\(ax^{ 2 }+bx+c=0\)
\( \left\{ x-\cfrac { 1 }{ 2a } \left( -b-\sqrt { b^{ 2 }-4ac } \right) \right\} \left\{ x-\cfrac { 1 }{ 2a } \left( -b+\sqrt { b^{ 2 }-4ac } \right) \right\} =0\)
\( x=\cfrac { 1 }{ 2a } \left( -b\pm \sqrt { b^{ 2 }-4ac } \right) =\cfrac { -b }{ 2a } \pm \sqrt { \left( \cfrac { b }{ 2a } \right) ^{ 2 }-\cfrac { c }{ a } } \) --- (1)
When,
\( b^{ 2 }-4ac\lt0\)
\( \left\{ x-\cfrac { 1 }{ 2a } \left( -b-i\sqrt { 4ac-b^{ 2 } } \right) \right\} \left\{ x-\cfrac { 1 }{ 2a } \left( -b+i\sqrt { 4ac-b^{ 2 } } \right) \right\} =0\)
\( x=\cfrac { -b }{ 2a } \pm i\cfrac { \sqrt { 4ac-b^{ 2 } } }{ 2a } =\cfrac { -b }{ 2a } \pm i\sqrt { \cfrac { c }{ a } -\left( \cfrac { b }{ 2a } \right) ^{ 2 } } \) --- (2)
Consider,
\( f(x)=ax^{ 2 }+bx+c\)
For the extrememum,
\( f^{'}(x)=0\)
\( 2ax+b=0\)
\( x=-\cfrac { b }{ 2a } \)
which is the real part of the roots of \(f(x)=0\).
At which the function, \(f(x)\) attains the value,
\(f(x)= ax^{ 2 }+bx+c=a\left( \cfrac { b }{ 2a } \right) ^{ 2 }-\cfrac { b^{ 2 } }{ 2a } +c\)
\(f(x)= a\left\{\cfrac { c }{ a } -\left( \cfrac { b }{ 2a } \right) ^{ 2 }\right\}\)
But from (2), given \(a\ne0\),
\( x=\cfrac { -b }{ 2a } \pm i\sqrt { \cfrac { c }{ a } -\left( \cfrac { b }{ 2a } \right) ^{ 2 } }=\cfrac { -b }{ 2a } \pm i\sqrt {\cfrac{f(x)}{a} } \)
When \(a\) is a constant, a given quadratic \(f(x)\), \(\sqrt {\left\lvert\cfrac{f(x)}{a}\right\rvert }\) attains extrememum as does \(f(x)\). So the complex part of the roots of \(f(x)=0\) attains extrememum as \(f(x)\) attains extrememum. And we have,
where \(B\) minimum is at the intersection of \(real(r_e)\) and \(complex(r_e)\).
Furthermore, from (1), at the extrememum (real roots only) for a given quadratic \(f(x)\),
\(\require{cancel}\)
\( x=\cfrac { -b }{ 2a } \pm\cancelto{0}{ \sqrt { \left( \cfrac { b }{ 2a } \right) ^{ 2 }-{\cfrac { c }{ a }} }} \)
\(\left( \cfrac { b }{ 2a } \right) ^{ 2 }-{\cfrac { c }{ a }} =0\)
\(b^2-4ac=0\)
This is why the plot of \(real(r_e)\) also passes through the point where \(\cfrac{d\,r_e}{d\,T}\rightarrow\infty\) in the graph above. In general, given,
\( x=\cfrac { -b }{ 2a } \pm{ \sqrt { \left( \cfrac { b }{ 2a } \right) ^{ 2 }-{\cfrac { c }{ a }} }} \)
where \(a\ne0\),
\(\cfrac{d\,x}{d\,{\{a,b,c\}}}=\left(\cfrac { -b }{ 2a }\right)^{'}_{\{a,b,c\}} \pm \cfrac{1}{2\sqrt { \left( \cfrac { b }{ 2a } \right) ^{ 2 }-{\cfrac { c }{ a }} }}\left({ \sqrt { \left( \cfrac { b }{ 2a } \right) ^{ 2 }-{\cfrac { c }{ a }} }}\right)^{'}_{\{a,b,c\}}\)
when,
\({ { \left( \cfrac { b }{ 2a } \right) ^{ 2 }-{\cfrac { c }{ a }} }}=0\)
\(b^2-4ac=0\)
we have,
\(\cfrac{d\,x}{d\,{\{a,b,c\}}}\rightarrow\infty\)
ie. a plot of all real roots \(x\) of the quadratic \(f(x)=0\), has a pole when,
\(b^2-4ac=0\)
and
\(\left({ \sqrt { \left( \cfrac { b }{ 2a } \right) ^{ 2 }-{\cfrac { c }{ a }} }}\right)^{'}_{\{a,b,c\}}\ne0\)
This pole must exist for a general plot of \(x\) in order that the graph curves backwards to give two real roots and at the pole, the plot gives a double root. For this reason,
\(\left({ \sqrt { \left( \cfrac { b }{ 2a } \right) ^{ 2 }-{\cfrac { c }{ a }} }}\right)^{'}_{\{a,b,c\}}\ne0\)
when
\(b^2-4ac=0\)
is always true.
Given a material, the condition to narrow its bandgap (if it is an insulator or semiconductor) or to maximize its electric conductivity (if it is a conductor) is the same,
\({ 1-AT\cfrac { q^{ 2 } }{ \pi (m_{ e }c^{ 2 })^{ 2 }\varepsilon _{ o } } } =0\)
where,
\(A=\cfrac{\tau_o}{2}\cfrac{qv_{\small{T}}}{\{r^2_{\small{T}}+r_{or}^2\}^{3/2}}.{r_{or}}\)
\(\tau_o\) is the analogue to \(\mu_o\). Next stop finding \(\tau_o\)...even if \(v_{\small{T}}=c\), \(r_{\small{T}}\), the orbital radius of the positive temperature particle that generates the weak field is still not available.
