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Rearrangement gives the quadratic inequality $x^2-3x+1\le 0$. The associated quadratic equation has roots $(3\pm \sqrt{5})/2$. Since the coefficient of $x^2$ is positive, the quadratic inequality holds between these roots, including the endpoints:

For the given differential form, set $P=x(y^4+1)$ and $Q=y(2x^2{y}^2+1)$. Their partial derivatives satisfyIn fact an explicit potential issince $F_x=P$ and $F_y=Q$. Thus the differential form is an exact differential form on all of ${\mathbb{R}}^2$.

Replacing $y^2$ by $u$ suggests the potentialDirect differentiation verifies that $dH/dx$ is exactly the left-hand side of the required ordinary differential equation. This direct verification remains valid even for negative $u$; it does not require a real square root $y=\sqrt{u}$. ThereforeThe initial condition gives $C=1$. Solving this quadratic equation in $u$ and selecting the branch that has $u(1)=0$ givesThe second expression also defines the removable value $u(0)=1$. The maximal open interval containing $1$ on which this is a differentiable real solution isAt either endpoint the square root vanishes and $2x^{2}u+1=0$, while $x(u^2+1)\ne 0$, so the original ordinary differential equation cannot hold there with a finite derivative.

Set $t=2^x>0$. The exponential function identity $4^x=(2^x{)}^2$ reduces the problem to the quadratic equationPositivity excludes $t=-1$, so $2^x=2$. Since the exponential function $2^x$ is strictly increasing, the unique answer is $\boxed{x=1}$.

Since $z\bar{z}=x^2+y^2$, the real quadratic equation isCompleting the square makes its geometry explicit:

A finite straightedge and compass construction repeatedly draws the line through two available distinct points or the circle with an available center and an available distance as radius, and adjoins isolated intersections of two such lines or circles. A finite set is constructible when its points occur after finitely many such steps, starting with the specified points.

If the current coordinate field is $F$, intersections of two lines solve linear equations over $F$. A line and a circle lead to a quadratic equation over $F$. For two distinct circles, subtraction of their equations gives a line, reducing to the previous case. Thus each step adjoins coordinates in an extension of degree one or two. The tower law gives a finite tower of quadratic-or-trivial field extensions containing all output coordinates. If the final coordinate field contains $K_0$, thenbecause its degree divides the power-of-two degree of that tower.

Two implicit qualifications are necessary for this degree statement. The output is obtained in finitely many steps; the aggregate of infinitely many individually constructible points need not generate a finite extension. Also, unless the output retains the original coordinates, $K$ need not contain $K_0$. In that case the degree statement applies to the compositum $K_{0}K$, rather than an undefined $[K:K_0]$. The heptagon application explicitly includes $S_0$, so neither issue obstructs it.

Here $K_0=\mathbb{Q}$, and a vertex adjacent to $(1,0)$ has first coordinate $c=\cos (2\pi /7)$, so $\mathbb{Q}(c)\subseteq K$. Let $\zeta =e^{2\pi i/7}$. The cyclotomic polynomial $1+X+\cdots +X^6$ is irreducible, giving $[\mathbb{Q}(\zeta ):\mathbb{Q}]=6$. Since $\zeta +{\zeta }^{-1}=2c$, it satisfies $X^2-2cX+1$ over $\mathbb{Q}(c)$. The element $\zeta$ is nonreal while $\mathbb{Q}(c)\subset \mathbb{R}$, so this relative degree is exactly two. By the tower law,If the regular regular heptagon were constructible, this degree would divide $[K:\mathbb{Q}]=2^r$, impossible. Therefore the regular heptagon is not constructible from the given unit segment.

The Backward Euler method uses the new point:Writing $a=h(x_n+h)>0$ gives the quadratic equation $a{y}_{n+1}^2+y_{n+1}-y_n=0$. Starting from $y_n>0$, its two polynomial roots have opposite signs. The positive root, which continues the solution as $h\to 0$, isThe other root is negative and does not approach $y_n$ as $h\to 0$. Rationalizing the numerator gives an equivalent form that avoids cancellation error when the step size is small:The defining quadratic equation gives $0<y_{n+1}<y_n$ for this root, so positivity is maintained inductively from $y_0=1$. The branch choice is part of specifying the Backward Euler method here: the algebraic update alone has two solutions.

Choose the Fourier series conventionFor a real potential energy, $V_{-1}=V_1^{\ast }$. The symbol $V(G_1)$ means the Fourier coefficient $V_1$, with units of energy, rather than the pointwise value of $V(x)$ at a position $G_1$.

