). “Details are left to the reader” (aka I can’t be bothered working through the algebra myself.)

What interests me though is that the nature of the these problems is so completely unlike what school students see today. There seems to me to be more of an emphasis on proof and results. Also, students seemed to be expected to manage reasonably complicated algebraic expressions. I would assume that students who passed this level of mathematics would arrive for tertiary studies quite well equipped. Unlike students I’ve had the pleasure of teaching, who are completely flummoxed by fractions.

Anyway, I shouldn’t indulge in the “it was better in the past”, because of course it always was, and has been for 2500 years.

Reply
tom says:

October 20, 2024 at 1:08 pm

I recall reading eons ago, maybe it was W. W. Sawyer, where the author stated that his interest in a complicated algebra problem was working out how the question arose. Here in Q5, I cannot imagine anyone constructing this from scratch. It smells a little like something to do with triangles?

Reply
1. marty says:
  
  October 20, 2024 at 1:10 pm
  
  Huh. Interesting thought. I doubt triangles were expected as part of the solution, but a good question where such an expression came from.
  
  Reply
Alasdair says:

October 20, 2024 at 8:50 pm

That it has 4 additive terms makes me wonder if it might originate from a Lagrange interpolating polynomial.

Reply
Red Five says:

October 23, 2024 at 8:00 pm

There seems to be a (little) bit more to Q7 than I first assumed. Which is good, it promotes reflection and deeper thinking.

My approach was to start with the various ways of getting the $ax^2$ and $yx^2$ terms and to go from there. Did anyone try a different method and, if so, how did it fare?

Reply
1. marty says:
  
  October 23, 2024 at 8:05 pm
  
  Thanks, RF. Been flat out and haven’t even glanced at TNDOT 11. (I did take a look at TNDOT 10, and am still trying to find a “natural” approach based on what you guys have done.)
  
  Reply
  1. Red Five says:
    
    October 23, 2024 at 9:09 pm
    
    All good. You will note I made an error in my previous comment: $by^2$ was what I should have written, but instead I wrote $yx^2$ . I blame the Clare Valley Shiraz…
    
    Reply
James says:

October 24, 2024 at 4:40 am

Question 7

I wonder how familiar students back then would have been with conics. Some may have noticed that $ax^2 + 2hxy + by^2 +2gx + 2fy + c$ is just another way of writing the general Cartesian form for a conic section. And that this question is asking for what condition will the conic be degenerate.

Which is when

$\begin{vmatrix} a & h & g\\h & b & f\\g & f & c\end{vmatrix} = abc + 2fgh - af^2 - ch^2 - bg^2 = 0$

Tom had the right idea here but I also suspect a different approach was probably expected back then. Possibly something along the lines of the following:

$\aligned ax^2 + 2hxy + by^2 +2gx + 2fy + c \endaligned$

Rewrite as a quadratic in x (or y)

$\aligned ax^2 + (2hy+2g)x + (by^2 + 2fy + c) \endaligned$

Find the discriminant

$(2hy+2g)^2 - 4a(by^2 + 2fy + c)$
$4h^2y^2+8ghy+4g^2 - 4aby^2 - 8afy - 4ac$
$4(h^2-ab)y^2 + 8(gh-af)y + 4(g^2-ac)$

Another quadratic, this time in y (or x). Again, find the discriminant. Will factor when the discriminant is zero.

$\aligned 8^2(gh-af)^2 - 4(4)(h^2-ab)(4)(g^2-ac) = 0\\ (gh-af)^2 - (h^2-ab)(g^2-ac) = 0\\ g^2h^2-2afgh+a^2f^2 - (g^2h^2-ach^2-abg^2+a^2bc) = 0\\ -2afgh + a^2f^2 + ach^2 + abg^2 - a^2bc = 0\\ abc + 2fgh - af^2 - ch^2 - bg^2 = 0 \endaligned$

Easy to check that this condition is satisfied for the second equation. Someone else can do the factoring.

Reply
1. marty says:
  
  October 24, 2024 at 8:28 am
  
  Thanks, James. Yes, it’s an odd one. There was plenty of focus upon Euclid at the time, but I don’t know that I’ve seen anything on conics. I suspect that some straight out algebraic hammering was expected, of the type you’ve done.
  
  I’ll look at what people done and browse through some books at the time, as soon as i can. (I still have TNDOT 10 to polish off, a million emails, a blog post, and life.)
  
  Reply
  1. Red Five says:
    
    October 24, 2024 at 9:21 am
    
    Euclid was tested in totally separate papers, aptly named Euclid Books 1 and 2 or similar. Cambridge papers at the time did a similar thing, so I can assume that is where Australia took the idea from.
    
    But your idea of using a matrix determinant is very interesting. As far as I can tell, Matrices were NOT in the curriculum but Determinants WERE, up to 3×3.
    
    Reply

UPDATE (25/10/24)

UPDATE (25/10/24)

UPDATE (25/10/24)

UPDATE (25/10/24)

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