Category Archives: Quadratics

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.

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November 29, 2024 · 8:11 am

Interesting Equation

I think this one is doing the rounds, I first saw it here.

    \begin{equation*}2^x3^{x^2}=6\end{equation}

x=1 is the obvious answer, 2^1\times 3^1=6, but are there more answers?

This was my approach

    \begin{equation*}ln(2^x3^{x^2})=ln(6)\end{equation}

    \begin{equation*}ln(2^x)+ln(3^{x^2})=ln(6)\end{equation}

    \begin{equation*}xln(2)+x^2ln(3)-ln(6)=0\end{equation}

    \begin{equation*}ln(3)x^2+ln(2)x-ln(6)=0\end{equation}

A quadratic equation.

Hence,

    \begin{equation*}x=\frac{-ln(2)\pm\sqrt{(ln(2))^2-4(ln(3))(ln(6))}}{2ln(3)}\end{equation}

I then used my calculator

Hence x=1 0r x=-1.631

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May 26, 2024 · 1:07 pm

Quadratic Rule from a Table of Values

How do you find a quadratic rule from a table of values?

For example,

x1 2 3 4
y061424

Find the first difference

First Difference6-0=614-6=824-14=10

Find the second difference (if the second difference is a constant, then it is quadratic)

Second Difference8-6=2 10-8=2

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

The second difference is 2a

Hence our equation is now y=x^2+bx+c

The c value is the vertical intercept (x=0). We can back track in the table

x01234
yc061424

As the first differences are 6, 8, 10, the one between 0 and 1 must be 4

0-c=4

c=-4

Our equation is now y=x^2+bx-4.

We can now use any other point to find the b value.

Let’s use the point (2, 6)

6=2^2+b(2)-4

6=4-2b-4

6=2b

b=3

The function is y=x^2+3x-4

Let’s try another one

x3456
y7173149

First differences

First difference101418

Second difference

Second Difference44

Hence 2a=4, therefore a=2

The equation is now y=2x^2+bx+c

Instead of back tracking, this time I am going to use two points and simultaneous equations.

Using points (3, 7) and (4, 17)

    \[7=2(3)^2+b(3)+c\]

(1)   \begin{equation*}3b+c=-11\end{equation*}

    \[17=2(4)^2+b(4)+c\]

(2)   \begin{equation*}4b+c=-15\end{equation*}

Equation 2 – Equation 1

b=-4

Substitute b=-4 into equation 1

3(-4)+c=-11

-12+c=-11

c=1

Hence the equation is y=2x^2-4x+1

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May 9, 2024 · 6:08 am

Intersections of Lines and Circles

The three possibilities: no intersection, one point of intersection, or two points of intersection

Once my students have studied circle and quadratic equations, I like to introduce lines intersecting with circles. It tests their quadratic equation solving skills, and if they can use the discriminant.

Example 1

Calculate the point(s) of intersection of

    \[3x+2y-6=0\ \textnormal{and}\ (x+4)^2+(y+2)^2=9\]

Rearrange the line equation

    \[y=\frac{6-3x}{2}\]

    \[y=3-\frac{3x}{2}\]

Substitute for y into the circle equation

    \[(x+4)^2+(3-\frac{3x}{2}+2)^2=9\]

    \[(x+4)^2+(5-\frac{3x}{2})^2=9\]

    \[x^2+8x+16+25-15x+\frac{9x^2}{4}-9=0\]

    \[\frac{13x^2}{4}-7x+32=0\]

    \[13x^2-28x+128=0\]

At this point I like to check the discrimnant

    \[\Delta=b^2-4ac\]

    \[\Delta=(-28)^2-4\times13\times128\]

    \[\Delta=-5872\]

As the discriminate is less than zero, there are no points of intersection.

Example 2

Calculate the point(s) of intersection of

    \[y=2x+1\ \textnormal{and}\ (x+5)^2+(y+3)^2=16\]

    \[(x+5)^2+(2x+1+3)^2=16\]

    \[(x+5)^2+(2x+4)^2=16\]

    \[x^2+10x+25+4x^2+16x+16-16=0\]

    \[5x^2+26x+25=0\]

Check the discriminant

    \[\Delta=26^2-4\times5\times25\]

    \[\Delta=176\]

Therefore there are two points of intersection

    \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    \[x=\frac{-26\pm\sqrt{176}}{10}\]

    \[x=\frac{-26\pm4\sqrt{11}}{10}\]

    \[x=\frac{-13\pm2\sqrt{11}}{5}\]

Substitute x into y

    \[y=2(\frac{-13\pm2\sqrt{11}}{5})+1\]

    \[\textnormal{The two points are}\ (\frac{-13+2\sqrt{11}}{5},\frac{-21+4\sqrt{11}}{5})\]

    \[\textnormal{and}\ (\frac{-13-2\sqrt{11}}{5},\frac{-21-4\sqrt{11}}{5})\]

Example 3

Find the value of k so that the line and the circle intersect at one point (i.e. the line is a tangent to the circle)

    \[-kx+y-5=0\ \textnormal{and}\ (x-2)^2+(y-3)^2=\frac{36}{5}\]

    \[y=kx+5\]

    \[(x-2)^2+(kx+5-3)^2=\frac{36}{5}\]

    \[(x-2)^2+(kx+2)^2=\frac{36}{5}\]

    \[x^2-4x+4+k^2x^2+4kx+4-\frac{36}{5}=0\]

    \[(1+k^2)x^2+(4k-4)x+\frac{4}{5}=0\]

Find the discriminant (remember for one solution we want the discriminant to be zero)

    \[\Delta=b^2-4ac\]

    \[\Delta=(4k-4)^2-4\times(1+k^2)\times\frac{4}{5}\]

    \[\Delta=16k^2-32k+16-\frac{16}{5}-\frac{16k^2}{5}\]

    \[0=16k^2-32k+16-\frac{16}{5}-\frac{16k^2}{5} \[0=80k^2-160k+80-16-16k^2\]

    \[0=64k^2-160k+64\]

    \[0=32(2k^2-5k+2)\]

    \[0=2k^2-5k+2\]

    \[0=2k^2-4k-k+2\]

    \[0=2k(k-2)-1(k-2)\]

    \[0=(2k-1)(k-2)\]

    \[k=\frac{1}{2}\ \textnormal{and}\ k=2\]

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April 2, 2024 · 7:10 am

Factorising Non-Monic Trinomials

I factorise trinomials by using the grouping method, I will show you what I mean

But one of my students showed me a different way

I have put in more steps than is necessary. I think this might be a quicker method.

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