NCEA Level 2 Mathematics, Algebra Exam 2024

A review of the excellence questions in this exam.

QUESTION ONE (d)

A cubic polynomial is given by `f(x) = 3x^3+ ax^2+ bx + c`, where `a`, `b`, and `c` are constants.

The graph of the polynomial has three roots, and cuts the x axis at `-2`, `1/3`, and `4`.

Using your knowledge of roots, find the value of the constants.

If we can factorise the cubic, each factor should correspond directly to one of the stated roots.

The coefficient `3` on the `x^3` term is a prime number, which is only factorisable one way (using whole numbers), so the coefficients on each x in the factors will be `3`, `1`, and `1`. These can be in any order, but if we keep them in the same order as the stated roots, the `3` will be in the middle factor.

The other factors are easy to get. Setting each factor equal to zero gives the three roots, and working backwards can be done just as easily. For example:

`x=4`
`implies x-4=0`
`implies (x-4)` is a factor

`(x + 2)(3x - 1)(x - 4)=0`

Expand to get the function with its coefficients.

`3x^3 - 7x^2 - 22x + 8 = 0`

Check each root to confirm the equation is correct. (Using the x variable memory on your calculator and your calculator history, you only need to enter the polynomial expression once.)

`a=-7`
`b=-22`
`c=8`

Comment: This doesn't seem hard enough to be an Excellence question.

QUESTION ONE (e)

Show that `((x^(3/2) + x^(1/2))(x^(1/2) - x^(-1/2)))/(x^(3/2) - x^(1/2))^2` can be simplified to `(x+1)/(x(x-1))`.

Use exponent rules to expand. The denominator is already a perfect square. The numerator becomes the difference of two squares, which can be factorised and simplified with the denominator. We start by using F.O.I.L. carefully (multiplying together first terms, outside terms, inside terms, last terms), then applying the exponent rules which say we add or multiply the exponents. This is one of those times that being able to expand a perfect square (directly) is really useful.

`(x^(3/2)x^(1/2) - x^(3/2)x^(-1/2) + x^(1/2)x^(1/2) - x^(1/2)x^(-1/2)) / ((x^(3/2))^2 - 2x^(3/2)x^(1/2) + (-x^(1/2))^2)`

`(x^(4/2) - x^(2/2) + x^(2/2) - x^0) / (x^3 - 2x^(4/2) + x^1)`

`(x^2 - x + x - 1)/(x^3 - 2x^2 + x)`

The denominator has a common factor of `x`. Taking that out leaves a perfect square.

`(x^2 - 1)/(x(x^2 - 2x + 1))`

`((x + 1)(x - 1))/(x(x-1)^2)`

`(x + 1)/(x(x-1))`

Comment: Intimidating fraction! But this is just a straight forward application of algebra, especially exponent rules and resist the temptation to do to much with each step of working; each step doesn't need to be a long stride. Students should be able to recognise the difference of two squares, and the simple perfect square used.

QUESTION TWO (c)

Find the value of `k` for which the equation `(x^2 - 2x)/(4x - 1) = (k - 1)/(k + 1)` has roots numerically equal but opposite signs (for example, `2` and `-2`).

I think this equation will be easiest to work with without denominators and all terms on the left.

`(x^2 - 2x)(k + 1) = (k - 1)(4x - 1)`

`kx^2 + x^2 - 2kx - 2x = 4kx - k - 4x + 1`

`kx^2 + x^2 - 2kx - 4kx - 2x + 4x + k - 1 = 0`

`(k + 1)x^2 + (- 6k + 2)x + (k - 1) = 0`

Good so far. This is now in the typical exanded form of quadratic, a good starting point for whatever we need to do next.

`ax^2 + bx + c = 0`

For one root to be the negative of the other, the vertex needs to be at `x = 0`; we want a parabola with no horizontal movement from the origin. With an expanded form quadratic equation this means the `x` term must be zero, so `b` must be zero.

