Racquel Sanderson

April 30, 2025 · 6:08 am

Factorising Non-Monic Quadratics

The general equation of a quadratic is $ax^2+bx+c$

Let’s explore different methods of factorising a non-monic quadratic (the $a$ term is not $1$ )

Factorise $6x^2+x-12$

We need to find two numbers that add to $1$ and multiply to $-72$ (i.e. add to $b$ and multiply to $a\times c)$

The two numbers are $9$ and $-8$

Method 1 – Splitting the middle term

This is the method I teach the most often

$\begin{equation*}6x^2+x-12\end{equation}$

Split the middle term (the $b$ term) into the two numbers

$\begin{equation*}6x^2-8x+9x-12\end{equation}$

The order doesn’t matter.

Find a common factor for the first term terms, and then for the last two terms.

$\begin{equation*}2x(3x-4)+3(3x-4)\end{equation}$

There is a common factor of $(3x-4)$ , factor it out.

$\begin{equation*}(3x-4)(2x+3)\end{equation}$

Method two – Fraction

$\begin{equation*}6x^2+x-12\end{equation}$

Put $6x$ into both factors and divide by $6$

$\begin{equation*}\frac{(6x-8)(6x+9)}{6}\end{equation}$

Factorise

$\begin{equation*}\frac{2(3x-4)3(2x+3)}{6}\end{equation}$

$\begin{equation*}\frac{6(3x-4)(2x+3)}{6}\end{equation}$

$\begin{equation*}(3x-4)(2x+3)\end{equation}$

Method 3 – Monic to non-monic

$\begin{equation*}y=6x^2+x-12\end{equation}$

Multiply both sides of the equation by $a$

$\begin{equation*}6y=6(6x^2+x-12)\end{equation}$

$\begin{equation*}6y=6^2x^2+6x-72\end{equation}$

$\begin{equation*}6y=(6x)^2+6x-72\end{equation}$

Let $A=6x$

$\begin{equation*}6y=A^2+A-72\end{equation}$

Factorise

$\begin{equation*}6y=(A+9)(A-8)\end{equation}$

Replace the $A$ with $6x$

$\begin{equation*}6y=(6x+9)(6x-8)\end{equation}$

$\begin{equation*}6y=3(2x+3)2(3x-4)\end{equation}$

$\begin{equation*}6y=6(2x+3)(3x-4)\end{equation}$

$\begin{equation*}y=(2x+3)(3x-4)\end{equation}$

Method 4 – Cross Method

$\begin{equation*}6x^2+x-12\end{equation}$

Place the two numbers in the cross

Place the two numbers that add to $1$ and multiply to $-72$ in the other parts of the cross.

Divide these two numbers by $6$ (i.e $a$ )

Simplify

Hence,

$\begin{equation*}(x-\frac{4}{3})(x+\frac{3}{2})\end{equation}$

Which is

$\begin{equation*}(3x-4)(2x+3)\end{equation}$

Method 5 – By Inspection

This is my least favourite method – although students get better with practice

$\begin{equation*}6x^2+x-12\end{equation}$

The factors of $a$ are $1, 2, 3,$ and $6$ and the factors of $12$ are $1, 2, 3, 4, 6, 12$

We know one number is positive and one number negative.

Which give us all of these possibilities

$\begin{equation*}(2x+3)(3x-4)\end{equation}$

With a bit of practice you don’t need to check all of the possibilties, but I find students struggle with this method.

Method 6 – Grid

$\begin{equation*}6x^2+x-12\end{equation}$

Create a grid like the one below


	$6x^2$
		$-12$

Find the two numbers that multiply to $-72$ and add to $1$ and place them in the other grid spots (see below)


	$6x^2$	$-8x$
	$9x$	$-12$

Find the HCF (highest common factor) of each row and put in the first column.

Row $1$ HCF= $2x$ , Row $2$ HCF= $3$


$2x$	$6x^2$	$-8x$
$3$	$9x$	$-12$

For the columns, calculate what is required to multiple the HCF to get the table entry.

