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Al-Khwarizmi wrote on algebra and treated quadratic equations using both algebraic procedures and geometry.

Place a converging thin lens at coordinate $x$ between the object at $0$ and the screen at $L$. In air and the paraxial approximation, the required object and image distances are $u=x$ and $v=L-x$. The thin lens equation gives a quadratic equation:Its discriminant is $L^2-4Lf$. For $L>4f>0$, the two roots and their separation areBoth roots lie inside $(0,L)$, and both object and image distances exceed $f$. The two roots therefore describe finite real images on the screen. Exchanging the roots exchanges $u,v$; from transverse magnification $m=-v/u$,The image-height ratio between the enlarged and reduced settings is $|m_{-}|/|m_{+}|=|m_{-}{|}^2$, rather than $|m_{-}|$.

At $L=4f$, the roots coincide: $d=0$, $x=L/2=2f$, $u=v=2f$ and $m=-1$. There is one setting, not two separately measurable settings. For $0<L<4f$ the discriminant is negative, so no position between the fixed object and screen gives a sharp real image there. This does not forbid a real image elsewhere if the screen is moved. Another short proof observes that $x(L-x)$ has maximum $L^2/4$ at $x=L/2$; the focusing requirement $x(L-x)=Lf$ is possible exactly when $L\ge 4f$.

Set $u=e^{iw}$ in $\cos w=z$. The resulting quadratic equation is $u^2-2zu+1=0$, with roots $z\pm i\sqrt{1-z^2}$. Their product is one, so taking a complex logarithm never encounters zero. Solving the exponential equation gives the displayed full inverse relation, with integer $n$. The principal complex logarithm and principal square root do not make both signs identical to the principal inverse cosine: at $z=0$ they yield $\pm \pi /2$, while principal $\arccos 0=\pi /2$. A single analytic branch requires compatible choices of sign and branches.

A complex number is $z=x+iy$, where $x,y$ are real numbers and the imaginary unit satisfies $i^2=-1$. The real part is $x$ and the imaginary part is $y$: for $3+4i$ they are $3$ and $4$, respectively. Equality requires both coordinates to agree. Thus the complex plane gives each complex number a concrete position, while ordinary addition and multiplication become geometric operations.

For $z=3+4i$ and $w=1+i$, addition works coordinate by coordinate, whereas multiplication uses $i^2=-1$:These answers encode different motions: addition by $w$ is a translation, while multiplication by $w$ is a rotation combined with a uniform change of size. Keeping these pictures distinct prevents a common beginner's confusion.

The complex numbers form a field: every nonzero $z$ has a multiplicative inverse, computable using its complex conjugate and modulus. They also supply the missing solutions of a quadratic equation such as $u^2+1=0$, namely $u=\pm i$. The surprising payoff is that extending arithmetic also makes Euclidean geometry easier to calculate. For a systematic introduction to these definitions, see Jeremy Orloff, MIT 18.04, topic 1, sections 1.3–1.5.

A constructible circle has a constructible point as its centre and radius equal to the distance between two constructible points. When those coordinates lie in a real field $F$, its equation $(x-a{)}^2+(y-b{)}^2=r^2$ has coefficients in $F$ because the squared radius is a sum of squared coordinate differences. Intersecting it with a constructible line reduces to a quadratic equation. Intersecting two distinct circles reduces to the same situation by subtracting their normalized equations to obtain the radical axis. Thus each isolated real intersection requires at most a real quadratic field extension.

A point of the Euclidean plane is constructible if it can be obtained from $(0,0)$ and $(1,0)$ by finitely many straightedge-and-compass operations: draw a line through two obtained distinct points, draw a circle with obtained centre and radius equal to an obtained segment length, and retain isolated intersection points. Its two coordinates are constructible numbers. Conversely, constructible real coordinates can be placed on perpendicular axes and their coordinate lines intersected, giving the point. Coordinates created at each isolated intersection require only solving linear or quadratic equations over the current coordinate field; hence they lie in a real field extension obtained by a finite tower of quadratic extensions. Coincident curves supply no isolated new intersection points.

Four mutually tangent circles with signed curvatures $k_i$ satisfy $(k_1+k_2+k_3+k_4{)}^2=2(k_1^2+k_2^2+k_3^2+k_4^2)$. Curvature is the reciprocal of the radius, with an enclosing circle assigned negative curvature. Given three curvatures, solving the quadratic equation gives $k_4=k_1+k_2+k_3\pm 2\sqrt{k_1{k}_2+k_2{k}_3+k_3{k}_1}$, providing the two completions of the configuration.

