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30/03/2023

Due to unforseen circumstances Burling Tuition will be closed until further notice.

15/02/2023

Stage 6 Extension I (Year 11 and 12 3- and 4-unit Mathematics)
Topics: Absolute Value, Inequalities

This is a later post than usual, because I'm typing this up after a session.

So, I was posed the following: |x - 4| + |x + 2| > 7. This is a pretty standard question, however trying to solve this using positive/negative method won't work, as it'll result in variables being removed. So I suggested squaring both sides.

Now, here's where I made my initial mistake. I squared the inequality as presented. Not a good idea because the resulting left hand side expression becomes unwieldy (specifically, it became a quartic). Especially since the student in question actually provided a way to circumvent this when we tried the positive/negative method: move one of the absolute values to the right hand side.

So the working is as follows:
|x - 4| > 7 - |x + 2|
(x - 4)² > (7 - |x + 2|)²
x² - 8x + 16 > 49 - 14|x + 2| + (x + 2)²
x² - 8x + 16 > 49 - 14|x + 2| + x² + 4x + 4
14|x + 2| > 12x + 37
196(x + 2)² > (12x + 37)²
196(x² + 4x + 4) > 144x² + 888x + 1369
196x² + 784x + 784 > 144x² + 888x + 1369
52x² - 104x - 585 > 0
x > (104 + √(104² + 4 × 52 × 585))/(2 × 52) or x < (104 - √(104² + 4 × 52 × 585))/(2 × 52)
x > (104 + √132496)/104 or x < (104 - √132496)/104
x > (104 + 364)/104 or x < (104 - 364)/104
x > 468/104 or x < -260/104
x > 4½ or x < -2½

Now, will this method work with all inequalities of the form |x + b| + |x + d| > n? Yes. Will it work for all inequalities of the form |ax + b| + |cx + d| > n? Not as easily. It may still result in a quartic, so you may need to work out the roots of it (which seems to be possible).

03/08/2022

Stages 5.2 and 5.3 (Years 9 and 10 Intermediate and Advanced)
Topics: Factorising Quadratic Expressions, Solving Quadratic Equations

I think most people who have encountered quadratics (i.e. expressions in the form ax² + bx + c) know about the PSF method. If you do not, then I'll do a brief overview for you with an example, 6x² + x - 15 (if you learnt cross method, then please read over this knowing that you have my deepest sympathy):
1. We multiply the coefficient of x² (i.e. the number in front of the x², which in this case is 6) by our constant term (which is -15); this gives us a total of -90, which we assign to P (which stands for “product”; you'll see why in a minute).
2. We now take the coefficient of x (as you can see in this example, there's nothing there in front except for a plus, which means the coefficient is 1) and assign that to S (which stands for “sum”; again, you'll see why in the next step).
3. Now, find two numbers that multiply together top give you P and add together to give you S. So, in this case, our P is -90, and our S is 1; the two numbers we are looking for are -9 and 10. These will be our F (standing for “factors” which will be used in the next step).
4. Write out two sets of brackets. In each one, put the coefficient of x² followed by an x, then one of the factors. So, in our example, we will write (6x – 9)(6x + 10).
5. Put these brackets as a fraction over the coefficient of x². (Put simply, in our example, we draw a vinculum under both sets of brackets, then a 6 under that.)
6. Fractorise out each set of brackets, if possible. In this case, we can take 3 out of the first bracket, and 2 out of the second, giving us 3(2x – 3)2(3x + 5)/6.
7. Now simplify; this'll give us (2x – 3)(3x + 5)
And there you have it. Despite how I've explained it, it's actually quite an easy method to use for factorising and finding the roots of quadratics.

Yes, I said that correctly. You can use PSF to find the roots of a quadratic. (Just to remind everyone, the roots of an equation, also called the zeroes or the x-intercepts, are the values of x which make the expression 0.) And the process for that is just as simple.

The first three steps are the same as above. Once we get our factors (our Fs), we then change their signs (so plus to minus, and vice versa), then divide them by the coefficient of x². So, for our example, we have -9 and 10. We switch their signs (9, and -10), then divide by 6, giving us 3/2 and – 5/3.

Extraordinarily simple.

25/05/2022

Stage 5.3 (Years 9 and 10 Advanced)
Topics: Area, Trigonometry

Helo, everyone. I know it's already a term and a half into the new year and I'm finally making my first post for the year; I had a lot of ideas for posts, and just didn't follow through on them mostly due to being busy. Then this question turned up, and I realised that this would be the perfect one to talk about. It's (supposedly) from a GED paper, and had a lot of people stumped. To be fair to them, it does require some knowledge that they may not have looked at.

Below you'll see a diagram of three circles, all the same radius (the question specified 4cm; we are generalising here with x), with centres P, Q and R. You are tasked with finding the total area of the pieces ACQ and BDQ.

The key behind this question is in the lens-shaped pieces APCQ and BQDR, which are each made up of two congruent segments. Since these'll be equal in size, we'll only look for the area of one.

The area for a segment requires two pieces of information: the radius (which we have: x), and the angle over which it subtends (which we do not). To find the angle, we'll use the radius and sagitta (the perpendicular "height" of the segment to the chord) to find the cosine ratio of half the angle; the reason for this is that the radius, sagitta and half the chord makes a right angle triangle, with the sagitta being adjacent to the half-angle. Working backwards from that, we find that half-angle to be 60°, π/3 radians (which I'll use from here on in); hence, our subtended angle is 2π/3 radians, or 120°.

From here, we use the area of a segment formula, A = r²(θ - sin θ)/2 to get the area of one segment (note that this is for use with radian measure; using degree measure adjusts this to r²(θπ/180 - sin θ)/2), which is x²(π/3 - √3/4), then multiply that by four (two for each lens, two lenses altogether), giving us x²(4π/3 - √3).

