# 2020 HSC Maths Ext 1 Exam Paper Solutions

In this post, we reveal the solutions to the 2020 HSC Maths Extension 1 Exam paper.

Our Mathematics team have been hard at work putting together the 2020 HSC Maths Extension 1 Exam Paper solutions for you. Here they are!

## 2020 HSC Maths Ext 1 Exam Paper Solutions

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## Section 1: Multiple Choice

 Question Answer Solution 1 A The quadratic factorises to $$x^2 -2x – 3 = (x-3)(x+1)$$, so the solution is $$x>3 \cup x<-1$$. 2 C The domain and range of $$f(x) = 1+\sqrt{x}$$ is $$D:x\geq 0, R: y\geq 1$$, so conversely the domain and range of $$f^{-1}(x)$$ is $$C:x\geq 1, R: y\geq 0$$ 3 D The standard antiderivative involved is $$\frac{1}{x^2+a^2} = \frac{1}{a}\tan^{-1}(\frac{x}{a})+c$$, and since $$\frac{1}{4x^2+1} = \frac{1}{4} \frac{1}{x^2 + \frac{1}{4}}$$, we have the antiderivative $$\frac{1}{4} \frac{1}{1/2}\tan^{-1}(\frac{x}{1/2})= \frac{1}{2} \tan^{-1}(2x)+c$$. 4 B First, we sum up all the vectors to get Maria’s final position, which is $$2i + 3j + 3i – 2j + 4i – 3j = 9i-2j$$. The magnitude of this is $$\sqrt{9^2 + 2^2} = \sqrt{85}$$. 5 C Since the polynomial is monic, its leading coefficient must be +1, making it overall concave up. This eliminates A and B. Next, $$x^2 + x + 1$$ has no solutions, hence the polynomial can only touch the x-axis at its double root (it cannot cross it). This leaves C as the correct answer. 6 D For $$a-b$$, we put $$a$$ and $$b$$ tip-to-tip and join the tails. 7 A Checking the quadrants, we expect the derivative to be positive in the 2nd and 4th quadrants and negative in the 1st and 3rd quadrants. This eliminates D (positive all quadrants) and B (some negative in 2nd quadrant). Next, we note that as we approach the $$x$$-axis, $$dy/dx$$ should become larger as the denominator tends to zero; rather than flattening out. Hence the solution is A. 8 C First we choose 6 from 10, giving us $$^{10} C_6 =\frac{10!}{4!6!}$$. Next, we choose 4 from the 6, this time caring about order, giving $$^6 P_4=\frac{6!}{2!}$$. In total this is $$\frac{10!}{6!4!} \frac{6!}{2!}=\frac{10!}{4!2!}$$. 9 B We can draw a diagram to show the situation: Hence, the answer should be $${4 \choose 8} + {4 \choose 8} – \left({6 \choose 7}-{4 \choose8}\right)={2 \choose 9}$$. 10 A We have \begin{align} &\frac{dP}{dt}=-l^2 \times \frac{dR}{dt}\\ &=-l^2 \times (-k^2)\\ &=l^2k^2\\ &>0\\ \end{align} So P is increasing. Now, \begin{align} &\frac{dQ}{dt}=\frac{1}{m^2}\times \frac{dP}{dt}\\ &=\frac{l^2k^2}{m^2}\\ &>0\\ \end{align} Which is also increasing.

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## Section 2: Long response

### Question 11a

(i)

 $$2^3 + 3(2^2) – 13(2) + 6 = 8 + 12 – 26 + 6 = 0$$.

(ii)

Since the quadratic polynomial is monic, the linear factor must be $$A(x) = x-2$$. Therefore

\begin{equation*}
P(x) = (x-2)(x^2 + bx + c)
\end{equation*}

By comparing constants, we get $$-2c = 6$$, so $$c = -3$$. Comparing the coefficients of $$x^2$$, we get $$b – 2 = 3$$, so $$b = 5$$. Therefore

\begin{equation*}
P(x) = (x-2)(x^2 + 5x – 3)
\end{equation*}

### Question 11b

Two vectors are perpendicular if and only if their dot product is 0. Hence we require

\begin{equation*}
{ a \choose -1 } \cdot  { {2a-3} \choose 2} = 0
\end{equation*}

Therefore

 \begin{equation*} a(2a – 3) – 2 = 0 \Rightarrow 2a^2 – 3a – 2 = 0 \Rightarrow (a – 2)(2a + 1) = 0 \\ \end{equation*}

We obtain the two solutions, $$a = 2, -\frac{1}{2}$$.