Retain the two plane waves $|1⟩=L^{-1/2}{e}^{ikx}$ and $|2⟩=L^{-1/2}{e}^{i(k-G_1)x}$ in the nearly-free electron model. Their kinetic energies are ${ϵ}_1={\hslash }^2{k}^2/(2m)$ and ${ϵ}_2={\hslash }^2(k-G_1{)}^2/(2m)$. The matrix representation of a linear map for the Quantum Hamiltonian projected onto this two-dimensional basis isThus $\det (H-EI)=0$ givesSolving this quadratic equation yieldsChoosing the zero of potential energy so that its spatial average is $V_0=0$ reproduces the requested expression. If that average has not been removed, both energy bands are shifted by $V_0$; no band gap, group velocity or effective mass is changed by this constant shift.

The two kinetic energies coincide at $k=G_1/2$. Their coupling gives an avoided crossing with band gap $2|V_1|$. The result is exact within the projected two-wave Quantum Hamiltonian. It is an approximation to the full crystal problem, appropriate near this crossing for a weak periodic potential whose other Fourier coefficients do not cause strong mixing with remote plane waves. Small $|V_1|$ alone would not justify ignoring arbitrarily large other Fourier coefficients.

Notation. The source uses $p_s$ for both the hypothetical dry surface pressure and the actual surface pressure after evaporation. Distinguish them as $P_d$ and $P$, respectively, and interpret the specified pressure thickness $\Delta p$ as fixed. This conserves the original hydrogen inventory and permits $P$ to vary. Reading both occurrences as one fixed number would be inconsistent with adding water vapor to a closed atmospheric column mass budget.

Let $E(T)$ denote the saturation pressure of water, let $x$ be the mole fraction of water vapor in the mixed layer, and let $M_d,M_v$ be the molar masses of hydrogen and water. Surface phase equilibrium and the ideal gas law giveBecause the layer has constant temperature, its relative humidity isThus every level strictly above the surface is subsaturated: hydrostatic balance makes $p<P$, whereas the saturation pressure stays constant. The result does not require a dilute water vapor approximation.

The Clausius-Clapeyron relation, neglecting the specific volume of liquid water and treating the water vapor as an ideal gas, givesHere $T_t,E_t$ are the triple point values, $L_v$ is the specific latent heat, and $R$ is the molar gas constant. All temperatures in these expressions are absolute. As throughout the question, this is a constant-latent heat, ideal gas model, not an extrapolation valid through the real critical point of water.

For a well-mixed ideal gas mixture, its mean molar mass and water vapor mass fraction areBy hydrostatic balance, the mixed layer has total mass per unit area $\Delta p/g$, of which $w_v\Delta p/g$ is water vapor. The remaining atmosphere contains no water vapor. Consequently conservation of hydrogen mass givesEquivalently the requested quadratic equation isFor each $T_s$, calculate $E$, take the positive root of this quadratic equation, and retain it only if $0\le x<1$ and $P>\Delta p$. The last condition leaves a nonempty dry layer. In particular the dry constituent's surface partial pressure is not generally $P_d$: $P_d$ measures its column weight, and the hydrogen and water vapor have different molar masses.

A convenient parameter is the water vapor mole fraction itself. The mass conservation calculation yieldsBoth functions increase for $0\le x\le 1$. At the endpoint,Therefore the physically admissible hydrogen–water vapor mixture requiresBeyond this bound, the formal positive quadratic equation root requires $x>1$, equivalently a negative hydrogen partial pressure. It is the physical constraints, rather than loss of real roots, that end the solution branch. For a concrete example, put $P_d=\Delta p=1$ in consistent pressure units, $ϵ=9$, and $E=3$. The positive root of $P^2+23P-51=0$ is about $2.037$, giving $x\simeq 1.473$, which is impossible.

The limiting temperature is obtained from $E(T_{\ast })=P_d+\Delta p$:This finite limit exists above $T_t$ if $E_t<P_d+\Delta p<E_t\exp [L_v/(R_v{T}_t)]$. The upper qualification matters mathematically: the stipulated constant-latent heat formula has a finite limiting saturation pressure as $T\to \infty$, so for an arbitrarily large unspecified $P_d+\Delta p$ the printed assertion of a finite breakdown temperature is not unconditional. For a concrete formal counterexample, choose $P_d+\Delta p=10^{15}\mathrm{Pa}$: with the usual triple-point constants the model has $E(\infty )\simeq 2.5\times 10^{11}\mathrm{Pa}$, so it never reaches the proposed inventory limit. This is a counterexample inside the stipulated equations, not a physically credible high-pressure ideal gas ocean. Actual liquid water also ceases to exist above its critical point, before such an extrapolation would make sense.

