heureka

Can vector AD - CD be simplified?

$\begin{array}{|rcll|} \hline AD - CD &=& AD + DC \\ \mathbf{AD-CD} &=& \mathbf{AC} \\ \hline \end{array}$

Thank you, CPhill !

Help please!

$\text{Let $\vec{AC}=\vec{a}$} \\ \text{Let $\vec{CB}=\vec{b}$} \\ \text{Let $\vec{CD}=(1-\lambda)\vec{b}$} \\ \text{Let $\vec{DB}=\lambda\vec{b}$} \\ \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{ABC}: \\ \hline \dfrac{1}{2}\vec{OB} &=& \vec{a} + \vec{b} \\ \mathbf{ \vec{OB} } &=& \mathbf{2(\vec{a} + \vec{b})} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{OAE}: \\ \hline \vec{OE} &=& \dfrac{1}{2}\vec{OB} + \dfrac{1}{2}\vec{a} \quad | \quad \mathbf{ \vec{OB}=2(\vec{a} + \vec{b})} \\\\ \vec{OE} &=& \dfrac{1}{2}\times 2(\vec{a} + \vec{b}) + \dfrac{1}{2}\vec{a} \\\\ \vec{OE} &=& \vec{a} + \vec{b} + \dfrac{1}{2}\vec{a} \\\\ \mathbf{ \vec{OE} } &=& \mathbf{\dfrac{3}{2}\vec{a} + \vec{b}} \\ \hline \end{array}$

$\text{Let $\vec{OD}=\mu\vec{OE}$} \\ \text{Let $\vec{OD}=\vec{OB}-\vec{DB}$} \\ \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{ODB}: \\ \hline \vec{OD} = \mu\vec{OE} &=& \vec{OB}-\vec{DB} \\ \mu\vec{OE} &=& \vec{OB}-\vec{DB} \quad | \quad \vec{OE}=\dfrac{3}{2}\vec{a} + \vec{b},~\vec{OB}=2(\vec{a} + \vec{b}),~ \vec{DB}=\lambda\vec{b} \\\\ \mu \left(\dfrac{3}{2}\vec{a} + \vec{b}\right) &=& 2(\vec{a} + \vec{b})-\lambda\vec{b} \\\\ \dfrac{3}{2}\vec{a}\mu + \mu \vec{b} &=& 2\vec{a} + 2\vec{b}-\lambda\vec{b} \\\\ \vec{a}\underbrace{ \left(\dfrac{3}{2}\mu-2\right)}_{=0} &=& \vec{b}\underbrace{\left( 2-\mu-\lambda\right)}_{=0} \\\\ \hline \dfrac{3}{2}\mu-2 &=& 0 \\ \dfrac{3}{2}\mu &=& 2 \\ \mathbf{\mu} &=& \mathbf{\dfrac{4}{3}} \\\\ \vec{OD} &=& \mu\vec{OE} \quad | \quad \vec{OE}=\dfrac{3}{2}\vec{a} + \vec{b}\\ \vec{OD} &=& \dfrac{4}{3} \left(\dfrac{3}{2}\vec{a} + \vec{b} \right)\\ \mathbf{\vec{OD}} &=& \mathbf{2\vec{a} + \dfrac{4}{3} \vec{b}} \\ \hline 2-\mu-\lambda &=& 0 \\ \lambda &=& 2-\mu \quad | \quad \mathbf{\mu=\dfrac{4}{3}} \\ \lambda &=& 2-\dfrac{4}{3} \\ \lambda &=& \dfrac{6-4}{3} \\ \mathbf{\lambda} &=& \mathbf{\dfrac{2}{3}} \\\\ \vec{CD} &=& (1-\lambda)\vec{b} \\ \vec{CD} &=& \left(1-\dfrac{2}{3}\right)\vec{b} \\ \mathbf{\vec{CD}} &=& \mathbf{\dfrac{1}{3}\vec{b}} \\\\ \vec{DB} &=& \lambda\vec{b} \\ \mathbf{\vec{DB}} &=& \mathbf{\dfrac{2}{3}\vec{b}} \\ \hline \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{\dfrac{1}{3}\vec{b}}{\dfrac{2}{3}\vec{b}} \\\\ \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} \\\\ \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{1}{3} \times \dfrac{3}{2} \\\\ \mathbf{\dfrac{\vec{CD}}{\vec{DB}}} &=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}$

Let u and v be the solutions to 3x^2 + 5x + 7 = 2x^2 - 8x + 1.  Find u/v + v/u.

$\begin{array}{|rcll|} \hline \mathbf{\dfrac{u}{v}+\dfrac{v}{u}} \\ &=& \dfrac{u^2+v^2}{uv} \\\\ &=& \dfrac{u^2+v^2}{uv} +2-2 \\\\ &=& \dfrac{u^2+v^2}{uv} +\dfrac{2uv}{uv}-2 \\\\ &=& \dfrac{u^2+2uv+v^2}{uv}-2 \\\\ &=& \dfrac{(u+v)^2}{uv}-2 \\\\ \mathbf{\dfrac{u}{v}+\dfrac{v}{u}}&=& \mathbf{\dfrac{(u+v)^2-2uv}{uv}} \\ \hline \mathbf{x^2+13x+6}&=&\mathbf{0} \\ \text{By Vieta: } \qquad u+v = -13, \quad uv = 6 \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{u}{v}+\dfrac{v}{u}} &=& \mathbf{\dfrac{(u+v)^2-2uv}{uv}} \quad | \quad u+v = -13, \quad uv = 6 \\\\ &=& \dfrac{(-13)^2-2*6}{6} \\\\ &=& \dfrac{169-12}{6} \\\\ &=& \mathbf{\dfrac{157}{6}} \\ \hline \end{array}$

Let u and v be the solutions to 3x^2 + 5x + 7 = 2x^2 - 8x + 1.  Find u/v + v/u.