An ellipse is the locus of points whose Euclidean distances from two fixed foci of a conic have constant sum $2a$. If the distance between the foci of a conic is $d<2a$, the semi-major and semi-minor axes are $a$ and $\sqrt{4a^2-d^2}/2$. Writing $c=d/2$ gives the conventional relation $b^2=a^2-c^2$.^[ref]

For an original calculation, choose $a=5$, $b=3$, and hence $c=4$. The ellipsehas foci of a conic $F_{-}=(-4,0)$ and $F_{+}=(4,0)$. The point $P=(3,12/5)$ lies on it since $9/25+16/25=1$. Its focal distances areIts eccentricity of a conic is $e=4/5$. The matching directrix of a conic for $F_{+}$ is $x=a/e=25/4$, and the perpendicular distance from $P$ to that straight line is $13/4$. Therefore the ratio of focal distance to directrix distance is $(13/5)/(13/4)=4/5$, agreeing with $e$.

The constant sum can be turned into a construction. Fix a pin at each focus of a conic and pull a closed string loop taut with a pencil. In this example the segment between the pins has length $8$, so a loop of length $18$ leaves $10$ for the two pencil-to-pin segments. Moving the pencil while keeping that loop taut traces the ellipse. This is a geometric prediction from the length constraint, not an assertion about a measured drawing.

The parametrization $x=5\cos t$, $y=3\sin t$ maps the unit circle onto this ellipse. It makes plotting easy, but is an affine map rather than an isometry: horizontal and vertical lengths have different scale factors. An ellipse is not a circle under the original Euclidean metric merely because this map exists. Its two semiaxes and eccentricity of a conic retain the metric information lost by that equivalence.

A useful proof outline begins with $|X-F_{-}|+|X-F_{+}|=2a$. Isolate one square root, square, and then repeat; the resulting quadratic equation is $x^2/a^2+y^2/(a^2-c^2)=1$. Since $a>c$, the quantities isolated before squaring have the required signs, so the reverse implication introduces no extra branch. The condition $a>c$ is essential: at equality the focal-distance locus becomes the line segment between the foci of a conic.

The ordinary generating function $C(x)={\sum }_{n\ge 0}{C}_n{x}^n$ packages the Catalan recurrence. Its Cauchy product gives the formal power series identityThe constant term is $C_0=1$. Solving the quadratic equation chooseswith the formal power series square root having constant term one. The apparent quotient at $x=0$ has constant term one after cancellation; the other quadratic solution has a pole and cannot be this ordinary generating function. The equation records the empty object or an outer unit with two ordered smaller objects.

Al-Khwarizmi presented geometric completions of squares in his treatment of quadratic equations. His exposition used verbal instructions, positive lengths and areas, rather than the modern algebraic notation used here. The area assembly above illustrates the same principle with a different expression and arrangement. The modern polynomial identity also covers negative values of the variable and roots of a polynomial in the complex numbers, which need no interpretation as positive side lengths.^[ref]

Consider a neutral monoprotic acid $\mathrm{HA}$ dissolved in water, with no added salts or other acids. Assume dilute ideal thermodynamic activities, fixed temperature, water thermodynamic activity one, and negligible self-ionization of water. Write $C_T$ for its analytical concentration, $c=C_T/c^{\circ }$ with $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, and $\alpha$ for its weak acid dissociation fraction. Conservation of acid units and electric charge givesHere ${\mathrm{A}}^{-}$ is the conjugate base, and ${\mathrm{H}}^{+}$ is conventional shorthand for hydrated hydrogen ions, represented by hydronium in this model. Substitution into the dimensionless acid dissociation constant givesSolving this quadratic equation and rejecting its negative root of a polynomial gives an exact result within the stated model:The second form avoids subtracting nearly equal numbers. Unlike the approximation $\alpha \approx \sqrt{K_a/c}$, it retains depletion of neutral weak acid and always satisfies $0<\alpha <1$ for positive $K_a,c$. The broader weak acid mass and charge balances restore self-ionization of water when that omitted contribution matters.

For an original worked example, stipulate $K_a=10^{-5}$ without identifying a real substance. At $C_T=10^{-2}\mathrm{mol}{\mathrm{L}}^{-1}$, the weak acid dissociation fraction is approximately $0.0311267$, or $3.11267\%$, and $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=3.11267\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$. A hundredfold dilution to $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ raises the fraction to $0.270156$, or $27.0156\%$, but lowers $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$ to $2.70156\times 10^{-5}\mathrm{mol}{\mathrm{L}}^{-1}$. A completely ionized strong acid at $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ would have $[{\mathrm{H}}_3{\mathrm{O}}^{+}]\approx 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$: less than the more concentrated weak acid in the first calculation, despite its greater converted fraction.