Physically, the imposed finite-pressure-thickness layer can hold no more than $\Delta p/g$ of water vapor per unit area. As $x\to 1$, it expels all its hydrogen into the dry upper layer. A still larger surface saturation pressure cannot coexist with this geometry. The layer must deepen, the temperature profile must change, or the assumed phase equilibrium must fail. If $P_d<\Delta p$, the additional restriction $P>\Delta p$ can also exclude the low-temperature end of the branch.

Use two plane waves, $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$, as a basis near their Bragg degeneracy. The potential has Fourier coefficients $V_{\pm \mathbf{G}}$; its zero Fourier coefficient vanishes. All other plane waves are off resonance and are neglected in the leading nearly-free electron model. With ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$, the Schrödinger equation reduces to the Hermitian matrix problemSetting the characteristic polynomial equal to zero givesSolving this quadratic equation givesFor the real cosine potential its coefficient $V_G$ is real; the absolute-value form also allows an arbitrary choice of origin and corresponding Fourier phase.

At the Brillouin-zone boundary, $\mathbf{k}\cdot \mathbf{G}=G^2/2$, the free energies are equal. The two standing waves have different probability densities relative to the lattice potential and are split byRepeating this avoided crossing at the reciprocal lattice boundaries joins smooth branches into energy bands. In one dimension a nonzero Fourier coefficient opens a gap at the corresponding crossing. In higher dimensions a direct gap at one wavevector need not give a common forbidden energy interval at every wavevector: different bands may still overlap in energy. The present two-dimensional example remains metallic even though the crossings are locally gapped.

For the rectangular lattice, choose real-space primitive vectors $a\hat{\mathbf{x}}$ and $b\hat{\mathbf{y}}$. The reciprocal lattice vectors areDefine the half-widths of the first Brillouin zone by $K=\pi /a$ and $L=\pi /b$. Their values and the zone areA primitive reciprocal cell can be drawn with a corner at the origin; its centred translate is the first Brillouin zone here. Its area is $4{\pi }^2/(ab)=37.9600\mathrm{n}{\mathrm{m}}^{-2}$.

Including both spin states, counting the occupied free-electron states per unit area givesConsequentlyThe occupied disk has precisely the area of the first Brillouin zone, but their shapes differ: $K<k_F<L$. It protrudes through both vertical zone boundaries and does not reach the horizontal ones. Thus two electrons per cell do not imply a filled first band in the absence of a separating band gap.

Fold the protruding caps by $\pm {\mathbf{G}}_x$. The original circle and the translated circle intersect at $k_x=K$, $k_y=\pm \sqrt{k_F^2-K^2}$. The potential mixes these states and replaces each crossing by an avoided crossing. The lower branch reconnects into two open sheets running across successive zones in the $k_x$ direction; the upper branch forms a closed lens-shaped pocket about a vertical zone boundary. In a reduced-zone sketch its two halves sit at opposite edges, but periodic identification makes them one closed sheet. This is rectangular-lattice Fermi-surface reconstruction.

The largest pocket in this weak-potential construction is the lens-shaped intersection of two disks enclosed by the intersecting free circles. Each has radius $k_F$ and their centres are separated by $2K$. If $\theta =\arccos (K/k_F)$, two sectors minus their two triangles give the lens bound on a reconstructed Fermi pocket:Opening a gap raises the upper branch at fixed chemical potential and shrinks its pocket. The reported quantum oscillation area satisfiesIt is therefore consistent with this closed sheet. Onsager quantization gives $F=\hslash A_k/(2\pi e)\simeq 601\mathrm{T}$. An orbit on this closed contour can return to itself under the Lorentz force and occupy quantized Landau levels. The open sheets do not enclose a finite orbit area for a perpendicular field in this two-dimensional model. The observation is consistent with the reconstructed pocket, without uniquely determining a completely arbitrary band structure.