`ax^2 + c = 0`

`=> - 6k + 2 = 0`

`6k = 2`

`k = 1/3`

Comment: I imagine this will stump a few students, but a good understanding of how quadratics work and what we're trying to get to makes the problem quite easy to solve.

QUESTION TWO (d)

The logo for an IT company, Doctor of Data, is shown in the diagram to the right.

The green background of the logo is the shape of a rectangle with length `x`. The letters D, O, D are formed using semi-circles for both Ds and a circle for the O. Each has the same radius.

Area of a circle = `pir^2`.

(i) Find an expression in terms of `x` for the area of the circle in the middle of the logo (i.e. the letter 'O').

radius of circle `= x/4`

area of circle `= pi(x/4)^2`

area of circle `= (pix^2)/16`

(ii) If the green background has an area of 10 cm², find the length of the rectangle.

Make an equation in `x` (the thing we're asked to find) using the expression for the area of the circle, and solve for `x`. Sometimes I like to start with a word equation; it gives me a plan of attack.

area of 2 circles + green area = area of rectangle

`(2pix^2)/16 + 10 = x(x/2)`

`(pix^2)/8 + 10 = x^2/2`

`pix^2 + 80 = 4x^2`

`4x^2 - pix^2 = 80`

`(4 - pi)x^2 = 80`

`x^2 = 80/(4 - pi)`

`x = sqrt(80/(4 - pi))`

`x = 9.654` cm (4sf)

Comment: Not too hard.

QUESTION THREE (d)

Consider the equation `3x^2 - 4kx + k^2 = 0`, where `k` is a constant.

Using the quadratic formula, find the fully simplified solutions to the quadratic equation in terms of `k`.

I find it's helpful to write down the formula I'm going to substitute stuff into. It means I can more easily see what I'm going to put where. I also like to clearly identify a, b, and c by underlining the coefficients in the given equation – including any negative signs!

`x = (-b +- sqrt(b^2 - 4ac))/(2a)`

`x = (4k +- sqrt((-4k)^2 - 4(3)(k^2)))/(2(3))`

`x = (4k +- sqrt(16k^2 - 12k^2))/6`

`x = (4k +- sqrt(4k^2))/6`

`x = (4k +- 2k)/6`

`x = (2k +- k)/3`

`x = k/3 or x = k`

Comment: Straight forward.

QUESTION THREE (e)

The level of sound (intensity) is measured on a logarithmic scale using a unit called a decibel. The formula for the decibel level, `d`, is given by

`d = 10log_10(P/P_0)`

where `P` is the intensity of the sound and `P_0` is the weakest sound that the human ear can hear. A cooling fan has a decibel level of 38 decibels. A heat pump has a decibel level of 30 decibels.

Show that the cooling fan sound intensity is more that six times that of the heat pump sound intensity.

We need to show the ratio `P_"fan"/P_"pump" > 6`, which means rearranging the given formula. We don't need to separate the ratio `P/P_0` because the `P_0` will cancel out.

`d/10 = log_10(P/P_0)`

`10^(d/10) = P/P_0`

This is what we need. Now we can make the ratio we want.

`P/P_0 = 10^(d/10)`

`(P_"fan"/P_0)/(P_"pump"/P_0) = (10^(d_"fan"/10))/(10^(d_"pump"/10))`

`P_"fan"/P_"pump" = 10^((d_"fan"/10)-(d_"pump"/10))`

`P_"fan"/P_"pump" = 10^(38/10-30/10)`

`P_"fan"/P_"pump" = 10^(3.8-3.0)`

`P_"fan"/P_"pump" = 10^0.8`

`P_"fan"/P_"pump" = 6.309... > 6`

Comment: Again, not very hard for well-prepared students who are clear on what they need to do.

Conclusion

The algebra questions that I looked at were not particularly difficult or unclear. On that basis it seems like an OK exam.

The algebra exam was held on 5 November. The NCEA Level 2 algebra exams in 2016 and 2017 were both held on 24 November. Why such a big loss of learning time?

I've always thought it extremely strange that the NCEA exam rules/guidelines on the front page of each exam do not mention the duration of the exam, or mention how long all three should take.