Category Archives: Algebra

December 10, 2024 · 6:00 am

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

December 3, 2024 · 6:54 am

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

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October 7, 2024 · 7:43 am

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

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September 23, 2024 · 3:11 am

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

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