The directions of both changes follow without relying on the plot. Put $x=[{\mathrm{H}}_3{\mathrm{O}}^{+}]/c^{\circ }=c\alpha$. Rearranging the chemical equilibrium relation givesThus smaller $c$ forces smaller $x$, which in turn makes $\alpha$ larger. The derivative proves that total hydronium concentration and converted fraction move in opposite directions during dilution. In this model, adding solvent also makes additional acid units ionize at fixed total acid amount of substance, even though their ions are spread over more volume. That distinction explains why “more ionized” alone does not predict greater electrical conductivity.

Do not extend the no-background model to unlimited dilution: its prediction $[{\mathrm{H}}_3{\mathrm{O}}^{+}]\to 0$ omits water's own ions. The weak acid mass and charge balances include that background. Added salts, other acids or bases, and nonideal activity coefficients also change the relation between analytical concentration, converted fraction and hydronium concentration.

For pure hydrogen illuminated by an isotropic central source with hydrogen-ionizing photon production rate $Q$, an empty cavity of physical radius $r_{\mathrm{in}}$ gives photoionization rate $q=Q⟨\sigma ⟩/(4\pi r_{\mathrm{in}}^2)$ per neutral atom. Local photoionization equilibrium is $(1-\zeta )q={\alpha }_B{n}_H{\zeta }^2$, so $\Theta ={\alpha }_B{n}_H/q$ and the physical quadratic equation root is the displayed rationalized expression. If an unextinguished total recombination-line observed radiative flux $F_{\ell }$ has photon yield $y_{\ell }$, then $Q=4\pi D^2{F}_{\ell }{\lambda }_{\ell }/(y_{\ell }hc)$ and $\Theta =y_{\ell }{\alpha }_B{n}_{H}hc(r_{\mathrm{in}}/D{)}^2/(⟨\sigma ⟩F_{\ell }{\lambda }_{\ell })$. The ratio $r_{\mathrm{in}}/D$, rather than a product of lengths, makes $\Theta$ dimensionless.

Every nonconstant entire function takes every complex value with at most one exception. Equivalently, its image is either $\mathbb{C}$ or $\mathbb{C}\setminus \{a\}$ for one complex number $a$. The hypothesis of being nonconstant matters: a constant function has only one value. The Little Picard theorem guarantees existence of preimages, without requiring infinitely many preimages for each attained value. ^[ref]

For a nonconstant polynomial $p$, the fundamental theorem of algebra applied to $p(z)-w$ supplies a preimage for every $w\in \mathbb{C}$. The complex exponential function has precisely one omitted value, zero, as the explicit formula for its preimages shows. In contrast, the complex sine function has no omitted value. To solve $\sin z=w$, set $u=e^{iz}$ and rearrange to the quadratic equationIts roots have product $-1$, so they are nonzero. Every such root has a preimage under the complex exponential function, giving a solution for $z$. Adding $2\pi k$ supplies further solutions by the periodicity of the complex sine.

The force of the Little Picard theorem is that it applies to every nonconstant entire function, without an explicit formula. The Liouville theorem already gives a dense image of a nonconstant entire function, but a dense subset could still leave out many individual points. The Little Picard theorem limits these missing points to one. In particular, establishing two omitted values proves that an entire function is constant.

For a nonconstant meromorphic function on the complex plane, the corresponding bound is two omitted values in the Riemann sphere $\mathbb{C}\cup \{\infty \}$, where poles take the value infinity. ^[ref] For example,omits zero and one, and has poles at $2\pi ik$ for $k\in \mathbb{Z}$. Every other finite value $w$ is attained by solving $e^z=1-1/w$. For an entire function, infinity is already an omitted value in the Riemann sphere, leaving at most one additional finite omitted value.

Proof outline: use the version of Montel's theorem for two fixed finite omitted values. ^[ref] It makes the rescaled holomorphic functions $g_n(\zeta )=f(a+n\zeta )$ a normal family on the unit disc. The fixed value $g_n(0)=f(a)$ gives pointwise anchoring of a holomorphic normal family, hence a subsequence with a finite holomorphic function as its locally uniform convergence limit. The Cauchy integral formula bounds its derivatives at zero, whereas $g_n^{\prime }(0)=n{f}^{\prime }(a)$. Consequently $f^{\prime }(a)=0$; arbitrary $a$ makes $f$ constant.

