heureka

Given right triangle ABC with altitude$\overline{BD}$ drawn to hypotenuse $\overline{AC}$.
If $\overline{AB}=2$ and $\overline{AC}=4$,
what is the length of $\overline{AD}$?

$\begin{array}{|rcll|} \hline \overline{AB}^2 &=& \overline{AD}\times \overline{AC} \\ 2^2 &=&\overline{AD}\times 4 \\ 4 &=&\overline{AD}\times 4 \\ \mathbf{\overline{AD}} &=& \mathbf{1} \\ \hline \end{array}$

Thank you Asinus:

Find all solutions to the system
a + b = 14
a^3 + b^3 = 812 + 3ab

$\begin{array}{|rcll|} \hline a+b &=& 14 \\ \mathbf{b} &=& \mathbf{14-a} \\ \hline (a+b)^3 &=& a^3+3a^2b+3ab^2+b^3 \\ (a+b)^3 &=& a^3+b^3 +3ab(a+b) \quad | \quad a^3 + b^3 = 812 + 3ab,~ a+b=14\\ 14^3 &=& 812 + 3ab+3*14*ab \\ \ldots \\ ab &=& \dfrac{1932}{45} \quad | \quad b = 14-a \\ a(14-a) &=& \dfrac{1932}{45}\\ 14a-a^2 &=& \dfrac{1932}{45}\\ \mathbf{a^2-14a+\dfrac{1932}{45} } &=& \mathbf{0} \\ a_1 &=& 9.4630604269 \\ a_2 &=& 4.5369395731\\\\ \qquad \mathbf{b} &=& \mathbf{14-a} \\ b_1 &=& 14-a_1 \\ b_1 &=& a_2 \\ b_1 &=& 4.5369395731 \\\\ b_2 &=& 14-a_2 \\ b_2 &=& a_1 \\ b_2 &=& 9.4630604269 \\ \hline \end{array}$

Find all solutions to the system
a + b = 14
a^3 + b^3 = 812 + 3ab

$\begin{array}{|rcll|} \hline a+b &=& 14 \\ \mathbf{b} &=& \mathbf{14-a} \\ \hline (a+b)^3 &=& a^3+3a^2b+3ab^2+b^3 \\ (a+b)^3 &=& a^3+b^3 +3ab(a+b) \quad | \quad a^3 + b^3 = 812 + 3ab,~ a+b=14\\ 14^3 &=& 812 + 3ab+3*14*ab \\ \ldots \\ ab &=& \dfrac{1932}{45} \quad | \quad b = 14-a \\ a(14-a) &=& \dfrac{1932}{45}\\ 14a-a^2 &=& \dfrac{1932}{45}\\ \mathbf{a^2-14a+\dfrac{1932}{45} } &=& \mathbf{0} \\ a_1 &=& 9.4630604269 \\ a_2 &=& 4.5369395731\\\\ \qquad \mathbf{b} &=& \mathbf{14-a} \\ b_1 &=& x_2 \\ b_2 &=& x_1 \\ \hline \end{array}$

Find the exact value of 2012-2002+1992-1982 +...+ 32 - 22 + 12 - 2

$\begin{array}{|l|rcl|} \hline 1.& -2+12 &=& 10 \\ 2.& -22+32 &=& 10 \\ 3.& -42+52 &=& 10 \\ \ldots & \ldots &\ldots & \ldots \\ n-1 & -1982+1992 &=& 10 \\ n & -2002+2012 &=& 10 \\ \hline & sum &=& 10*n \\ & sum &=& 10*101 \\ & \mathbf{sum} &=& \mathbf{1010} \\ \hline \end{array} \begin{array}{|rcll|} \hline &&\text{ finite arithmetic progression} \\ \hline a_n &=& a_1+(n-1)d \quad | \quad a_n = 2012,~a_1 = 12,~ d = 20 \\ 2012 &=& 12+(n-1)*20 \\ 2000 &=& (n-1)*20 \\ 100 &=& n-1 \\ \mathbf{n} &=& \mathbf{101} \\ \hline \end{array}$

7x+5y-3z=16
3x-5x+2z=-8
5x+3x-7z=0

$\begin{array}{|lrcll|} \hline (1) & 7x+5y-3z &=& 16 \\ (2) & 3x-5y+2z &=& -8 \\ (3) & 5x+3y-7z &=& 0 \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (1) & 7x+5y-3z&=& 16 \\ (2) & 3x-5y+2z &=& -8 \\ (1)+(2): & 10x-z &=& 8 \\ & \mathbf{z} &=& \mathbf{10x-8} \qquad (4) \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (2) & 3x-5y+2z &=& -8 \quad | \quad \mathbf{z=10x-8} \\ & 3x-5y+2(10x-8) &=& -8 \\ & 3x-5y+20x-16 &=& -8 \\ & 23x-5y &=& 8 \\ & 5y &=& 23x-8 \\ & \mathbf{y} &=& \mathbf{\dfrac{23x-8}{5}}\qquad (5) \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (3) & 5x+3y-7z &=& 0 \quad | \quad \mathbf{z=10x-8} \\ & 5x+3y-7(10x-8) &=& 0 \\ & 5x+3y-70x+56 &=& 0 \\ & \mathbf{65x-3y} &=& \mathbf{56} \quad | \quad \mathbf{y=\dfrac{23x-8}{5}} \\ & 65x-3\left(\dfrac{23x-8}{5}\right) &=& 56 \quad | \quad *5 \\ & 5*65x-3*(23x-8) &=& 56*5 \\ & 325x-69x +24 &=& 280 \\ & 256x &=& 265 \\ & \mathbf{x} &=& \mathbf{1} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (5) & \mathbf{y} &=& \mathbf{\dfrac{23x-8}{5}} \quad | \quad \mathbf{x=1} \\ & y &=& \dfrac{23-8}{5} \\ &y &=& \dfrac{15}{5} \\ & \mathbf{y} &=& \mathbf{3} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (4) & \mathbf{z} &=& \mathbf{10x-8} \quad | \quad \mathbf{x=1} \\ & z &=& 10-8 \\ & \mathbf{z} &=& \mathbf{2} \\ \hline \end{array}$