### Question 11c

In the following graph, the original $$f(x)$$ is shown in blue and $$\frac{1}{f(x)}$$ is in red.

### Question 11d

Observe that

 \begin{equation*} A \sin(x + \alpha) = A(\sin x \cos \alpha + \cos x \sin \alpha) \end{equation*}

Equating this with the expression $$\sqrt{3} \sin x + 3 \cos x$$ gives

\begin{equation*}
\begin{cases}
A \cos \alpha = \sqrt{3} \quad &(1) \\ A \sin \alpha = 3 \quad &(2)
\end{cases}
\end{equation*}

By considering $$(2)$$ divided by $$(1)$$, we deduce $$\tan \alpha = \sqrt{3}$$, therefore $$\alpha = \frac{\pi}{3}$$. Squaring both equations and summing up, we obtain

\begin{equation*}
A^2(\cos^2\alpha + \sin^2\alpha) = 3 + 9 = 12
\end{equation*}

Therefore $$A = \sqrt{12} = 2\sqrt{3}$$, and we get

\begin{equation*}
\sqrt{3} \sin x + 3 \cos x = 2\sqrt{3} \sin \left( x + \frac{\pi}{3} \right)
\end{equation*}

The equation to be solved can now be rewritten as

 \begin{align*} 2\sqrt{3} \sin \left( x + \frac{\pi}{3} \right) &= \sqrt{3} \\ \sin \left( x + \frac{\pi}{3} \right) &= \frac{1}{2} \\ x + \frac{\pi}{3} &= \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \ldots \\ \Rightarrow x &= -\frac{\pi}{6}, \frac{3\pi}{6} = \frac{\pi}{2}, \frac{11\pi}{6} \end{align*}

The first solution is invalid (as $$\frac{-\pi}{6}$$ is outside the domain of interest), so we have the solutions

\begin{equation*}
x = \frac{\pi}{2}, \frac{11\pi}{6}
\end{equation*}

### Question 11e

This is a separable differential equation. To obtain $$x$$ as a function of $$y$$, we separate variables and integrate both sides:

\begin{align*}
\int \frac{dy}{e^{2y}} &= \int \,dx \\
\int e^{-2y} \,dy &= x \\
\Rightarrow x &= -\frac{1}{2}e^{-2y} + C
\end{align*}

### Question 12a

When $$n = 1$$,

\begin{align*}
\text{LHS} &= 1 \times 2 \\
&= 2 \,
\end{align*}

and

\begin{align*}
\text{RHS} &= 1^2 \left(1 + 1\right) \\
&= 1 \times 2 \\
&= 2 \,
\end{align*}

so the statement is true for $$n = 1$$. Assume that the statement is true for $$n = k$$, i.e.

 \begin{equation*} 1 \times 2 + 2 \times 5 + 3 \times 8 + \dots + k \left(3k – 1\right) = k^2 \left(k + 1\right) \,. \end{equation*}

When $$n = k + 1$$,

 \begin{align*} \text{LHS} &= 1 \times 2 + 2 \times 5 + 3 \times 8 + \dots + k \left(3k – 1\right) + \left(k + 1\right) \left[3 \left(k + 1\right) – 1\right] \\ &= k^2 \left(k + 1\right) + \left(k + 1\right) \left[3 \left(k + 1\right) – 1\right] \quad \text{(applying the inductive assumption)} \\ &= k^2 \left(k + 1\right) + \left(k + 1\right) \left(3k + 2\right) \\ &= \left(k + 1\right) \left(k^2 + 3k + 2\right) \\ &= \left(k + 1\right) \left(k + 1\right) \left(k + 2\right) \\ &= \left(k + 1\right)^2 \left[\left(k + 1\right)+1\right] \,. \end{align*}

Now, since the statement is true for $$n=1$$ and $$k+1$$ given $$k$$, by the principle of mathematical induction, the statement is true for all integers $$n \ge 1$$.