For example, what do you need to multiple $2x$ and $3$ by to get $6x^2$ and $9x$ ? In this case it is $3x$ . It’s always going to be the same thing, so just use one value to calculate it,

	$3x$	$-4$
$2x$	$6x^2$	$-8x$
$3$	$9x$	$-12$

The factors are column $1$ and row $1$
$(2x+3)(3x-4)$

The two methods I use the most are splitting the middle term, and the cross method, but I can see value in the grid method.

Leave a Comment

Filed under Algebra, Factorising, Factorising, Polynomials, Quadratic, Quadratics

Tagged as Factorising, factorising non-monic trinomials, non-monic quadratics

April 23, 2025 · 9:30 am

Problem Solving

I am came across this problem and was fascinated. It’s from this book

At first I went straight to the 14-sided polygon, and tried to draw the diamonds (parallelograms), but then I thought let’s start smaller and see if there is a pattern.

Clearly a square contains 1 diamond (itself).

Pentagon

It’s not possible with a pentagon.

Hexagon

A hexagon has 6 diamonds

Septagon

I am guessing it’s not possible to fill a regular 7-sided shape with diamonds

It’s not possible with odd numbers of sides. Regular polygons with an odd number of sides have no parallel sides, so we can’t cover it with rhombi (which have opposite sides parallel).

Octagon

An octagon has 6 diamonds.

We know a decoagon has 10 diamonds (from the question)

Let’s put together what we know

$n$	$4$	$6$	$8$	$10$
Diamonds	$1$	$3$	$6$	$10$

These are the triangular numbers, so when $n=12$ the number of diamonds is $15$ , and for $n=14$ it’s $21$ .

We can work out a rule for calculating the number of diamonds given the number of sides.

Because the difference in the $n$ values is not $1$ , I am going to get $n$ and $D$ in terms of $k$ and then combine the two equations.

From the above table, $n=2k+2$

We know this rule is quadratic as the second difference is constant, hence

$D=\frac{1}{2}k^2+bk+c$

$\begin{equation*}1=\frac{1}{2}+b+c\end{equation}$

(1) $\begin{equation*}\frac{1}{2}=b+c\end{equation*}$

$\begin{equation*}3=\frac{1}{2}2^2+2b+c\end{equation}$

(2) $\begin{equation*}1=2b+c\end{equation*}$

Solve simultaneously, subtract equation $1$ from equation $2$

(3) $\begin{equation*}\frac{1}{2}=b\end{equation*}$

Substitute for $b=\frac{1}{2}$ into equation $1$

$\begin{equation*}\frac{1}{2}=\frac{1}{2}+c\end{equation}$

$c=0$ , therefore $D=\frac{1}{2}k^2+\frac{1}{2}k$

We know $n=2k+2$ hence $k=\frac{n-2}{2}$

Hence $D=\frac{1}{2}(\frac{n-2}{2})^2+\frac{1}{2}(\frac{n-2}{2})$

$D=\frac{1}{8}(n^2-4n+4)+\frac{1}{4}(n-2)=\frac{n^2}{8}-\frac{n}{2}+\frac{1}{2}+\frac{n}{4}-\frac{1}{2}=\frac{n^2}{8}-\frac{n}{4}$

$\begin{equation*}D=\frac{n^2}{8}-\frac{n}{4}\end{equation}$

Let’s test our rule for $n=14$

$\begin{equation*}D=\frac{14^2}{8}-\frac{14}{4}=\frac{49}{2}-\frac{7}{2}=\frac{42}{2}=21\end{equation}$

Leave a Comment

Filed under Area, Geometry, Interesting Mathematics, Puzzles, Quadratics

Tagged as Finding a quadratic rule, regular polygons, rhombus, tessellations

April 21, 2025 · 11:08 am

Area/Geometry Problem

This problem is from The Geometry Forum Problem of the Week June 1996

In triangle ABC, AC=18 and D is the point on AC for which AD=5. Perpendiculars drawn from D to AB and CB have lengths of 4 and 5 respectively. What is the area of triangle ABC?