Variables a, b, and c, are positive real numbers.
Prove that:
sqrt(a^2-ab+b^2) + sqrt(a^2-ac+c^2) is greater than or equal to sqrt(b^2+bc+c^2)

Find the number of terms in the sequence.
$-18,~ 36,~ -72,~\ldots ~,~9437184$

My attempt:

$\begin{array}{|r|rl|} \hline \text{sequence} & : 9 \\ \hline -18 & -2 & = (-2)^1 \\ \hline 36 & 4 & = 2^2 \\ \hline -72 & -8 & = (-2)^3 \\ \hline \ldots & \ldots & \ldots \\ \hline 9437184 & 1048576 & = 2^{20} \\ \hline \end{array}$

$\text{Formula:}\\ \begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{9*(-2)^n} \\ a_{20} &=& 9*(-2)^{20} \\ a_{20} &=& 9*2^{20} \\ a_{20} &=& 9*1048576 \\ \mathbf{a_{20}} &=& \mathbf{9437184} \\ \hline \end{array}$

There are 20 terms in the sequence.

Find all complex numbers $z$ such that $z^4 = -4$.

$\begin{array}{|rcll|} \hline z^4 &=& -4 \quad | \quad \text{sqrt both sides} \\ z^2 &=& \pm \sqrt{-4} \\ z^2 &=& \pm \sqrt{(-1)*4} \\ z^2 &=& \pm \sqrt{-1}*\sqrt{4} \quad | \quad \sqrt{-1} = i \\ z^2 &=& \pm 2i \quad | \quad \text{sqrt both sides} \\ \mathbf{z} &=& \mathbf{\pm \sqrt{\pm 2i}} \\ &&\boxed{z_1 = \sqrt{2i}\\z_2 = \sqrt{-2i}\\ z_3 = -\sqrt{2i}=-z_1\\z_4 = -\sqrt{-2i}=-z_2 } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline (1+i)^2 &=& 1+2i+i^2 \quad | \quad \mathbf{i^2 =-1} \\ (1+i)^2 &=& 1+2i-1 \\ \mathbf{(1+i)^2} &=& \mathbf{2i} \\ \hline \end{array} \begin{array}{|rcll|} \hline (1-i)^2 &=& 1-2i+i^2 \quad | \quad \mathbf{i^2 =-1} \\ (1-i)^2 &=& 1-2i-1 \\ \mathbf{(1-i)^2} &=& \mathbf{-2i} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline z_1 &=& \sqrt{2i} \quad | \quad \mathbf{2i = (1+i)^2} \\ z_1 &=& \sqrt{(1+i)^2}\\ \mathbf{z_1} &=& \mathbf{1+i} \\\\ z_2 &=& \sqrt{-2i} \quad | \quad \mathbf{-2i = (1-i)^2} \\ z_2 &=& \sqrt{(1-i)^2}\\ \mathbf{z_2} &=& \mathbf{1-i} \\\\ z_3 &=& -z_1 \\ z_3 &=& -(1+i) \\ \mathbf{z_3} &=& \mathbf{-1-i} \\\\ z_4 &=& -z_2 \\ z_4 &=& -(1-i) \\ \mathbf{z_4} &=& \mathbf{-1+i} \\ \hline \end{array}$

Suppose z and w are complex numbers.  
Prove that $z + \overline{z}$ and $w \overline{w}$ are real.

$\begin{array}{|rcll|} \hline \text{Let $z=a+bi$} \\ \text{Let $\overline{z}=a-bi$} \\ \begin{array}{|rcll|} \hline z + \overline{z} &=& a+bi +a-bi \\ \mathbf{z + \overline{z}} &=& \mathbf{2a} \\ \hline \end{array}\\ \\ \text{Let $w=a+bi$} \\ \text{Let $\overline{w}=a-bi$} \\ \begin{array}{|rcll|} \hline w \overline{w} &=& (a+bi)(a-bi) \\ w \overline{w} &=& a^2-(bi)^2 \\ w \overline{w} &=& a^2-b^2i^2 \quad | \quad \mathbf{i^2 = -1} \\ \mathbf{w \overline{w}} &=& \mathbf{a^2+b^2} \\ \hline \end{array}\\ \hline \end{array}$

Let

$x = 2001^{1002} -2001^{-1002}$ and

$y = 2001^{1002} + 2001^{-1002}$.

Find $x^2 - y^2$.

$\text{Let $a =2001^{1002}$} \\ \text{Let $b =2001^{-1002}$} \\ \text{Let $ab \\\qquad~=2001^{1002} *2001^{-1002}\\\qquad~=2001^{1002-1002}\\\qquad~=2001^0\\\qquad~=1 $} \\ \text{Let $x =a-b$} \\ \text{Let $y =a+b$}$

$\begin{array}{|rcll|} \hline x^2-y^2 &=& (a-b)^2-(a+b)^2 \\ x^2-y^2 &=& a^2-2ab+b^2-(a^2+2ab+b^2) \\ x^2-y^2 &=& a^2-2ab+b^2-a^2-2ab-b^2 \\ x^2-y^2 &=& -2ab-2ab \\ x^2-y^2 &=& -4ab \quad | \quad \mathbf{ab=1} \\ x^2-y^2 &=& -4*1 \\ \mathbf{x^2-y^2} &=& \mathbf{-4} \\ \hline \end{array}$