### Question 12b

(i)

Recall that the expectation of a binomial random variable is given by $$E(X) = np$$, $$n$$ is the number of trials, and $$p =$$ probability of success. In this experiment, let “heads” be considered a success, and $$n=100$$. Then $$E(X) = np = 100 \times \frac{3}{5} = 60$$.

(ii)

Recall that the variance of a binomial random variable is $$V(X) = np(1-p)$$. In our experiment, we have

 \begin{equation*} V(X) = (100 \times \frac{3}{5}) \times \frac{2}{5} = 60 \times \frac{2}{5} = 24 \end{equation*}

Hence the standard deviation is $$\sigma = \sqrt{V(X)} = \sqrt{24} \approx 5$$.

(iii)

Given the standard deviation is approximately 5 and the expected value is 60, the bounds 55 and 65 correspond to $$z$$-scores of $$-1$$ and $$+1$$ respectively.

For 100 coin tosses we can use a normal approximation of the binomial distribution; and a normal distribution has $$68 \%$$ of values lying between $$z=-1$$ and $$z=+1$$.

Hence the approximate probability is 0.68.

### Question 12c

There are exactly $${8 \choose 3} = 56$$ ways to choose a subset of 3 from a set of 8 courses. Now, with 400 students as pigeons and 56 topic combinations as pigeonholes, notice that

\begin{equation*}
\frac{400}{56} = 7.142\ldots > 7 \\
\end{equation*}

Hence, by the pigeonhole principle, there are at least 8 (the next integer up) students who passed exactly the same three topics.

### Question 12d

Using one of the product-to-sum formulae, we have

 \begin{align*} \sin \left(3x\right) \cos \left(5x\right) &= \frac{1}{2} \left[\sin \left(3x + 5x\right) + \sin \left(3x – 5x\right)\right] \\ &= \frac{1}{2} \left[\sin \left(8x\right) + \sin \left(-2x\right)\right] \\ &= \frac{1}{2} \left[\sin \left(8x\right) – \sin \left(2x\right)\right] \,. \end{align*}

Therefore,

 \begin{align*} \int_0^{\pi/2} \cos \left(5x\right) \sin \left(3x\right) \, dx &= \int_0^{\pi/2} \frac{1}{2} \left[\sin \left(8x\right) – \sin \left(2x\right)\right] \, dx \\ &= \frac{1}{2} \int_0^{\pi/2} \left[\sin \left(8x\right) – \sin \left(2x\right)\right] \, dx \\ &= \frac{1}{2} \left[-\frac{1}{8} \cos \left(8x\right) + \frac{1}{2} \cos \left(2x\right)\right]_0^{\pi/2} \\ &= \frac{1}{2} \left\{\left[-\frac{1}{8} \cos \left(4 \pi\right) + \frac{1}{2} \cos \left(\pi\right)\right] – \left[-\frac{1}{8} \cos 0 + \frac{1}{2} \cos \left(0\right)\right]\right\} \\ &= \frac{1}{2} \left\{ \left[-\frac{1}{8} \left(1\right) + \frac{1}{2} \left(-1\right)\right] – \left[-\frac{1}{8}\left(1\right) + \frac{1}{2} \left(1\right)\right]\right\} \\ &= -\frac{1}{2} \,. \end{align*}

### Question 12e

Separating variables and integrating both sides gives

\begin{equation*}
\int y \, dy = – \int x \, dx \,,
\end{equation*}

so

\begin{equation*}
\frac{y^2}{2} = – \frac{x^2}{2} + C \,,
\end{equation*}

where $$C$$ is a constant. Substituting $$x = 1$$ and $$y = 0$$ into the general solution above gives

\begin{equation*}
\frac{0^2}{2} = – \frac{1^2}{2} + C \,,
\end{equation*}

so

\begin{equation*}
C = \frac{1}{2} \,.
\end{equation*}

Therefore, the curve required is

\begin{equation*}
\frac{y^2}{2} = – \frac{x^2}{2} + \frac{1}{2} \,,
\end{equation*}

which can be written as

\begin{equation*}
y^2 = -x^2 + 1
\end{equation*}

after multiplying both sides of the equation by $$2$$.