I put together a diagram (in Geogebra)

Add points P and Q

Triangle APD and triangle DQC are right angled. Using pythagoras, $AP=3$ and $QC=12$

$BQDP$ is a cyclic quadrilateral and $BD$ is the diameter. I am not sure if this is useful, but it is good to notice.

$\begin{equation*}sin(A+B+C)=sin(180)=0\end{equation}$

$\begin{equation*}sin((A+C)+B)=sin(A+C)cosB+sinBcos(A+C)=0\end{equation}$

$\begin{equation*}cosB(sinAcosC+sinCcosA)+sinB(cosAcosC-sinAsinC)=0\end{equation}$

$\begin{equation*}cosB(\frac{4}{5}\times\frac{12}{13}+\frac{5}{13}\times\frac{3}{5})+sinB(\frac{3}{5}\times\frac{12}{13}-\frac{4}{5}\times\frac{5}{13})=0\end{equation}$

$\begin{equation*}cosB(\frac{48}{65}+\frac{15}{65})+sinB(\frac{36}{65}-\frac{20}{65})=0\end{equation}$

$\begin{equation*}\frac{63}{65}cosB+\frac{16}{65}sinB=0\end{equation}$

$\begin{equation*}63cosB+16sinB=0\end{equation}$

$\begin{equation*}63+16tanB=0\end{equation}$

$\begin{equation*}tanB=\frac{-63}{16}\end{equation}$

If $tanB=\frac{-63}{16}$ then $sinB=\frac{63}{65}$

Now,

$\begin{equation*}\frac{y+12}{sinA}=\frac{18}{sinB}\end{equation}$

$\begin{equation*}y+12=\frac{4}{5}(18)\frac{65}{63}\end{equation}$

$\begin{equation*}y+12=\frac{104}{7}\end{equation}$

Hence the Area is

$\begin{equation*}A=\frac{1}{2}(18)(\frac{104}{7})sinC\end{equation}$

$\begin{equation*}A=\frac{1}{2}(18)(\frac{104}{7})\frac{5}{13}\end{equation}$

$\begin{equation*}A=\frac{360}{7}=51.43\end{equation}$

Leave a Comment

Filed under Area, Finding an area, Geometry, Identities, Non-Right Trigonometry, Pythagoras, Trigonometry

Tagged as area, area of triangles, problem solving

April 16, 2025 · 8:26 am

Binomial Expansion – deriving the formula for Variance

We have seen how the formula for mean (expected value) was derived, and now we are going to look at variance.

In general variance of a probability distribution is

(1) $\begin{equation*} Var(X)=E(X^2)-(E(X))^2\end{equation*}$

We are going to start by calculating $E(X^2)$

$\begin{equation*}E(X^2)=\sum_{x=0}^nx^2p(x)\end{equation}$

$\begin{equation*}E(X^2)=\sum_{x=0}^nx^2\begin{pmatrix}n\\x\end{pmatrix}p^x(1-p)^{n-x}\end{equation}$

$\begin{equation*}E(X^2)=\sum_{x=0}^n x^2\frac{n!}{(n-x)!x!}p^x(1-p)^{n-x}\end{equation}$

The $x^2$ cancels with the $x!$ to leave $x$ on the numerator and $(x-1)!$ on the denominator.

Also, when $x=0, x^2=0$ and we can start the sum at $x=1$

$\begin{equation*}E(X^2)=\sum_{x=1}^n x\frac{n!}{(n-x)!(x-1)}!p^x(1-p)^{n-x}\end{equation}$

Let $y=x-1$ and $m=n-1$ , when $x=n, y=n-1$ and hence $y=m$ and when $x=1, y=0$

Our equation is now

$\begin{equation*}E(X^2)=\sum_{y=0}^m (y+1)\frac{(m+1)!}{(m+1-(y+1))!y!}p^{y+1}(1-p)^{m+1-(y+1)}\end{equation}$

Simplify

$\begin{equation*}E(X^2)=\sum_{y=0}^m(y+1)\frac{(m+1)!}{(m-y)!y!}p^{y+1}(1-p)^{m-y}\end{equation}$

$\begin{equation*}E(X^2)=\sum_{y=0}^m(y+1)(m+1)\frac{m!}{(m-y)!y!}p\times p^y(1-p)^{m-y}\end{equation}$

