About the Quadratic Equation

Derivation of the solutions ::

$x^2+px+q=0=\left({x-u}\right)\left({x-v}\right)=x^2-\left({u+v}\right)x+uv$

$p=2a$

$x^2+2ax+a^2-a^2+q=0$

$x+a=±\sqrt{a^2-q}$

$a=\frac p2$

$x=-\frac p2±\sqrt{{\left({\frac p2}\right)}^2-q}$

$x_1=-\frac p2-\sqrt{{\left({\frac p2}\right)}^2-q}$

$x_2=-\frac p2+\sqrt{{\left({\frac p2}\right)}^2-q}$

$x_2-x_1=2\sqrt{{\left({\frac p2}\right)}^2-q}$

$x_2+x_1=-p$

$x_2=-p-x_1$

$x_1=-p-x_2$

$x+p+\frac qx=0$

$\displaystyle{x=-p-\frac qx\quad \rightarrow \quad x_n=x_n+x_\overline{n}-\frac q{x_n}}$

$\displaystyle{x_n=\frac q{x_\overline{n}}\ \equiv\ x_nx_\overline{n}=q\ |\ about:\ q=uv}$

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Example ::

$\mathbf{p}$	$\mathbf{q}$	$\mathbf{x_n}$	$\mathbf{x_1}\\ -p-x_2\\ \displaystyle{\frac q{x_\overline{n}}}$	$\mathbf{x_2}\\ -p-x_1\\ \displaystyle{\frac q{x_\overline{n}}}$
$-1$	$-1$	$-\frac{-1}2±\sqrt{{\left({\frac{-1}2}\right)}^2-\left({-1}\right)}=$ $\frac{1±\sqrt{5}}2=$ $=\cases{-\frac{\sqrt{5}-1}2\\ \frac{\sqrt{5}+1}2}$ $=\cases{-\frac{5-1}{2\left({\sqrt{5}+1}\right)}\\ \frac{5-1}{2\left({\sqrt{5}-1}\right)}}$ $=\cases{\frac{-2}{\sqrt{5}+1}\\ \frac2{\sqrt{5}-1}}$	$\frac{-2}{\sqrt{5}+1}\mathbf{=}$ $\mathbf{=}-\left({-1}\right)-\frac{2}{\sqrt{5}-1}=$ $=1-\frac{\sqrt{5}+1}2=$ $=-\frac{-2+\sqrt{5}+1}2=$ $=-\frac{\sqrt{5}-1}2=$ $=-\frac2{\sqrt{5}+1}$ $\mathbf{=}\frac{-1}{\left({\frac2{\sqrt{5}-1}}\right)}=$ $=-\frac{\sqrt{5}-1}2=$ $=\frac{-2}{\sqrt{5}+1}$	$\frac2{\sqrt{5}-1}\mathbf{=}$ $\mathbf{=}-\left({-1}\right)-\frac{-2}{\sqrt{5}+1}=$ $=1+\frac{\sqrt{5}-1}2=$ $=\frac{2+\sqrt{5}-1}2=$ $=\frac{\sqrt{5}+1}2=$ $=\frac2{\sqrt{5}-1}$ $\mathbf{=}\frac{-1}{\left({\frac{-2}{\sqrt{5}+1}}\right)}=$ $=\frac{\sqrt{5}+1}2=$ $=\frac2{\sqrt{5}-1}$

Dynamic Formulas for the Capacitor and for the Inductor

Charging the capacitor from the CV(=Vs) src. ::

The voltage drop on the series resistor $$V_R=V_S-V_C$$

$$t_{0_{↑}^{↑}→V_C}=-R·C·ln\left({1-\frac{V_C}{V_S}}\right)=R·C·ln\frac{V_S}{V_R}$$

$$t_{V_{1_C}{}_{↑}^{↑}→V_{2_C}}=R·C·ln\frac{V_S-V_{1_C}}{V_S-V_{2_C}}=R·C·ln\frac{V_{1_R}}{V_{2_R}}$$

Dis-charging the capacitor through a fixed value resistor ::

$$t_{V_{2_C}{}_{↓}^{↓}→V_{1_C}}=-R·C·ln\frac{V_{1_C}}{V_{2_C}}=R·C·ln\frac{V_{2_C}}{V_{1_C}}$$

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"Charging" the inductor from the CV(=Vs) src. ::

The peak current $$I_{MAX}=\frac{V_S}R$$

$$t_{0_{↑}^{↑}→I_L}=-\frac LR·ln\left({1-\frac{I_L}{I_{MAX}}}\right)$$

$$t_{I_{1_L}{}_{↑}^{↑}→I_{2_L}}=\frac LR·ln\frac{I_{MAX}-I_{1_L}}{I_{MAX}-I_{2_L}}$$

"Dis-charging" the inductor through a fixed value resistor ::

$$t_{I_{2_L}{}_{↓}^{↓}→I_{1_L}}=\frac LR·ln\frac{I_{2_L}}{I_{1_L}}$$

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about how to get linear I U ramps on the inductor and on the capacitor : simulation

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[Eop]

