Category Archives: Algebra

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

Binomial Expansion Theorem

My Year 11 Mathematics Methods students are working on the Binomial Expansion Theorem.

But before we get onto that, remember Pascal’s triangle

First 8 rows of Pascal’s triangle

Now we can use combinations to find the numbers in each row. For example, 1 4 6 4 1 is \begin{pmatrix}4\\0\end{pmatrix}=1, \begin{pmatrix}4\\1\end{pmatrix}=4, \begin{pmatrix}4\\2\end{pmatrix}=6,  \begin{pmatrix}4\\3\end{pmatrix}=4, \begin{pmatrix}4\\4\end{pmatrix}=1

ExpressionExpansionCo-efficients
(x+y)^2x^2+2xy+y^21, 2, 1
(x+y)^3x^3+3x^2y+3xy^2+y^31, 3, 3, 1
(x+y)^4x^4+4x^3y+6x^2y^2+4xy^3+y^41, 4, 6, 4, 1

As you can see, the coefficients are the row of pascal’s triangle corresponding to the power. So (x+y)^6 would have co-efficients from the sixth row of the table 1, 6, 15, 20, 15, 6, 1.

To generalise

(x+y)^n=\begin{pmatrix}n\\0\end{pmatrix}x^ny^0+\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}y^1+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}y^2+ ...+\begin{pmatrix}n\\n-1\end{pmatrix}x^1{y^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}x^0y^n

Which we can condense to

(x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i

Worked Examples

(1) Expand (2x-3)^4

(2x-3)^4=\begin{pmatrix}4\\0\end{pmatrix}(2x)^4(-3)^0+\begin{pmatrix}4\\1\end{pmatrix}(2x)^3(-3)^1+\begin{pmatrix}4\\2\end{pmatrix}(2x)^2(-3)^2+\begin{pmatrix}4\\3\end{pmatrix}(2x)^1(-3)^3+\begin{pmatrix}4\\4\end{pmatrix}(2x)^0(-3)^4
(2x-3)^4=16x^4-96x^3+216x^2-216x+81

(2) Find the co-efficient of the x^3 term in the expansion of (2-5x)^5.

Remember (x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i, the x^3 is when i=3
\begin{pmatrix}5\\3\end{pmatrix}(2)^2(-5)^3=10\times 2\times -125=-5000

(3) Find the constant term in the expansion of (x^2+\frac{3}{x^4})^6

We need to find the term where the x‘s cancel out. Each term is \begin{pmatrix}6\\i\end{pmatrix}(x^2)^{6-i}(\frac{3}{x^4})^i.
\begin{pmatrix}6\\i\end{pmatrix}(x^{12-2i})(3^ix^{-4i}).
We need 12-2i-4i=0, hence i=2
Therefore, the co-efficient is \begin{pmatrix}6\\2\end{pmatrix}\times3^2=135

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Filed under Algebra, Binomial Expansion Theorem, Counting Techniques, Year 11 Mathematical Methods

Puzzle Page 2

If x^2-3x+1=0, then find x^5+\frac{1}{x^5}.

My first thought was to solve for x, but it doesn’t factorise easily, and I didn’t want to find the fifth power of an expression involving surds (x=\frac{3\pm \sqrt{5}}{2}), there must be an easier way.

Because x\neq0, we can divide by x

    \begin{equation*}x-3+\frac{1}{x}=0\end{equation}

Hence

(1)   \begin{equation*}x+\frac{1}{x}=3\end{equation*}

What is the expansion of (x+\frac{1}{x})^5?

Using the binomial expansion theorem

    \begin{equation*}(x+\frac{1}{x})^5=x^5+5x^4(\frac{1}{x})+10x^3(\frac{1}{x^2})+10x^2(\frac{1}{x^3})+5x(\frac{1}{x^4})+\frac{1}{x^5}\end{equation}

    \begin{equation*}(x+\frac{1}{x})^5=x^5+\frac{1}{x^5}+5(x^3+\frac{1}{x^3})+10(x+\frac{1}{x})\end{equation}

Therefore

(2)   \begin{equation*}x^5+\frac{1}{x^5}=(x+\frac{1}{x})^5-5(x^3+\frac{1}{x^3})-10(x+\frac{1}{x})\end{equation*}

Let’s do it again for x^3+\frac{1}{x^3}

    \begin{equation*}(x+\frac{1}{x})^3=x^3+3x^2(\frac{1}{x})+3x(\frac{1}{x^2})+\frac{1}{x^3}\end{equation}

(3)   \begin{equation*}x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x})\end{equation*}

Substitute 3 into 2

    \begin{equation*}x^5+\frac{1}{x^5}=(x+\frac{1}{x})^5-5((x+\frac{1}{x})^3-3(x+\frac{1}{x}))-10(x+\frac{1}{x})\end{equation}

Remember x+\frac{1}{x}=3

Therefore

    \begin{equation*}x^5+\frac{1}{x^5}=3^5-5(3^3)+15\times3-10\times3\end{equation}

    \begin{equation*}x^5+\frac{1}{x^5}=243-135+45-30=123\end{equation}

This would be a good extension question for students learning the binomial expansion theorem. We also use this technique for trigonometric identities using complex numbers.