$\begin{equation*}E(X^2)=\sum_{y=0}^m p(y+1)(m+1)\frac{m!}{(m-y)!y!} p^y(1-p)^{m-y}\end{equation}$

$\begin{equation*}E(X^2)=\sum_{y=0}^m p(y+1)(m+1)p(y)\end{equation}$

$\begin{equation*}E(X^2)=p(m+1)(\sum_{y=0}^myp(y)+\sum_{y=0}^mp(y))\end{equation}$

$\sum_{y=0}^mp(y)=1$ and $\sum_{y=0}^myp(y)=E(Y)$

$\begin{equation*}E(X^2)=p(m+1)(E(Y)+1)\end{equation}$

$E(Y)=mp$

$\begin{equation*}E(X^2)=p(m+1)(mp+1)\end{equation}$

$m+1=n$

$\begin{equation*}E(X^2)=pn((n-1)p+1)\end{equation}$

$\begin{equation*}E(X^2)=n^2p^2-np^2+np\end{equation}$

Now from equation $1$

$\begin{equation*}Var(X)=E(X^2))-(E(X))^2\end{equation}$

$\begin{equation*}=Var(X)=n^2p^2-np^2+np-n^2p^2\end{equation}$

$\begin{equation*}Var(X)=-np^2+np\end{equation}$

(2) $\begin{equation*}Var(X)=np(1-p)\end{equation*}$

and the standard deviation is

(3) $\begin{equation*}\sigma_X=\sqrt{np(1-p)}\end{equation*}$

1 Comment

Filed under Algebra, Binomial, Probability Distributions, Standard Deviation

Tagged as binomial distribution, standard deviation of a binomial distribution, variance of a binomial distribution

April 13, 2025 · 9:23 am

Binomial Distribution – deriving the equation for mean (expected value)

The mean, $\mu$ of a binomial distribution is

(1) $\begin{equation*}\mu=np\end{equation*}$

where $n$ is the number of trials and $p$ is the probability of success.

For any discrete probability distribution , the expected value or mean is

(2) $\begin{equation*}E(X)=\sum_{x=0}^nxp(x)\end{equation*}$

For example, if a coin is tossed $3$ times and the number of heads is recorded, the distribution is

$X$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

$\begin{equation*}E(X)=0\times\frac{1}{8}+1\times\frac{3}{8}+2\times\frac{3}{8}+3\times{1}{8}\end{equation}$

$\begin{equation*}E(X)=\frac{12}{8}=\frac{3}{2}\end{equation}$

I want to show how the $\mu=np$ formula is derived from the general formula (equation $(2)$ ).

$\begin{equation*}E(X)=\sum^n_{x=0}xp(x)\end{equation}$

For a binomial distribution, $p(x)=\begin{pmatrix}n\\x\end{pmatrix}p^x(1-p)^{n-x}$

$\begin{equation*}E(X)=\sum_{x=0}^nx\begin{pmatrix}n\\x\end{pmatrix}p^x(1-p)^{n-x}\end{equation}$

$\begin{equation*}E(X)=\sum^n_{x=0}x\frac{n!}{(n-x)!x!}p^x(1-p)^{n-x}\end{equation}$

The $x$ can cancel with the $x!$ to leave $(x-1)!$ on the denominator.

$\begin{equation*}E(X)=\sum^n_{x=0}\frac{n!}{(n-x)!(x-1)!}p^x(1-p)^{n-x}\end{equation}$

Also, when $x=0 \Rightarrow xp(x)=0$ , hence the sum can start at $x=1$ .

