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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[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 . . .

\(\displaystyle{{e}^{{e}^{{e}^{{e}^{e}}}}}\)

over-/under-line :
\[\begin{array}{|c|c|c|} \hline \overline{A}+\overline{B}=\overline{AB} & \overline{A}\ \overline{B}=\overline{A+B} & A\overline{B}+\overline{A}B=A⊕B\\ \hline {R\\ _{NORMAL}} & {\mathbf{R}\\ _{BOLD}} & {\mathbb{R}\\ _{OUTLINE}}\\ \hline \underline{UNDERLINE} & \overrightarrow{VECTOR} & {A·\cancel{B}\\ _{CANCELING\ ^{1)}\ AN\ OPERAND}} \\ \hline \end{array} \] about ¹⁾ MathJax TeX and LaTeX Support — MathJax 2.7 documentation - TeX and LaTeX extensions