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### Question 13a

(i)

\begin{equation*}
\dfrac{d}{d \theta} \left(\sin^3 \theta\right) = 3 \sin^2 \theta \cos \theta
\end{equation*}

(ii)

Let $$x = \tan \theta$$. Then $$\tfrac{dx}{d \theta} = \sec^2 \theta$$, so

\begin{equation*}
dx = \sec^2 \theta \, d\theta \,.
\end{equation*}

When $$x = 0$$, we have $$\tan \theta = 0$$ so $$\theta = 0$$; when $$x = 1$$, we have $$\tan \theta = 1$$ so $$\theta = \tfrac{\pi}{4}$$. Therefore,

 \begin{align*} \int_0^1 \frac{x^2}{\left(1 + x^2\right)^{5/2}} \, dx &= \int_0^{\pi/4} \frac{\tan^2 \theta}{\left(1 + \tan^2 \theta\right)^{5/2}} \left(\sec^2 \theta\right) \, d\theta \\ &= \int_0^{\pi/4} \frac{\tan^2 \theta \sec^2 \theta}{\left(\sec^2 \theta\right)^{5/2}} \, d\theta \\ &= \int_0^{\pi/4} \frac{\tan^2 \theta}{\sec^3 \theta} \, d\theta \\ &= \int_0^{\pi/4} \left(\frac{\sin^2 \theta}{\cos^2 \theta}\right) \cos^3 \theta \, d\theta \\ &= \int_0^{\pi/4} \sin^2 \theta \cos \theta \, d\theta \\ &= \left[\frac{1}{3} \sin^3 \theta\right]_0^{\pi/4} \quad \text{by (a)(i)} \\ &= \frac{1}{3} \left(\frac{\sqrt{2}}{2}\right)^3 \\ &= \frac{\sqrt{2}}{12} \,. \end{align*}

### Question 13b

Let us first find the point of intersection between the two graphs. Of course, we can simply make the educated guess $$x = \frac{\pi}{6}$$ (which is correct), but we include a more systematic method too. We solve $$\cos (2x) = \sin x$$, which is equivalent to

 \begin{equation*} 1 – 2\sin^2 x = \sin x \Rightarrow 2 \sin^2 x + \sin x – 1 = 0 \\ \end{equation*}

Notice that we obtain quadratic in $$\sin x$$, which may be factorised to give

\begin{equation*}
(2 \sin x – 1)(\sin x + 1) = 0
\end{equation*}

The second factor gives $$\sin x = -1 \Rightarrow x = \pi$$, which is outside the domain. Hence the point of intersection must be given by $$2 \sin x = 1 \Rightarrow x = \frac{\pi}{6}$$.

Now we can compute the required volume. Observe that it is obtained by subtracting the volume generated by the curve $$y = \sin x$$ from the volume generated by the $$y = \cos (2x)$$ curve. Hence,

 \begin{align*} V &= \pi \int_0^{\pi/6} \cos^2 (2x) \,dx – \pi \int_0^{\pi/6} \sin^2 x \,dx \\ &= \frac{\pi}{2} \int_0^{\pi/6} (1 + \cos(4x)) \,dx – \frac{\pi}{2} \int_0^{\pi/6} (1 – \cos(2x)) \,dx \\ &= \frac{\pi}{2} \int_0^{\pi/6} \cos(4x) + \cos(2x) \,dx \\ &= \frac{\pi}{2} \left[ \frac{1}{4} \sin(4x) + \frac{1}{2} \sin(2x) \right]^{\pi/6}_0 \\ &= \frac{\pi}{2} \left( \frac{1}{4} \sin \frac{2\pi}{3} + \frac{1}{2} \sin \frac{\pi}{3} \right) \\ &= \frac{\pi}{2} \cdot \frac{3}{4} \sin \frac{\pi}{3} \\ &= \frac{3\sqrt{3}\pi}{16} \text{ units}^3 \end{align*}

### Question 13c

(i)

We calculate the derivative of $$f$$:

 \begin{align*} f'(x) &= \sec^2 \left( \cos^{-1}(x) \right) \frac{d}{dx}(\cos^{-1}(x)) \\ (*) \quad &= \frac{1}{x^2} \cdot \left( \frac{-1}{\sqrt{1 – x^2}} \right) \\ &= \frac{-1}{x^2 \sqrt{1-x^2}} \end{align*}