$\begin{equation*}E(X)=\sum^n_{x=1}\frac{n!}{(n-x)!(x-1)!}p^x(1-p)^{n-x}\end{equation}$

Let $y=x-1$ and $m=n-1$

When $x=n \Rightarrow y=n+1=m$

$\begin{equation*}E(X)=\sum^{m}_{y=0}\frac{(m+1)!}{((m+1)-(y+1))!(y)!}p^{y+1}(1-p)^{(m+1)-(y+1)}\end{equation}$

Simplify

$\begin{equation*}E(X)=\sum^{m}_{y=0}\frac{(m+1)!}{((m-y))!(y)!}p^{y+1}(1-p)^{(m-y)}\end{equation}$

$\begin{equation*}E(X)=\sum^{m}_{y=0}\frac{(m+1)m!}{((m-y))!(y)!}p^yp^1(1-p)^{(m-y)}\end{equation}$

We can move $(m+1)$ and $p$ out of the sum.

$\begin{equation*}E(X)=(m+1)p\sum^{m}_{y=0}\frac{m!}{((m-y))!(y)!}p^y(1-p)^{(m-y)}\end{equation}$

$\begin{equation*}\sum^{m}_{y=0}\frac{m!}{((m-y))!(y)!}p^y(1-p)^{(m-y)}=\sum^{m}_{y=0}\begin{pmatrix}m\\y\end{pmatrix}p^y(1-p)^{(m-y)}\end{equation}$

$\begin{equation*}\sum^{m}_{y=0}\begin{pmatrix}m\\y\end{pmatrix}p^y(1-p)^{(m-y)}=1\end{equation}$

As it is the sum of the probabilities of a binomial distribution with $m$ trials.

Hence $E(X)=(m+1)p=np$

Next, deriving the variance formula for a binomial distribution.

1 Comment

Filed under Binomial, Mean, Probability Distributions

Tagged as binomial distribution, expected value, mean

April 7, 2025 · 9:09 am

The vector projection (vector resolution or vector component) of $\mathbf{a}$ onto a non-zero vector $\mathbf{b}$ is splitting $\mathbf{a}$ into two vectors, one is parallel to $\mathbf{b}$ (the vector projection) and one perpendicular to $\mathbf{b}$

In the above diagram $\mathbf{a_1}$ is the vector projection of $\mathbf{a}$ onto $\mathbf{b}$ and $\mathbf{a_2}$ is perpendicular to $\mathbf{b}$ .

How do we find $\mathbf{a_1}$ and $\mathbf{a_2}$ ?

Using right trigonometry,

$cos(\theta)=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}$

Remember the scalar product (dot product) of vectors is

(1) $\begin{equation*}\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|cos(\theta)\end{equation*}$

Hence $cos(\theta)=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$

$\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}$

and, $|\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\times|\mathbf{a}|$

$|\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$

This is the scalar projection of $\mathbf{a}$ onto $\mathbf{b}$

To find the vector projection we need to multiply by $\mathbf{\hat{b}}$ , that is find a vector with the same magnitude as $\mathbf{a_1}$ in the direction of $\mathbf{b}$ .

The vector projection is

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \mathbf{\hat{b}}$

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \frac{\mathbf{b}}{|\mathbf{b}|}$

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{(|\mathbf{b}|)^2}\times \mathbf{b}$

Now for $\mathbf{a_2}$ , we know $\mathbf{a}=\mathbf{a_1}+\mathbf{a_2}$

Hence, $\mathbf{a_2}=\mathbf{a}-\mathbf{a_1}$

Example

From Cambridge Year 11 Specialist Mathematics (Chapter 3)

(a) $\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}$

$=\frac{\begin{pmatrix}4\\1\end{pmatrix}\cdot\begin{pmatrix}1\\-1\end{pmatrix}}{2}\times(\mathbf{i}-\mathbf{j})$

$=\frac{3}{2}{(\mathbf{i}-\mathbf{j})$

(b) $4\mathbf{i}+\mathbf{j}-\frac{3}{2}{(\mathbf{i}-\mathbf{j})=\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}$

(c)

The shortest distance (green vector) is the vector component of $\mathbf{a}$ perpendicular to $\mathbf{b}$ , i.e. $\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}$

$|\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}|=\sqrt{\frac{25}{4}+\frac{25}{4}}=\sqrt{\frac{50}{4}}=\frac{5\sqrt{2}}{2}$

Leave a Comment

Filed under Right Trigonometry, Trigonometry, Vector Projection, Vectors, Year 11 Specialist Mathematics

Tagged as dot product, scalar product, scalar projection, vector projection

April 6, 2025 · 1:51 am

Using a Vector Method to Find an Angle Bisector

Points $A$ and $B$ are defined by the position vectors $\mathbf{a}=3\mathbf{i}+4\mathbf{j}$ and $\mathbf{b}=12\mathbf{i}+5\mathbf{j}$ .