For a quantitative rough coupling estimate, use the supplied fixed-chemical-potential approximation and retain the two waves coupled by ${\mathbf{G}}_x$. WriteTheir upper eigenvalue isSetting it equal to $C{k}_F^2$ gives the pocket boundaryIts intersection with $k_y=0$ occurs at $q=\pm q_m$, whereThe two-wave estimate of a Bragg-gap coupling from a pocket area therefore usesAt $v=0$ this reproduces the lens bound on a reconstructed Fermi pocket; the pocket disappears at $v=C(k_F^2-K^2)=0.23786\mathrm{eV}$. Numerical integration gives $A(0.075\mathrm{eV})=6.22\mathrm{n}{\mathrm{m}}^{-2}$ and $A(0.100\mathrm{eV})=5.54\mathrm{n}{\mathrm{m}}^{-2}$. Matching the measured area gives $v\simeq 0.0931\mathrm{eV}$, so the justified rough result isThis estimate neglects remote plane-wave admixtures and other Fourier coefficients. It also holds the chemical potential fixed as requested; an exact fixed-density calculation would determine it self-consistently after reconstruction. The numerical root is a way to implement the approximate model, not a measurement of $V_G$ to four significant figures.

With a fixed budget of $n$ independent proposal trials, the accepted count $K_n$ has binomial distribution with success probability $1/M$. HenceFor $M>1$, the central limit theorem gives the one-sided approximationUse $z=2.33$, since now only the lower tail of the accepted count matters. Requiring $n-MN\ge z\sqrt{n(M-1)}$ and solving the resulting quadratic equation in $t=\sqrt{n}$ gives the one-sided normal approximation for a rejection proposal budgetEquivalently, before rounding, this is $MN+z^2(M-1)/2+(z/2)\sqrt{(M-1)[z^2(M-1)+4MN]}$. For the value in (a)(iii),When $M=1$, every proposal is accepted and $\boxed{n_0=N=66,564}$ exactly. Otherwise the positive safety margin above $MN$ allows for random fluctuations in the number accepted. The average cost in (b)(ii) describes stopping at the $N$th acceptance; this answer chooses a deterministic budget that reaches that acceptance with approximately $99\%$ probability.

The approximation requires a sufficiently accurate normal distribution approximation to the binomial distribution. An exact $99\%$ budget can instead be found from its tail probabilities. Also, a $99\%$ chance of enough accepted draws and approximate $99\%$ precision conditional on obtaining them are separate claims: their intersection is not automatically a $99\%$ event.

Put $A={\sum }_i{x}_i$, $B={\sum }_i{x}_i^2$ and $M_2=B/n$. The normal distribution model givesand, up to a constant independent of $\mu$, its log-likelihood function isThe score function and its zero equation areFor $B>0$, this quadratic equation has exactly one positive root, giving the maximum-likelihood estimatorThe numerator $B-A\mu -n{\mu }^2$ of the score function is positive between its negative and positive roots and negative above its positive root. Thus the likelihood function increases up to $\hat{\mu }$ and decreases afterwards: this is its unique global maximum on $(0,\infty )$. Also, $\ell (\mu )\to -\infty$ at both ends. The all-zero observation vector has probability zero; on that exceptional vector $L(\mu )\propto {\mu }^{-n}$ and there is no attained maximum.

The second derivative and the relevant moments areTherefore the expected Fisher information isThe support of a probability distribution is the whole real line, independent of the positive statistical parameter. The law of large numbers gives $\overline{X}\to \mu$ and $M_2\to 2{\mu }^2$ in probability, so the displayed maximum-likelihood estimator is a consistent estimator. The regular asymptotic normality of a maximum likelihood estimator now yieldsThus the requested leading asymptotic variance on the unscaled estimator's scale is ${\mu }^2/(3n)$. The normal mean proportional to standard deviation model provides a reusable version of this calculation.

The algebraic identitysuggests $f(t)=t^3-3t$ for every $t>0$, and substitution verifies the functional equation.

In fact $x+1/x\ge 2$, since $(x-1{)}^2/x\ge 0$. Conversely, for each $t\ge 2$, the quadratic equation $x^2-tx+1=0$ has a positive root $(t+\sqrt{t^2-4})/2$. Thus the functional equation forces $f(t)=t^3-3t$ exactly on $[2,\infty )$; its values on $(0,2)$ are arbitrary. The displayed function is one convenient choice on the full required domain.

The $n$th pentagonal number is $P_n=n(3n-1)/2$ for a positive integer $n$, giving $1,5,12,22,\dots$. A positive integer $x$ is pentagonal exactly when $24x+1$ is a perfect square and $(1+\sqrt{24x+1})/6$ is an integer, by the defining quadratic equation.