In the step marked ($$*$$), note carefully(!) that $$\sec^2 \theta = (\sec \theta)^2 = \dfrac{1}{(\cos \theta)^2}$$ by definition, and hence, for $$\theta = \cos^{-1}(x)$$, we find that

\begin{equation*}
\sec^2 \left( \cos^{-1}(x) \right) = \frac{1}{x^2}
\end{equation*}

Now we turn to $$g$$, and use the quotient rule carefully:

 \begin{align*} g'(x) &= \frac{ \frac{-x}{\sqrt{1-x^2}}x – \sqrt{1-x^2} }{x^2} \\ &= \frac{1}{x^2 \sqrt{1-x^2}} ( -x^2 – (1-x^2)) \\ &= \frac{-1}{x^2 \sqrt{1-x^2}} \end{align*}

This proves that $$f'(x) = g'(x)$$.

(ii)

Define the function $$h(x) = f(x) – g(x)$$. Then by part (i), we deduce

\begin{equation*}
h'(x) = f'(x) – g'(x) = 0
\end{equation*}

Therefore $$h(x) = c$$ where $$c$$ is a constant! But what is the value of this constant? We need to plug in a value that lies within the domain of $$h$$, which in turn depends on the domains of $$f$$ and $$g$$.

Now the inverse cosine function is defined only for $$x \in [-1, 1]$$, and $$g(x)$$ is defined for $$x \in [-1, 1]$$ excluding $$x = 0$$. Thus we can plug in $$x = 1$$ to discover

 \begin{equation*} h(1) = f(1) – g(1) = \tan(\cos^{-1}(1)) – 0 = \tan(0) = 0 \end{equation*}

Hence $$h(x) = 0 \Rightarrow f(x) = g(x)$$ for all $$x \in [-1, 1], x \ne 0$$.

Remark: This was a rather tricky question, requiring careful observation and understanding of the relationship between functions.

### Question 14a

(i)

We have the identity

\begin{align*}
(1+x)^{2n} = (1+x)^n (1+x)^n
\end{align*}

Note that $${2n \choose n}$$ is the coefficient of the $$x^n$$ term in the expansion of $$(1+x)^{2n}$$; and from the RHS we notice that there are n ways of constructing the $$x^n$$ term as:

 \begin{align*} {2n \choose n}x^n = {n \choose 0}x^0 \cdot {n \choose n} x^n +{n \choose 1}x^1 \cdot {n \choose n-1} x^{n-1} + \cdot\cdot\cdot +{n \choose n}x^n \cdot {n \choose 0} x^0 \end{align*}

Now, since $${n \choose k}= {n \choose n-k}$$, we can equate coefficients to obtain:

 \begin{align*} {2n \choose n} x^n&= {n \choose 0} \cdot {n \choose 0} x^n +{n \choose 1} \cdot {n \choose 1} x^n + \cdot\cdot\cdot +{n \choose n}\cdot {n \choose n} x^n \\ {2n \choose n} &= {n \choose 0}^2 +{n \choose 1}^2 + \cdot\cdot\cdot +{n \choose n}^2 \end{align*}

as required.

(ii)

Assume $$2k$$ people are chosen for the group, with $$k$$ men and $$k$$ women. There are $${n \choose k}$$ ways of choosing the men, multiplied by $${n \choose k}$$ ways of choosing the women.

Since $$k$$ can be from $$0$$ to $$n$$, we have a total of

 \begin{align*} {n \choose 0}^2 +\binom{n}{1}^2 + \cdot\cdot\cdot +{n \choose n}^2 \end{align*}

total ways of choosing an even-sized group; and from part i) this is equal to $${2n \choose n}$$ as required.

(iii)

In the group of $$k$$ men, there are $$k$$ ways of choosing a leader; and similarly in the group of $$k$$ women there are $$k$$ ways of choosing a leader. Hence, for a group of size $$2k$$, there are

\begin{align*}
k^2 {n \choose k}^2
\end{align*}

ways of choosing groups with a leader. Again since $$k$$ can range from 1 to $$n$$ (as there cannot be a leader in an empty group), the total number of ways is

 \begin{align*} 1^2 {n \choose 1}^2 +2^2{n \choose 2}^2 + \cdot\cdot\cdot +n^2 {n \choose n}^2 \end{align*}

as required.

