Algebra

See also: Calculus | Index | MathJax

Exercises

Q1: \(i^i=? \)

A1:

\[ \begin{align} \color{red}i^i & = (\color{red}{e^{\frac {i \pi }{2}}})^i \\ & = e^{\frac{\pi}{2}i^2} \\ & = e^{\frac{\pi}{2}(-1)} \\ & = e^{-\frac{\pi}{2}} \\ & = \mathbf{0.208} \end{align} \]


Q2: There are 600 pupils in the school, with 30 more girls than boys. How many girls are in the school?

A2:

If \(g = \text{Girls}\) and \(b = \text{Boys}\) then \( \begin{align} g+b & = 600 && \text{(600 pupils)} \tag{1}\\ g & = b+30 && \text{(30 more girls)} \tag{2} \end{align} \) Using (2) in (1): \( \begin{align} (b+30)+b & = 600 \\ 2b+30 & = 600 &&|-30 \\ 2b & = 570 &&|\div 2 \\ b & = 285 \end{align} \) And with (2): \( \begin{align} g & = b+30 \\ g & = 285+30 \\ g & = \mathbf{315} \end{align} \) Double-check: \( \begin{align} g+b & = 315+285=600 \checkmark \\ \end{align} \) Answer: There are 315 girls in the school.


Q3: What is \(x=1+1/(1+1/(1+1/(1+\cdots)))\)?

A3:

Code (estimate numerically)

Algebraic answer

\[ \begin{align} x &= 1+1/(\color{red}{1+1/(1+1/(1+\cdots))}) \\ x &= 1+\frac{1}{\color{red}x} && |\times x \\ x^2 &= x+1 && |-x-1\\ x^2-\color{blue}{1}x-1 &=0 &&| \text{ C.T.S. (Completing the square)}\\ x^2-2\color{blue}{\frac 12}x+\left(\color{blue}{\frac 12}\right)^2 -\left(\frac 12\right)^2-1 &=0 &&| x^2-2\color{blue}ax+\color{blue}a^2=(x-\color{blue}a)^2\\ \left(x-\frac 12\right)^2 -\frac 14-1 &=0 \\ \left(x-\frac 12\right)^2 -\frac 54 &=0 && \left\vert +\frac 54 \right. \\ \left(x-\frac 12\right)^2 &= \frac 54 &&| \sqrt{\phantom{a}}\\ x-\frac 12 &= \pm\sqrt{\frac 54} &&|+\frac 12\\ x &= +\frac 12 \pm\frac{\sqrt{5}}{2} \\ x &= \frac{1\pm\sqrt{5}}{2} \\ x_1&=1.618034=\phi \\ x_2&=-0.618034=-(\phi-1)=-\frac{1}{\phi} \end{align} \]


Triangles in coordinate systems

Q4:

\(AB\) of a triangle \(ABC\) lies on the line defined by \(y=2x+1\). \(BC\) is on a line defined by \(y=-x+7\). \(CA\) is on a line defined by \(y=x-1\). What is the area of that triangle?

A4:

Calculate Points

\( \begin{align} y &= \color{red}{2x+1} && \tag{AB} \\ y &= \color{green}{-x+7} && \tag{BC} \\ y &= \color{blue}{x-1} && \tag{CA} \\ \end{align} \)


Since \(A\) (and only \(A\)) lies on BOTH \(AB\) and \(CA\), both equations apply: \[ \begin{align} y&= \color{red}{2x+1} &&\tag{1} \\ y&= \color{blue}{x-1} &&\tag{2} \\ \text{substitute y in (1) into (2)}: \\ \hline \color{red}{2x+1}&=\color{blue}{x-1} &&|-x-1 \\ x &=-2 \\ \text{use this in (2):} \\ y&=-2-1 \\ y&=-3 \\ A=\mathbf{(-2,-3)} \end{align} \]


Since \(B\) (and only \(B\)) lies on BOTH \(AB\) and \(BC\), both equations apply: \[ \begin{align} y&= 2x+1 && \tag{1} \\ y&=-x+7 && \tag{2} \\ \text{substitute y in (1) into (2)}: \\ 2x+1&=-x+7 &&|+x-1 \\ 3x&=6 &&|\div3 \\ x&=\mathbf{2} &&| \text{use this in (2):} \\ y&= -2+7 \\ y&= \mathbf{5} \\ B=\mathbf{(2,5)} \end{align} \]


Since \(C\) (and only \(C\)) lies on BOTH \(AC\) and \(BC\), both equations apply: \[ \begin{align} y &=x-1 && \tag{1} \\ y &=-x+7 && \tag{2} \\ \text{substitute y in (1) into (2)}: \\ x-1&=-x+7&&|+x+1 \\ 2x&=8 &&|\div2 \\ x&=\mathbf{4} \\ \text{use this in (1):} \\ y&=4-1 \\ y&=\mathbf{3} \\ C=\mathbf{(4,3)} \\ \end{align} \]

\( \text{I.e. } A=\mathbf{(-2,-3)}, B=\mathbf{(2,5)}, C=\mathbf{(4,3)} \)