$\begin{array}{|rcll|} \hline \mathbf{\dfrac{u}{v}+\dfrac{v}{u}} \\ &=& \dfrac{u^2+v^2}{uv} \\ &=& \dfrac{u^2+v^2}{uv} +2-2 \\ &=& \dfrac{u^2+v^2}{uv} +\dfrac{2uv}{uv}-2 \\ &=& \dfrac{u^2+2uv+v^2}{uv}-2 \\ &=& \dfrac{(u+v)^2}{uv}-2 \\ &=& \mathbf{\dfrac{(u+v)^2-2uv}{uv}} \\ \hline \mathbf{x^2+13x+6}&=&\mathbf{0} \\ \text{By Vieta: } \qquad u = -13, \quad v = 6 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{\dfrac{u}{v}+\dfrac{v}{u}} &=& \mathbf{\dfrac{(u+v)^2-2uv}{uv}} \\\\ &=& \dfrac{(-13)^2-2*6}{6} \\\\ &=& \dfrac{169-12}{6} \\\\ &=& \mathbf{\dfrac{157}{6}} \\ \hline \end{array}$

Find the sum of the squares of the solutions to $2x^2+4x-1=0$.

$\begin{array}{|rcll|} \hline 2x^2+4x-1 &=& 0 \quad | \quad :2 \\ x^2+2x-\dfrac{1}{2} &=& 0 \\ \hline \end{array}$

By Vieta:

$2 = -(x_1+x_2) \\ -\dfrac{1}{2} = x_1*x_2$

$\begin{array}{|rcll|} \hline x_1^2+x_2^2 &=& (x_1+x_2)^2 - 2x_1*x_2 \\ x_1^2+x_2^2 &=& (-2)^2 - 2*\left(-\dfrac{1}{2}\right) \\ x_1^2+x_2^2 &=& 4 + 1 \\ \mathbf{x_1^2+x_2^2} &=& \mathbf{5} \\ \hline \end{array}$

Solve for $z$ in the following equation: $2-iz = 1 + 3iz$.

$\begin{array}{|rcll|} \hline 2-iz &=& 1 + 3iz \\ 2-1-iz-3iz &=& 0 \\ 1-4iz &=& 0 \quad | \quad \times i\\ 1*i-4i^2z &=& 0 \\ i-4i^2z &=& 0 \quad | \quad \mathbf{i^2 = -1} \\ i-4(-1)z &=& 0 \\ i+4z &=& 0 \\ 4z &=& -i \\ \mathbf{z} &=& \mathbf{-\dfrac{i}{4}} \\ \hline \end{array}$

Find a cubic polynomial with integer coefficients that has $\sqrt[3]{2}+\sqrt[3]{4}$ as a root.

What is $3^{-1}+3^{-2} \pmod {25}$?
Express your answer as an integer from 0-24 inclusive.

My attempt:

$\begin{array}{|rcll|} \hline && 3^{-1}+3^{-2} \pmod {25} \\ &\equiv& 3^{-1}+3^{-1}3^{-1} \pmod {25} \\ &\mathbf{\equiv}& \mathbf{3^{-1}\left( 1+3^{-1}\right) \pmod {25}} \\ \hline \end{array}$

$3^{-1} \pmod {25} = \ ?$

$\begin{array}{|rcll|} \hline 3^{-1} \pmod {25} &\equiv& \dfrac{1}{3} \pmod {25} \\ &\equiv& 3^{\phi(25)-1} \pmod {25} \quad | \quad \phi(25)=20 \\ &\equiv& 3^{20-1} \pmod {25} \\ &\equiv& 3^{19} \pmod {25} \\ &\equiv& 3^{3*6+1} \pmod {25} \\ &\equiv& 3^{3*6}*3 \pmod {25} \\ &\equiv& \left(3^3\right)^6*3 \pmod {25} \quad | \quad 3^3 =27 \equiv 2 \pmod{25} \\ &\equiv& 2^6*3 \pmod {25} \\ &\equiv& 192 \pmod {25} \\ \mathbf{3^{-1} \pmod {25}} &\equiv& \mathbf{17 \pmod {25}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 3^{-1}+3^{-2} \pmod {25} &\mathbf{\equiv}& \mathbf{3^{-1}\left( 1+3^{-1}\right) \pmod {25}} \quad | \quad \mathbf{3^{-1} \pmod {25}} \equiv \mathbf{17 \pmod {25}} \\ &\equiv& 17 ( 1+17 ) \pmod {25} \\ &\equiv& 17*18 \pmod {25} \\ &\equiv& 306 \pmod {25} \\ \mathbf{3^{-1}+3^{-2} \pmod {25}} &\equiv& \mathbf{6 \pmod {25}} \\ \hline \end{array}$

If
$\sqrt{x+\!\sqrt{x+\!\sqrt{x+\!\sqrt{x+\cdots}}}}=10$
solve for x.

My attempt:

$\begin{array}{|rcll|} \hline y &=& \sqrt{x+\!\sqrt{x+\!\sqrt{x+\!\sqrt{x+\cdots}}}} \\ y &=& \sqrt{x+\!y} \quad | \quad \text{square both sides} \\ y^2 &=& x+y \quad | \quad y = 10 \\ 100 &=& x + 10 \\ \mathbf{x} &=& \mathbf{90} \\ \hline \end{array}$