(iv)

If we choose the leaders first, we have $$n$$ ways of selecting a male leader and $$n$$ ways of selecting a female leader. We now have $$2(n-1)$$ people left to choose from, and we have to choose up to $$n-1$$ males and up to $$n-1$$ females. This is exactly the same scenario as part (ii) with fewer members; so this is equal to

 \begin{align*} n^2 \textrm{(Pick the leaders)} \times {2(n-1) \choose n-1} = n^2 {2n-2 \choose n-1} \end{align*}

### Question 14b

(i)

Expanding $$\sin(3\theta)$$, we have:

 \begin{align*} \sin (3\theta) &= \sin (2\theta) \cos \theta + \cos(2\theta) \sin \theta\\ &=2 \sin \theta \cos^2 \theta + (1-2\sin^2 \theta)\sin \theta\\ &= 2 \sin \theta (1-\sin^2\theta) + \sin \theta – 2 \sin^3 \theta\\ &= 2 \sin\theta – 2 \sin^3 \theta + \sin \theta – 2 \sin^3 \theta \\ &= 3 \sin^3 \theta – 4 \sin^3 \theta \end{align*}

Substituting this into the entire expression above, we have:

 \begin{align*} \sin^3 \theta – \frac{3}{4}\sin\theta + \frac{\sin(3\theta)}{4} &= \sin^3 \theta – \frac{3}{4}\sin \theta + \frac{3 \sin^3 \theta – 4 \sin^3\theta}{4}\\ &= \sin^3 \theta-\sin^3 \theta+\frac{3}{4}\sin\theta-\frac{3}{4}\sin\theta\\ &=0 \\ \end{align*}

as required.

(ii)

Let $$x = 4 \sin \theta$$. Then the equation $$x^3 – 12x + 8 = 0$$ becomes

 \begin{align*} \left(4 \sin \theta\right)^3 – 12 \left(4 \sin \theta\right) + 8 &= 0 \\ 64 \sin^3 \theta – 48 \sin \theta + 8 &= 0 \\ 4 \sin^3 \theta – 3 \sin \theta + \frac{1}{2} &= 0 \quad \text{(dividing the whole equation by $16$)} \\ 3 \sin \theta – 4 \sin^3 \theta &= \frac{1}{2} \,. \end{align*}

Rearranging the equation from part (b)(i) gives

 \begin{align*} \frac{\sin \left(3 \theta\right)}{4} &= \frac{3}{4} \sin \theta – \sin^3 \theta \\ \sin \left(3 \theta\right) &= 3 \sin \theta – 4 \sin^3 \theta \end{align*}

Finally, substituting the RHS from above, we have

\begin{align*}
\sin \left(3 \theta\right) &= \frac{1}{2}
\end{align*}

as required.

(iii)

We note that if $$\theta = \frac{\pi}{18}$$, then $$\sin(3\theta) = \frac{1}{2}$$, so $$\sin \theta$$ where $$\theta = \frac{\pi}{18}$$ is a solution to the equation in part (ii).

Similarly,

\begin{align*}
\sin(3\cdot\frac{5\pi}{18}) &=
\sin{\frac{15\pi}{18}} \\
&= \sin{\frac{5\pi}{6}} \\
&=\frac{1}{2}
\end{align*}

and

\begin{align*}
\sin(3 \cdot \frac{25\pi}{18}) &= \sin{\frac{25\pi}{6}}\\
&=\sin({2\pi +\frac{25\pi}{6}}) \\
&=\frac{1}{2}
\end{align*}

so

\begin{align*}
\alpha &= 4\sin \frac{5\pi}{18}\\
\beta &= 4\sin\frac{5\pi}{18} \\
\gamma&=4\sin{\frac{25\pi}{18}}
\end{align*}

are also roots of the polynomial $$x^3 – 12x + 8$$ from part (ii).

Reframing it as a polynomial question, we need to find

 \begin{align*} \frac{\alpha^2 + \beta^2 + \gamma^2}{4^2} &= \frac{1}{16}((\alpha + \beta + \gamma)^2 – 2(\alpha\beta + \beta\gamma + \alpha\gamma))\\ &= \frac{1}{16}(\frac{-b}{a}^2 – 2 \frac{c}{a})\\ &= \frac{1}{16}(0 – 2 \times \frac{-12}{1})\\ &= \frac{24}{16}\\ &=\frac{3}{2} \end{align*}

as required.

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