Find a vector that bisects $\angle{AOB}$ .

If we think about how we add vectors using the parallelogram rule

Finding the resultant vector using the parallelogram rule

we can take advantage of the geometric properties of parallelograms (or of a rhombus).

If $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, then the parallelogram is a rhombus, and the diagonal (i.e the resultant) bisects the angle.

We need to find the sum of the unit vectors.

$|\mathbf{a}|=\sqrt{3^2+4^2}=5$

$\therefore \hat{\mathbf{a}}=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}$

$|\mathbf{b}|=\sqrt{12^2+5^2}=13$

$\therefore \hat{\mathbf{b}}=\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}$

The vector that bisects $\angle{AOB}$ is

$\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}+\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}=\frac{99}{65}\mathbf{i}+\frac{64}{65}\mathbf{j}$

Leave a Comment

Filed under Geometry, Vectors, Year 11 Specialist Mathematics

Tagged as bisecting an angle, vectors, Year 11 Specialist Mathematics

April 3, 2025 · 7:28 am

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Find the area of an $n-$ sided regular polygon if you know the side length, $l$ .

Find the $h$ of the triangle in terms of $l$ and theta.

$tan(\theta)=\frac{\frac{l}{2}}{h}$

$h=\frac{l}{2tan(\theta)}$

Remember the area of a triangle is $A=\frac{1}{2}bh$

Hence, $A=\frac{1}{2} l \times \frac{l}{2tan(\theta)}=\frac{l^2}{4tan(\theta)}$

And $\theta=\frac{360}{2n}=\frac{180}{n}$

Therefore $A=\frac{l^2}{4tan(\frac{180}{n})}$

There are $n$ triangles in an $n-$ sided polygon

(1) $\begin{equation*}A=\frac{nl^2}{4tan(\frac{180}{n})}\end{equation*}$

A=\frac{6\times10^2}{4tan(\frac{180}{6})}

Finding the area of a polygon if you know the inradius or the apothem

The apothem and the inradius are the same. It is the radius of the incircle.

Find the area of the triangle in terms of $a$ and theta.

$tan(\theta)=\frac{\frac{l}{2}}{a}$

$l=2atan(\theta)$

$A=\frac{1}{2}bh$

$A=\frac{1}{2}2atan(\theta)a=a^2tan(\theta)$

And $\theta=\frac{180}{n}$

Hence for an $n-$ sided polygon

(2) $\begin{equation*}A=na^2tan(\frac{180}{n})\end{equation*}$

Finding the area of a regular polygon given the circumradius

The circumradius is the radius of the circumscribed circle ( $R$ in the diagram above)

Remember the area of triangle formula

$A=\frac{1}{2}absin(\theta)$

$A=\frac{1}{2}R^2sin(\theta)$

$\theta=\frac{360}{n}$

Hence, $A=\frac{1}{2}R^2sin(\frac{360}{n})$

Hence, for an $n-$ sided polygon

(3) $\begin{equation*}A=\frac{nR^2sin(\frac{360}{n})}{2}\end{equation*}$

Leave a Comment

Filed under Area, Area of Triangles (Sine), Finding an area, Non-Right Trigonometry, Regular Polygons, Right Trigonometry, Year 11 Mathematical Methods

Tagged as area of regular polygons, area of triangles, trig

March 31, 2025 · 7:14 am

Deriving the Quadratic Equation formula

My year 10 students have been learning how to complete the square with the idea of then deriving the quadratic equation formula.

The general equation for a quadratic is $y=ax^2+bx+c$

Completing the square,

$\begin{equation*}ax^2+bx+c\end{equation}$

Factorise out the leading coefficient (i.e. $a$ )

$\begin{equation*}a(x^2+\frac{bx}{a}+\frac{c}{a})\end{equation}$

Half the second term (i.e $\frac{b}{a}$ ) and subtract the square of the second term.