For the logistic map $F(u)=\lambda u(1-u)$, factor $F(F(u))-u$ and remove its two fixed points. The remaining quadratic equation is ${\lambda }^2{u}^2-\lambda (\lambda +1)u+\lambda +1=0$. Its distinct roots form a periodic orbit of period two when $\lambda >3$. The periodic-orbit multiplier is $F^{\prime }(u_{-})F^{\prime }(u_{+})=4+2\lambda -{\lambda }^2$, so this orbit is asymptotically stable for $3<\lambda <1+\sqrt{6}$. It is born at a period-doubling bifurcation at $\lambda =3$.

For a dry background inventory of column weight $P_d$, a mixed layer of fixed pressure thickness $\Delta p$, and saturated surface vapor partial pressure $E$, put $x=E/P$ and let $ϵ$ be the vapor-to-background molar mass ratio. The vapor mass fraction is $ϵx/[1+(ϵ-1)x]$, so hydrostatic balance and mass conservation give the displayed quadratic equation. Physical solutions require $x\le 1$ and a nonnegative upper-layer pressure. At $x=1$, $P=E=P_d+\Delta p$. A real positive algebraic root beyond that value need not be physically admissible.

For steady uniaxial extension of the Giesekus model, the component-first velocity gradient is diagonal, $G=\mathrm{diag}(\gamma ,-\gamma /2,-\gamma /2)$. For a steady spatially uniform diagonal stress, the material derivative vanishes and the convected terms are $-2\gamma {\sigma }_{11}$ and $+\gamma {\sigma }_{22}$. The off-diagonal equations vanish identically, and symmetry permits equal transverse components. The remaining equations areSet $W=\lambda \gamma \ge 0$, $b=\alpha p_T$ and $s_i={\sigma }_{ii}/p_T$, with $\mu =\lambda p_T$. For $b>0$, the branches continuously connected to zero stress at zero rate areA sufficient nondegenerate range for physical solutions at every nonnegative rate is $0<b<1$. Indeed $M=(1-b)I+bA$ with $A=HC/(k_{B}T)$, so positive conformation implies positive mobility. The longitudinal quadratic equation is nonpositive at $s_1=0$ and has a nonnegative root of a polynomial. The transverse quadratic equation has values $b-1<0$ at $s_2=-1$ and $W\ge 0$ at $s_2=0$, so its larger root of a polynomial lies in $(-1,0]$. Thus the displayed branches satisfy $s_i>-1$, giving $A_i=1+s_i>0$. The other algebraic roots do not provide a physical branch connected to the equilibrium distribution. For constant positive diagonal $A$, the associated stationary configuration distribution is a Gaussian with covariance $C=(k_{B}T/H)A$.

For positive $\alpha$ on the physical branch, the stretching stress remains finite through $W=1/2$. At that rate,At large extension rate,Thus the stress continues to grow, but the uniaxial extensional viscosity ${\eta }_E=({\sigma }_{11}-{\sigma }_{22})/\gamma$ approaches the finite plateau $2\lambda /\alpha$ rather than developing a finite-rate pole. The transverse stress approaches $-p_T$ as its conformation tensor variance becomes small. At small rate the two stresses are $2\mu \gamma$ and $-\mu \gamma$, so ${\eta }_E\to 3\mu$.

When $\alpha =0$, the physical steady branch instead isThe stretching stress diverges as $W\to 1/2$ from below. Above that threshold its algebraic continuation has $1+s_1=1/(1-2W)<0$ and cannot be a conformation covariance. There is no physical finite steady stretching state. The transient equationgives linear growth at $W=1/2$ and exponential growth at larger rates, the Hookean coil-stretch failure of the upper-convected Maxwell limit.

Parameter qualification. Positivity of $\alpha$ alone, without a restriction on $b=\alpha n{k}_{B}T$, does not ensure real physical steady solutions for the entire tensor. For example $b=2$, $W=1$ makes the transverse equation $2s_2^2+2s_2+1=0$, which has no real root. Its diffusion/resistance interpretation is already inadmissible for some conformations. The physically positive-mobility regime stated above avoids that counterexample; the source's positive-$\alpha$ case must not be taken as a guarantee for arbitrary parameters.

Let $q_{\mathrm{in}}$ be the photoionization rate per neutral hydrogen atom just outside the empty cavity. With an isotropic central source and no attenuation inside that cavity,The cross section is zero below threshold; this permits the lower integration limit zero. Pure hydrogen with hydrogen ionization fraction $\zeta$ has $n_{\mathrm{HI}}=n_H(1-\zeta )$ and $n_e=n_p=n_H\zeta$. Local photoionization equilibrium is thereforeDefine the dimensionless quantity ${\Theta }_{\mathrm{phys}}={\alpha }_r{n}_H/q_{\mathrm{in}}$. Solving the resulting quadratic equation, ${\Theta }_{\mathrm{phys}}{\zeta }_{\mathrm{in}}^2+{\zeta }_{\mathrm{in}}-1=0$, gives the physical rootThe rationalized form avoids numerical cancellation for small ${\Theta }_{\mathrm{phys}}$. Strong illumination gives ${\zeta }_{\mathrm{in}}\simeq 1-{\Theta }_{\mathrm{phys}}$; weak illumination gives ${\zeta }_{\mathrm{in}}\simeq {\Theta }_{\mathrm{phys}}^{-1/2}$.