Finally, after doing all that, we subtract that value from the area of the disc (πx²), and we get x²(√3 - π/3) as our area. For the radius in the original question (4cm), this came to about 10.96cm².

26/01/2022

Stage 3 (Years 5 and 6)
Topic: Factors

Let's talk about abundance.

Say you have a positive integer n. This number has factors; σ(n) represents the sum of these factors. The abundance of n is the result of σ(n) - 2n. (I'm not sure why they chose the term "abundance".)

Abundant numbers have a positive abundance. All multiples of perfect numbers (discussed below), save for the perfect numbers themselves, are abundant numbers. If the abundance is 1, they are called quasiperfect; we have not yet found a quasiperfect number.

A perfect number has an abundance of 0. A number of the form (2^n - 1)2^(n - 1) is perfect if 2^n - 1 is prime.

A deficient number has a negative abundance. (As such, the sign switch of abundance is referred to as the "deficiency".) All prime numbers are deficient, as are their powers. An abundance of -1means the number is almost perfect; the powers of 2 (i.e. 2^n) are examples of almost perfect numbers.

17/12/2021

Now for the solution to the final quiz for the year. And I will admit to not completely thinking it through when I wrote it, as it could be solved by a Year 7 or 8 student.

We can start by equating two of the sides, 2x+y+6=2x+4y+3. Rearranging this equation gives us the result y=1, which we can now put into all three equations.

Having done that, we can equate another pair of sides, 3x+5=2x+7. (Remember, at this point, we have substituted y for 1.) Rearranging again, we get x=2, which we again substitute into all equations. This gives us a result of 11.

Thus, the length of each side of the triangle is 11cm.

12/12/2021

Year 9 and 10 Immediate and Advanced
Topic: Equations

Final quiz for the year. The question is pretty simple: Assuming the measurements are in centimetres, and the triangle is equilateral, find the length of each side.

Solution Friday.

16/11/2021

Stages 5 and 6, All Levels
Topic: Simultaneous Equations

(For reference: https://www.news.com.au/lifestyle/parenting/school-life/hsc-maths-exam-question-about-emu-and-goanna-legs-leaves-students-fuming/news-story/ecd87d92b27bbaec90da8342fe774eaa)

Let's chat about this one.

I didn't need to see the question to know what they were talking about, since it's actually a pretty standard question in maths. You have two kinds of animals in an enclosure (usually one has two feet, the other four). There are a certain number of animals (let's call it c, just to keep things open) in the yard, and they have a certain number of feet altogether (d). How many of each are there altogether?

(Now, just as a note, d is always even, and 2c

High school students were left reeling after the HSC maths exam on Monday, with one question in particular leaving many completely stumped.

29/10/2021

Time for Monday's quiz, and we'll start off by looking at the angles for each triangle (since each needs to gave an internal angle sum of 180°).

For the first triangle, the angles total 6x. So 6x = 180°, or x = 30°. Thus, the angles in that are 30°, 60° and 90°. (This already looks promising for us in terms of future working out.)

Similarly, in the second triangle, the angles add up to 8x + 60. So 8x + 60° = 180°, or 8x = 120°, or x =15°. And thus the angles are 30°, 60° and 90°.

So, we've proven they're similar, but we need the sides equal in order to be congruent. Well, it's made easier by our triangles, for which we know the ratios of sides (being the basis is two sets of standard trig ratios). To make this easier for us, the hypotenuse (longest side) of each triangle is 16, the shortest side is 8, and that gives use enough information to use one of our congruency tests (you can pick which one) to thus prove congruency.

24/10/2021

Stages 5 and 6, all levels.
Topics: Congruency, Trigonometry

Time for the Monday quiz. And this one is pretty simple: prove the two triangles below are congruent.

Solution Friday. Good luck.

15/10/2021

Time for the solution to Monday's quiz...well, solutions. As I said, I found five, and I should thank one of my students for one of them, as it was the prompt for this quiz. (Diagrams for the respective solutions are given below.)

The first two solutions are the solutions we expect as teachers. The first one, we construct an interval through x parallel to the other two parallel lines. Then we can use alternate angles to work out the size of x. In the second solution, the parallel construction is opposite to x, and we use cointerior angles to find the reflex angle to x.

The third solution is courtesy of my student. You construct an interval between the parallel lines through the vertex of x, giving us two triangles, and a pair of cointerior angles. Angle sum of a triangle gives us the two angles surrounding x, and angles on a line gives us the solution for x. We need to be careful, as the line we draw in should not become an extension of either arm of x, otherwise it becomes solution four. In that case, once we produce the arm, we can use alternate angles to get a second angle for the triangle, then use external angle of a triangle to find x.

The final solution forms a triangle with the vertices of a and b. This forms a pair of cointerior angles a + c and b + d; rearranging the equality gives us c + d = 180 - a - b. We can then use internal angle sum of a triangle to work out x.

In all of these cases, the solution becomes x = a + b.

Find a different solution? If you didn't share it in Monday's quiz post, feel free to put it below; I'd love to know what other solutions you guys come up with.

10/10/2021

Stage 4 (Years 7 and 8)
Topics: Angle Relationships

So I have been busy during the latest lockdown here in NSW, sorting out Data Dumps (by the way, current counts are two completed, one being written, and one pending) and managing things from home; so the Monday quizzes have been put on the backburner in the meantime.

Now that we're coming out, it's time for a Monday quiz. And this one's going to be less "find the solution" and more "find as many ways to solve it as possible". And I'm going to do it with a often-posed question.

Below is the diagram of the question. The quiz: find x. (And just to clarify, since I didn't mark it on the diagram: the horizontal lines are parallel.)

I found five solutions, which I'll put up on Friday. Feel free to add your solutions below.