For a quadratic equation $a{x}^2+bx+c=0$ with $a\ne 0$, completing the square givesOver the complex numbers both roots exist, with a repeated root of a polynomial when the discriminant $b^2-4ac$ is zero. Over the real numbers the discriminant determines whether there are two distinct roots, one repeated root, or no real root.

A quadratic polynomial in one variable over a field has polynomial degree two and the displayed form. Its roots of a polynomial satisfy a quadratic equation. The polynomial and its associated polynomial function are distinct objects: over a finite field, different polynomials can define the same function.

Removing any interior point $a$ from the Euclidean closed unit ball allows a retraction onto its unit sphere. For $v=x-a\ne 0$, the positive root of $|a+tv{|}^2=1$ isThen $r(x)=a+\tau (x)v$ is a continuous map on the punctured closed unit ball, has norm one, and fixes its boundary because $\tau (x)=1$ there. The quadratic equation has one positive and one negative root since $|a|<1$, so the chosen root is unambiguous.

A solubility product constrains a product of thermodynamic activities; the available inventories determine which combinations of concentrations are possible. Their intersection determines the molar solubility. Consider an original, abstract AB solid, with the preceding one-to-one dissolution chemical equation. Stipulate $c^{\circ }=1\mathrm{M}$, ${\gamma }_{\mathrm{A}}={\gamma }_{\mathrm{B}}=1$ and $K_{\mathrm{sp},c}=10^{-4}{\mathrm{M}}^2$ at fixed temperature and pressure. Thus $K_{\mathrm{sp}}=10^{-4}$ is dimensionless. These are chosen model parameters, not data for an identified material.

Initially, a completely dissociated soluble XB salt supplies B anions at molar concentration $c_0\ge 0$ and X cations at the same molar concentration; A is absent. All three ions have unit electric charge magnitude. X remains a spectator ion, and only AB can form a solid. Assume constant volume, no additional chemical reactions, and negligible contributions from self-ionization of water. Dissolving AB raises both reacting-ion concentrations by the same $s$:Charge neutrality is satisfied: $c_{\mathrm{A}}+c_{\mathrm{X}}-c_{\mathrm{B}}=s+c_0-(s+c_0)=0$. Adding B alone without its accompanying cation would omit the initial electric charge balance.

In the plane of the two reacting-ion concentrations, this inventory constraint is the line $c_{\mathrm{B}}=c_{\mathrm{A}}+c_0$. Saturation supplies the hyperbola $c_{\mathrm{B}}=K_{\mathrm{sp},c}/c_{\mathrm{A}}$ for $c_{\mathrm{A}}>0$. Their positive intersection solves a quadratic equation:The quadratic formula supplies two roots of a polynomial; only the positive one is a physical concentration. Uniqueness also follows because $s(s+c_0)$ strictly increases for $s\ge 0$, starting at zero and growing without bound. The second expression, obtained by multiplying by the conjugate, avoids subtraction of nearly equal numbers at large $c_0$.

For pure water, $c_0=0$ gives $s_{\ast }=\sqrt{K_{\mathrm{sp},c}}=0.010\mathrm{M}$. For $c_0=0.030\mathrm{M}$,Their product is $10^{-4}{\mathrm{M}}^2$, although neither concentration equals the pure-water value. This is the useful geometrical distinction: an inventory line and an equilibrium constant specify different constraints.

Check the solid inventory before imposing saturation. If the initially added AB has amount of substance $n_0$ in volume $V$, the saturated branch requires $n_0>V{s}_{\ast }$ for excess solid to remain. For example, choose $n_0=0.0020\mathrm{mol}$ and $V=0.100\mathrm{L}$. In pure water, $0.0010\mathrm{mol}$ dissolves and the same amount of substance remains as solid. At $c_0=0.030\mathrm{M}$, approximately $0.0003027756\mathrm{mol}$ dissolves, leaving $0.0016972244\mathrm{mol}$ as solid. In this model, an insufficient supply instead givesIf $n_0/V<s_{\ast }$, all AB dissolves and the ion product is below the solubility product. At $n_0/V=s_{\ast }$, the last solid disappears exactly at saturation.