Eliminating $Q$ with the H-alpha luminosity, $Q=2.2(4\pi D^2{F}_{\mathrm{H}\alpha }){\lambda }_{\mathrm{H}\alpha }/(hc)$, yieldsThe printed definition of $\Theta$ is inconsistent with the defined radii. The source calls $r_{\ast }$ a physical stellar radius, but its factor $(2\pi r_{\ast }D{)}^2$ gives $\Theta$ units of length to the fourth power, rather than a dimensionless value. It also omits the physically necessary cavity radius $r_{\mathrm{in}}$. For a concrete counterexample, double $r_{\mathrm{in}}$ while holding the source, $D$, and $n_H$ fixed: the incident photoionization rate falls by four, so ${\Theta }_{\mathrm{phys}}$ rises by four; the printed expression does not change. The root has the requested form with the corrected ${\Theta }_{\mathrm{phys}}$ above. There is no supplied relation allowing $r_{\mathrm{in}}$ to be replaced by $r_{\ast }$.

Assume uncompensated donor impurities, negligible intrinsic electron holes, and a nondegenerate conduction band. Then charge neutrality equates the conduction electron number density to the ionized-donor number density:The low-temperature condition relative to the band gap supports neglect of intrinsic carriers; one also needs $n_i\ll n$ and the Boltzmann approximation for the conduction electrons. Combining part (b)(i) with the effective density of states result givesEliminate the chemical potential by writing $e^{\beta (\mu -{ϵ}_D)}=(n/N_c)e^{\beta ({ϵ}_c-{ϵ}_D)}$. With the donor binding energy $E_i={ϵ}_c-{ϵ}_D>0$,Writing $A(T)=N_c{e}^{-\beta E_i}$, the physical root of the resulting quadratic equation isThis lies between zero and $N_D$. At semiconductor carrier freeze-out, $A\ll N_D$, it becomes $n\simeq \sqrt{N_D{N}_c}{e}^{-E_i/(2k_{B}T)}$; at essentially complete ionization, $A\gg N_D$, it tends to $N_D$. Including donor charge-state degeneracy would replace $A$ by $A{g}_{\mathrm{empty}}/g_{\mathrm{full}}$.

The function $F$ is an odd function, passes through $(0,0)$, and has the horizontal asymptote $F=0$ at both infinities. Its derivative isIt decreases for $|x|>1$ and increases for $|x|<1$, with its lowest value $F(-1)=-1/2$ and highest value $F(1)=1/2$. These features determine the graph of a function shown below.

For $0<ϵ<1/2$, the equation is the quadratic equation $ϵx^2-x+ϵ=0$. The quadratic formula gives exactly two real equation roots:The binomial series gives $\sqrt{1-4{ϵ}^2}=1-2{ϵ}^2-2{ϵ}^4-4{ϵ}^6+O({ϵ}^8)$, hence the two requested asymptotic expansions areTheir product is exactly one. Both are positive, with $x_{-}<1<x_{+}$; there are no negative equation roots because $F(x)<0$ for $x<0$. At $ϵ=1/2$ the two real equation roots coincide at one, and for $ϵ>1/2$ there are none. In the stated limit $ϵ\to 0^{+}$, these two branches are therefore exhaustive.

The radial retraction of a punctured ball gives a retraction: project away from the removed interior point until reaching the unit sphere. Write $a=x_0$, $v=x-a\ne 0$, and setThe equation $|a+tv{|}^2=1$ is the quadratic equationIts constant term is negative, so it has exactly one positive root, namely $\tau (x)$. Since $|a+v|=|x|\le 1$, that positive root satisfies $\tau (x)\ge 1$. Thus $r(x)\in \partial {\mathbb{B}}^n$ for every $x\ne a$ in the closed unit ball.

The displayed expression is a continuous map on the punctured closed unit ball: its denominator is nonzero there, and its square root has strictly positive argument. If $|x|=1$, $t=1$ is the unique positive root, so $r(x)=x$. These are exactly the requirements for a retraction. The missing interior point is essential: the formula need not extend continuously to $a$, in agreement with the no-retraction theorem.