The following Algebra problem is taken from one of the American Mathematics Competitions (AMC 10) Preparation books. It is notable that this is one of many books helping 10th grade students get ready for math olympics competition.
The problem is, in my estimation, of medium difficulty. It is really an ‘exercise’ in algebraic manipulation:
\begin{align}
\text{If }\, \mathbf{x = \frac{3 + \sqrt{13}}{2}} \text{, find the value of }\, \mathbf{x -\, \frac{1}{x}}
\end{align}
This is a multiple-choice question in which 5 answers are given; all of them positive integers and only ONE is correct.
My own solution consisted in using the given value of \(\boldsymbol{x}\) and subtract from it its reciprocal
\begin{align*}
\text{Given: } \quad \qquad \; \, x &= \frac{3 + \sqrt{13}}{2} \\
\text{then: } \qquad x\, – \frac{1}{x} &= \frac{3 + \sqrt{13}}{2} -\, \frac{2}{3 + \sqrt{13}} \\
\text{For simplicity, let’s put the radicals first: } \qquad &= \frac{\sqrt{13} + 3}{2} -\, \frac{2}{\sqrt{13} + 3} \\
\text{Normalize the denominator: } \qquad &= \frac{\sqrt{13} + 3}{2} -\, \left(\frac{2}{\left(\sqrt{13} + 3\right)} \cdot \frac{\left(\sqrt{13}\; -\, 3\right)} {\left(\sqrt{13}\; -\, 3\right)}\right) \\
&= \frac{\sqrt{13} + 3}{2} -\, \frac{2 \cdot \left(\sqrt{13}\; -\, 3\right)}{13\; -\, 9} \\
\text{Common denominator and simplify: } \qquad &= \frac{2 \cdot \left(\sqrt{13} + 3\right)\; -\, 2 \cdot \left(\sqrt{13}\; -\, 3\right)}{4} = \frac{6 + 6}{4} = 3
\end{align*}
That answer turned out to be correct! The AMC-10 Preparation book contains worked out answers to all the problems. In most instances, using more than one method. For this specific problems it has two possible methods:
Method 1:
\begin{align*}
\text{From the given value: } \qquad \qquad \qquad x &= \frac{3 + \sqrt{13}}{2} \\
2x\, – 3 &= \sqrt{13} \\
\text{Square both sides: } \quad 4x^{2} -\, 12x + 9 &= 13 \\
\text{Divide both sides by } 4x \text{: } \left(\frac{1}{4x}\right) \cdot \left(4x^{2} -\, 12x\; -\, 4\right) &= 0 \left(\frac{1}{4x}\right) \\
\text{I was skeptical of this last step but then I realized } \boldsymbol{x} \text{ couldn’t be zero: } \quad x -\, 3 -\, \frac{1}{x} &= 0 \\
\text{And the solution emerges like magic: } \qquad x -\, \frac{1}{x} &= 3
\end{align*}
Method 2:
The second method is similar to the way I answered but a bit more sophisticated.
\begin{align*}
\text{From the given value: } \qquad x &= \frac{3 + \sqrt{13}}{2} \\
\text{Then } \mathbf{\frac{1}{x}} \text{ is the reciprocal: } \quad \frac{1}{x} &= \frac{2}{3 + \sqrt{13}} \\
\text{Normalize the denominator: } \qquad &= \frac{2}{\left(\sqrt{13} + 3\right)} \cdot \frac{\left(\sqrt{13}\, -\, 3\right)}{\left(\sqrt{13}\, -\, 3\right)} = \frac{2 \cdot \left(\sqrt{13}\, -\, 3\right)}{13\, -\, 9} = \frac{\sqrt{13}\, -\, 3}{2}
\end{align*}
With the given: \(\mathbf{x = \frac{3 + \sqrt{13}}{2}}\) and the computed: \(\mathbf{\frac{1}{x} = \frac{\sqrt{13}\, -\, 3}{2}}\) ……….
\begin{align*}
\text{We can now find the value of: } \qquad \qquad x\, -\, \frac{1}{x} &= \; \boldsymbol{?} \\
\frac{3 + \sqrt{13}}{2}\, -\, \frac{\sqrt{13}\, -\, 3}{2} &= \frac{6}{2} = 3
\end{align*}
Quiz #5
This week’s quiz is ‘short and sweet’.
Find \(\mathbf{\tan\left(75^{o}\right)}\) without using a calculator.
Answer to Quiz #4
The question was:
A particle is moving along the x-axis. The position of the particle at time \(\boldsymbol{t}\) is given by:
\[\mathbf{x\left(t\right) = t^{3} -\, 6t^{2} + 9t\, -\, 2, \quad 0 \le t \le 5}\]
Find the total distance the particle travels in 5 units of time.
We have the position function. To find the distance the particle travels we need to know how fast the particle travels (the velocity function): \[\mathbf{\frac{d}{dt} [x\left(t\right)] = 3t^{2} -\, 12t + 9, \quad 0 \le t \le 5} \] and find when it is positive or negative (traveling to the right or to the left):
\begin{align*}
3t^{2} -\, 12t + 9 &= 0 \\
\left(t -\, 1\right) \cdot \left(t -\, 3\right) &= 0 \\
t &= 1, 3 \\
\text{Find the sign of } \boldsymbol{t} \text{ on the interval } [0,1] \text{: } x’ \left(0.5\right) &> 0 \\
\text{on the interval } [1,3] \text{: } x’ \left(2\right) &< 0 \\
\text{on the interval } [3,5] \text{: } x’ \left(4\right) &> 0
\end{align*}
To find the total distance travelled, we integrate the velocity function over three integrals, one for each interval, being careful noting when the particle was traveling to the left (negative velocity):
\begin{align*}
\text{Distance } &= \underbrace{\int_{0}^{1} \left(t^{2} -\, 4t +3\right) dt}_{\text{moving to the right}} \underbrace{\; – \int_{1}^{3} \left(t^{2} -\, 4t +3\right) dt}_{\text{moving to the left}} \underbrace{\; + \int_{3}^{5} \left(t^{2} -\, 4t +3\right) dt}_{\text{moving to the right}} \\
&= \bigg[\frac{1}{3}t^{3} -\, 6t^2 + 9t\bigg]\bigg|_{0}^{1} -\, \bigg[\frac{1}{3}t^{3} -\, 6t^2 + 9t\bigg]\bigg|_{1}^{3} + \bigg[\frac{1}{3}t^{3} -\, 6t^2 + 9t\bigg]\bigg|_{3}^{5} \\
&= \left(4 -\, 0\right) -\, \left(0 -\, 4\right) + \left(20 -\, 0\right) \\
&= 4 + 4 + 20 = \mathbf{28} \, \Leftarrow \text{Total distance travelled.}
\end{align*}
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