Diluting the solution

If there is solution with the concentration $x=\frac b{b+a}$ and we need to get the solution with concentration $y=\frac b{b+c}$ then from $x\left({b+a}\right)=b=y\left({b+c}\right)$ we get the ratio of masses $\displaystyle{\frac xy=\frac{b+c}{b+a}=\frac{m_{FINAL}}{m_{INITIAL}}}$ . . . so - if the $\displaystyle{m_{FINAL}}$ is given we need the $\displaystyle{m_{INITIAL}=\frac yx·m_{FINAL}}$ of the initial solution , the amount of water needed to be added is $c-a=\mathbf{d}$ , from $\displaystyle{\frac xy=\frac{m_F}{m_I}=1+\frac d{m_I}=\frac1{1-\frac d{m_F}}}$ we see that $\displaystyle{d=\left({\frac xy-1}\right)·m_I=\left({1-\frac yx}\right)·m_F}$ . . . checking
$\displaystyle{d=\left({\frac xy-1}\right)·m_I=\left({\frac xy-1}\right)·\frac yx·m_F}$ and
$\displaystyle{d=\left({1-\frac yx}\right)·m_F=\left({1-\frac yx}\right)·\frac xy·m_I}$

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PS! -- the below DEMO works and "in reverse" for the initial percentage being less than the final ... the result is given as a negative amount of the water to be "added" --e.g.-- describing the amount of water to be (evaporated/)extracted from the solution ... Now updated! to match the possible concentrating option ...

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. . . a Q.C. (in MSO Excel) for some non-intuitive cases ::

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Physics

Linear	Circular
Time $t\ \left({s}\right)$	Time $t\ \left({s}\right)$
Mass $m\ \left({kg}\right)$	Moment of inertia $I=\sum{m_i·r_i}\ \left({kg·m^2}\right)$
Distance $s\ \left({m}\right)$	Angle $\overrightarrow{φ}\ \left({rad=1}\right)$
Velocity $v\ \left({\frac ms}\right)$	Angular velocity $\overrightarrow{ω}=\frac{∂φ}{∂t}=\frac vR\ \left({\frac {rad}s=\frac1s}\right)$
Acceleration $a\ \left({\frac m{s^2}}\right)$	Angular_acceleration $\overrightarrow{ɛ}=\frac{∂ω}{∂t}=\frac{a_{\tau}}R\ \left({\frac{rad}{s^2}=\frac1{s^2}}\right)$
Impulse $p=m·v=F·dt\ \left({N·s}\right)$	Angular momentum $N=I·\overrightarrow{ω}=\overrightarrow{R}×\overrightarrow{p}\ \left({\frac{kg·m^2}s}\right)$
Force $F=\frac{∂p}{∂t}=m·a\ \left({N}\right)$	Torque $M=I·\overrightarrow{ɛ}=\overrightarrow{R}×\overrightarrow{F}\ \left({N·m}\right)$
Energy $E=\frac{m·v^2}2=m·a·s\ \left({J}\right)$	Energy $W=\frac{I·ω^2}2=I·ɛ·φ\ \left({J}\right)$
Work $A=F·∂s·Cos\ φ\ \left({J}\right)$	Work $A=M·∂φ\ \left({J}\right)$
Power $P=\frac{∂A}{∂t}\ \left({W}\right)$	Power $P=\frac{∂W}{∂t}\ \left({W}\right)$
$d\left({m·\overrightarrow{v}}\right)=\overrightarrow{F}·dt$	$d\left({I·\overrightarrow{ω}}\right)=\overrightarrow{M}·dt$

p+ , p– , x(T) , f(x)

Form the $x^2-xT-1=0$ where $T\in\mathbb{R}^{+}\ ;\ T≠0$
$x-\frac1x=T\ →\ \frac xT=\frac{xT+1}{xT}=\frac{x+\frac1T}x$
Def.: $x≡p^+$ and $\frac1x≡p^-$
$p^+=\frac{\sqrt{T^2+4}+T}2\ →\ p^-=\frac1{p^+}=\frac2{\sqrt{T^2+4}+T}=\frac{\sqrt{T^2+4}-T}2$

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Def.: $x\left({T}\right)=T+p^-\left({\frac1x}\right)=T+\frac{\sqrt{T^2+4}-T}2=\frac{\sqrt{T^2+4}+T}2$ Def.: $f\left({x}\right)≡f\left({x_T}\right)≡x\left({T}\right)$

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The properties of the $f\left({x}\right)$ :

1. $f\left({-x}\right)=\frac1{f\left({x}\right)}=f\left({x}\right)-x$
   $\frac1{f\left({-x}\right)}=f\left({x}\right)=f\left({-x}\right)+x$
2. $f\left({-x}\right)+x=\frac1{f\left({x}\right)-x}$
3. $x·f\left({x}\right)-x·f\left({-x}\right)=x^2$
   $x·f\left({-x}\right)-x·f\left({x}\right)=-x^2$

NULL

testing it out . . .
embedded super and subscript

$\displaystyle{e^{e^{e^{e^e}}}}\\\displaystyle{a_{1_{2_{3_4}}}}$

over-/under-line :
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