Consider an original ideal model: $V_A=10.00\mathrm{mL}$ of hydrochloric acid at $c_A=0.100\mathrm{mol}{\mathrm{L}}^{-1}$, titrated with sodium hydroxide at $c_B=0.100\mathrm{mol}{\mathrm{L}}^{-1}$. Treat these numbers as exact model inputs. Assume complete dissociation of an ionic compound, complete strong acid ionization, additive volumes, uniform mixing, and activity coefficients equal to one. Use stipulated $K_w=10^{-14}$, approximately appropriate near $25{}^{\circ }\mathrm{C}$. These are calculated pH values, not measurements of these materials.

The chemical equation for neutralization isInitial acid inventory is $n_A=1.000\mathrm{mmol}$, so the equivalence point is $v=10.00\mathrm{mL}$. Added water does not change this inventory. But the final volume $V_A+v$, including the titrant, determines all concentrations.

Keep self-ionization of water even at the equivalence point. With $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, define the dimensionless acid excess $d$, hydronium level $h$, and hydroxide level $o$ byThe sodium cation and chloride anion inventories fix the electric charge balance $h-o=d$, while chemical equilibrium gives $ho=K_w$. Eliminating $o$ yields a quadratic equation $h^2-dh-K_w=0$ with exactly one positive root. For stable calculation on either side, writeThe second expression follows by multiplying by $s-d$; it avoids subtracting two nearly equal large quantities on the alkaline side. A logarithm acts on the dimensionless $h$, as required by the pH definition.

At $v=5.00\mathrm{mL}$, the excess is $0.500\mathrm{mmol}$ in $15.00\mathrm{mL}$, giving $d=1/30$ and pH $1.477121\dots$. At $v=15.00\mathrm{mL}$, excess hydroxide is $0.500\mathrm{mmol}$ in $25.00\mathrm{mL}$, giving $d=-1/50$ and pH $12.301030\dots$. Neglecting dilution would give the wrong values even far from the equivalence point.

Exactly at $v=10.00\mathrm{mL}$, $d=0$ and $h=o=\sqrt{K_w}=10^{-7}$, giving pH $7$ in this model. At $9.999\mathrm{mL}$ and $10.001\mathrm{mL}$, the same smooth expression gives pH $5.300835\dots$ and $8.699121\dots$. The familiar excess-only formulas would predict zero hydronium or zero hydroxide at the equivalence point; retaining self-ionization of water removes that failure. A steep curve is still continuous.

The sensitivity is especially revealing. Differentiating $h-K_w/h=d$ giveswhen $c_A=c_B$. At this model's equivalence point, the derivative is about $1.086\times 10^4$ pH units per millilitre. This is a local derivative, not a linear prediction over a whole drop. Real endpoint accuracy also depends on dosing, mixing, measurement uncertainty, temperature, extra reacting chemical species, and nonideal thermodynamic activities.

Choose a signed angular coordinate along the source-lens line. The point-mass gravitational lens becomesFor $\beta \ne 0$, multiplication by $\theta$ gives the quadratic equation ${\theta }^2-\beta \theta -{\theta }_E^2=0$. Its positive discriminant yields the two displayed roots. Their product is ${\theta }_{+}{\theta }_{-}=-{\theta }_E^2$, so they lie on opposite sides of the lens. With $\beta >0$, ${\theta }_{+}>{\theta }_E$ and $-{\theta }_E<{\theta }_{-}<0$: the outer image lies on the source side. For $\beta <0$, the outer image is instead the minus root. These are image directions seen by the observer, not two emitting objects.^[ref]

For a dimensionless example, set $u=\beta /{\theta }_E=0.6$ and $x=\theta /{\theta }_E$. Solving $u=x-1/x$ gives $x_{+}\simeq 1.344$ and $x_{-}\simeq -0.744$, with separation $\sqrt{u^2+4}\simeq 2.088$. The lensing magnification follows from image parity in gravitational lensing: ${\mu }_{+}\simeq 1.442$, ${\mu }_{-}\simeq -0.442$, and total amplification $\simeq 1.884$. Changing $M$ or the distances rescales the angles through the Einstein angular radius, while these dimensionless values remain the same.

For $u\gg 1$, the quadratic equation gives $x_{+}\simeq u+1/u$ and $x_{-}\simeq -1/u$. Using $\mu =1/(1-x^{-4})$, the outer image tends to unit lensing magnification and the inner image has $|{\mu }_{-}|\simeq u^{-4}$. Two mathematical images can therefore look like one at finite sensitivity or angular resolution. Furthermore, the inner impact parameter eventually becomes too small for the stated weak-field approximation or passes through the opaque lens: the ideal formula must then be replaced by an appropriate physical model.