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Filed under Algebra, Binomial Expansion Theorem, Puzzles

Puzzle Page 1

If \frac{a+b+2c}{a+b-c}=\frac{31}{15}, what does \frac{a+b}{c} equal?

    \begin{equation*}15(a+n+2c)=31(a+b-c)\end{equation}

    \begin{equation*}15a+15b+30c=31a+31b-31c)\end{equation}

    \begin{equation*}61c=16a+16b)\end{equation}

    \begin{equation*}\frac{61}{16}=\frac{a+b}{c}\end{equation}

Two positive numbers are such that their difference, their sum, and their product are in the ratio 2:5:21. What is the smaller of the two numbers?

Let x and y be the two numbers. Then

(1)   \begin{equation*}x-y=2k\end{equation*}

(2)   \begin{equation*}x+y=5k\end{equation*}

(3)   \begin{equation*}xy=21k\end{equation*}

Add equation 1 and 2 together to eliminate the y

    \begin{equation*}2x=7k\end{equation}

(4)   \begin{equation*}x=\frac{7k}{2}\end{equation*}

From 2 =5k-x, substitute for y into equation 3.

(5)   \begin{equation*}x(5k-x)=21k\end{equation*}

Substitute x=\frac{7k}{2} into equation 5.

    \begin{equation*}\frac{7k}{2}(5k-\frac{7k}{2})=21k\end{equation}

    \begin{equation*}\frac{35k^2}{2}-\frac{49k^2}{4}=21k\end{equation}

    \begin{equation*}\frac{70k^2}{4}-\frac{49k^2}{4}=\frac{84k}{4}\end{equation}

    \begin{equation*}21k^2-84k=0\end{equation}

    \begin{equation*}21k(k-4)=0\end{equation}

Hence, k=0 or k=4.

When k=4, x=\frac{7\times 4}{2}=14 and y=5\times 4-14=6

Therefore the smaller number is 6.

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Filed under Algebra, Puzzles, Ratio, Solving Equations

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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Filed under Algebra, Index Laws, Interesting Mathematics, Quadratics, Solving

Arithmetic Sequence

I did this question with on of my year 11 students. I think the algebra and the subscripts can be a bit tricky.

If T_m=n and T_n=m, then prove that T_{m+n}=0. Here where T_n and T_m are terms of an arithmetic sequence.
Mathematics Methods Units 1&2 – Exercise 15B Question 19

If T_m=n then,

(1)   \begin{equation*}n=a+(m-1)d\end{equation*}


And if T_n=m then,

(2)   \begin{equation*}m=a+(n-1)d\end{equation*}


Subtract equation (2) from equation (1)

    \begin{equation*}n-m=(m-1)d-((n-1)d)\end{equation*}


    \begin{equation*}n-m=md-nd\end{equation*}


(3)   \begin{equation*}n-m=d(m-n)\end{equation*}


Therefore d must equal -1
Substitute d=-1 into equation (1)

    \begin{equation*}n=a+(m-1)(-1)\end{equation*}


(4)   \begin{equation*}n=a-m+1\end{equation*}


Therefore a=n+m-1


(5)   \begin{equation*}T_{m+n}=a+(m+n-1)d\end{equation*}


Substitute a=n+m-1 and d=-1 into equation (5)

    \begin{equation*}$T_{m+n}=n+m-1+(m+n-1)(-1)$\end{equation*}


    \begin{equation*}$T_{m+n}=n+m-1-m-n+1$\end{equation*}


(6)   \begin{equation*}$T_{m+n}=0$\end{equation*}

As you can see from equation (6), T_{m+n}=0

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Filed under Algebra, Arithmetic, Sequences, Year 11 Mathematical Methods

Cats and Dogs

In my town 10% of the dogs think they are cats and 10% of the cats think they are dogs. All the other cats and dogs are perfectly normal. When all the cats and dogs in my town were rounded up and subjected to a rigorous test, 20% of them thought they were cats. What percentage of them really were cats?
Hamilton Olympiad 2003 B4 – The Ultimate Mathematical Challenge

Let x be the number of cats and y be the number of dogs.
Then 0.9x+0.1y think they are cats.
But we also know 20% of the total think they are cats.
0.2(x+y)
Therefore, 0.9x+0.1y=0.2(x+y)
0.9x+0.1y=0.2x+0.2y
0.7x=0.1y
7x=y
Percentage of cats is \frac{x}{x+y}\times100
Substitute 7x for y
\frac{x}{x+7x}\times100=\frac{x}{8x}\times100=\frac{1}{8}\times100=12.5%
\therefore 12.5% of the animals are cats

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Filed under Algebra, Arithmetic, Percentages, Simplifying fractions, UK Mathematics Challenge