$\begin{equation*}a((x+\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a})\end{equation}$

$\begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a})\end{equation}$

Simplify

$\begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2})\end{equation}$

$\begin{equation*}a((x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a^2})\end{equation}$

$\begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}\end{equation}$

Now let’s solve

$\begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}=0\end{equation}$

$\begin{equation*}a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})=\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

$\begin{equation*}x=-\frac{b}{2a}\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

Which is the quadratic equation formula.

Leave a Comment

Filed under Algebra, Quadratic, Quadratics, Solving, Solving, Solving Equations

Tagged as completing the square, quadratic equation formula

March 26, 2025 · 12:23 pm

Logistic Growth Worked Example

A brumby is a free-roaming wild horse found in large number in parts of Australia. The culling of brumbies was banned in the year 2000. At this time the estimated population of brumbies in
Kosciuszko National Park was 1600. Scientists have modelled the population, P(t), of brumbies in
Kosciuszko National Park t years since the ban, by

$\begin{equation*}P(t)=\frac{18000}{10.25e^{0.15t}+1}\end{equation}$

(a) Use the model to determine how long it will take the brumbies to increase to a number that is triple the number when the ban came into effect.

(b) From this model, determine the estimated long run number of brumbies in Kosciuszko National Park.

It can be shown that the growth rate of the population of brumbies can be expressed as

$\begin{equation*}\frac{dP}{dt}=\frac{1}{r}P(k-P)\end{equation}$

(d) Determine the greatest growth rate for the population of brumbies.

ATAR 2024 Specialist Mathematics Question 13

(a) $1600\times 3=4800$
$4800=\frac{18000}{10.25e^{0.15t}+1}$
$t=8.8$ years.

(b) $\lim_{\limits_{t \to \infty}\frac{18000}{10.25e^{0.15t}+1}=18000$

$\begin{equation*}\frac{dP}{dt}=rP(k-P)\Longleftrightarrow P=\frac{KP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}$

We have $\frac{1}{r}$ instead of $r$ .

Therefore,

$\begin{equation*}0.15=\frac{1}{r}\times 18000\end{equation}$

$r=120000$

(d) The greatest growth rate occurs when $P=\frac{k}{2}=9000$

$\begin{equation*}\frac{dP}{dt}=\frac{1}{r}P(k-P)\end{equation}$

$\begin{equation*}\frac{dP}{dt}=\frac{1}{120000}(9000)(9000)\end{equation}$

The greatest growth rate is 675 Brumbies per year.

1 Comment

Filed under Differential Equations, Logistic Growth, Year 12 Specialist Mathematics

Tagged as differential equations, logisitic growth, worked example, Year 12 Specialist Mathematics

Possible factorisations	$b$ term of expansion
$(x-1)(6x+12)$	$12x-6x=6x$	No
$(x-2)(6x+6)$	$6x-12x=-6x$	No
$(x-3)(6x+4)$	$4x-18x-14x$	No
$(2x-1)(3x+12)$	$24x-3x=21x$	No
$(2x-2)(3x+6)$	$12x-6x=6x$	No
$(2x-3)(3x+4)$	$8x-9x=-1x$	Almost, switch the signs
$(2x+3)((3x-4)$	$-8x+9x=1x$	Yes

Factorising Non-Monic Quadratics

Method 1 – Splitting the middle term

Method two – Fraction

Method 3 – Monic to non-monic

Method 4 – Cross Method

Method 5 – By Inspection

Method 6 – Grid

Problem Solving

Pentagon

Hexagon

Septagon

Octagon

Area/Geometry Problem

Binomial Expansion – deriving the formula for Variance

Binomial Distribution – deriving the equation for mean (expected value)

Vector and Scalar Projection

Example

Using a Vector Method to Find an Angle Bisector

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Finding the area of a polygon if you know the inradius or the apothem

Finding the area of a regular polygon given the circumradius

Deriving the Quadratic Equation formula

Logistic Growth Worked Example

Recent Posts

Categories

Archives