Calculate Area

Calculate lengths of sides of triangle: \[ \begin{align} a^2=\overline{AB}^2 &=(2-(-2))^2+(5-(-3))^2&&=16+64=80 \\ a &=\sqrt{80}=4\sqrt5=8.944272 \\ b^2=\overline{BC}^2 &=(4-2)^2+(5-3)^2 &&=4+4=8 \\ b &=\sqrt8=2\sqrt2=2.828427 \\ c^2=\overline{CA}^2 &=(4-(-2))^2+(3-(-3))^2&&=36+36=72 \\ c &=\sqrt{72}=6\sqrt2=8.485281 \\ \end{align} \] Angle C from cosine rule: \[ \begin{align} a^2 &=b^2+c^2-bc\cos{C} &&|+bc\cos{C} -a^2 \\ bc\cos{C}&=b^2+c^2-a^2 &&|\div{bc} \\ cosB &=\frac{b^2+c^2-a^2}{bc} &=\frac{8+72-80}{\sqrt{80}\cdot\sqrt{72}} &=\frac{0}{\sqrt{80}\cdot\sqrt{72}} &=0 \\ B &=\underline{\mathbf{90°}} \\ \end{align} \]

This is a generalization of the fact that \(a^2=b^2+c^2\), i.e. \(b\) and \(c\) are at a right angle.

Another proof of showing that \(BC\) and \(CA\) are at a right angle is the fact, that:

\[ \begin{align} \text{slope of }BC&=-\frac{1}{\text{slope of }CA} \\ (-1) &=-\frac{1}{(1)} \\ -1 &=-1 \\ \end{align} \]

Area with sine:

\[ \begin{align} Area &=\frac12bc\sin{B} \\ &=\frac12\cdot(2\sqrt{2})(6\sqrt{2})\sin90° \\ &=\frac12\cdot2\cdot6\sqrt{2}\sqrt{2})\cdot1 \\ Area &=\underline{\mathbf{12}} \\ \end{align} \]


Generic question (one corner at origin): What is area X?

\( \begin{align} X & = Square-A-B-C \\ & = ad-\frac12 ab-\frac12 (a-c)(d-b)-\frac12 cd \\ & = ad-\frac12 ab-\frac12 (a-c)(d-b)-\frac12 cd \\ & = ad-\frac12 ab- ( \frac12 ad -\frac12 ab -\frac12 cd +\frac12 cb ) -\frac12 cd \\ & = \color{red}{ad} \require{cancel} \cancel{-\frac12 ab} \color{red}{-\frac12 ad} \require{cancel} \cancel{+\frac12 ab} \require{cancel} \bcancel{+\frac12 cd} \color{blue}{-\frac12 cb} \require{cancel} \bcancel{-\frac12 cd} \\ & = \color{red}{+\frac12 ad} \color{blue}{-\frac12 cb} \\ & = +\frac12(ad-cb) \\ \end{align} \)

With Linear Algebra

\( \text{Area}=\frac12 \begin{vmatrix} a & b \\ c & d \end{vmatrix} =\frac12(ab-cd) \\ \text{ With } \mathbf{b}=\begin{bmatrix}a\\b\end{bmatrix} =\begin{bmatrix}-2\\2\end{bmatrix} \text{ and } \mathbf{c}=\begin{bmatrix}c\\d\end{bmatrix} =\begin{bmatrix}-6\\-6\end{bmatrix} \\ \text{ Area }=\frac12 \begin{vmatrix} -2 & 2 \\ -6 &-6 \end{vmatrix} =\frac12((-2)\cdot(-6)-(-6)\cdot2) =\frac12(24)=12 \)

Parametric line and quadratic formula

\( \text{A line has equation } y=mx-1 \text{, where } m \text{ is a constant.} \\ \text{A curve has equation } y=x^2-5x+3 \text{. } \\ \text{a) For what values of } m \text{ are there repeated roots when you solve} \\ \text{the simultaneous equations given above?} \\ \text{b) Describe geometrically the situation where } m \text{takes either of those values.} \\ \)

\( \begin{align} y&=mx-1\\ y&=x^2-5x+3\\ \text{Therefore:}\\ mx-1&=x^2-5x+3\\ 0&=x^2-mx-5x+3+1\\ x^2-(m+5)x+4&=0\\ \left(x-\frac{m+5}{2}\right)^2-\left(\frac{m+5}{2}\right)^2+4&=0\\ \left(x-\frac{m+5}{2}\right)^2&=\left(\frac{m+5}{2}\right)^2-4\\ x&=+\frac{m+5}{2} \pm \sqrt{\left(\frac{m+5}{2}\right)^2-4}\\ \text{These roots become equal if:}\\ \sqrt{\left(\frac{m+5}{2}\right)^2-4}&=0\\ \left(\frac{m+5}{2}\right)^2-4&=0\\ \left(\frac{m+5}{2}\right)^2&=4\\ \frac{m+5}{2}&=\pm \sqrt{4} = \pm 2\\ m&=\pm \sqrt{4} = \pm 4 -5\\ \end{align} \\ \underline{\mathbf{m_1=-1}}\\ \underline{\mathbf{m_2=-9}}\\ \) \( \text{Therefore the lines } y=-x-1 \text{ and } y=-9x-1 \\ \text{just touch the graph (are tangent to) } y=x^2-5x+3 \)