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Completing the square givesThus the circle has

Completing the square in the determinant givesfor every real $\alpha$. A square matrix is invertible exactly when its determinant is nonzero, so $A_{\alpha }$ is nonsingular for every real $\alpha$.

Completing the square transforms the equation intoComparison with the standard circle equation gives

Divide by four and use completing the square:This is the standard equation of a circle, so

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

Completing the square directly with the complex modulus givesFor $|b{|}^2>c$, this is the circle with centre $-b$ and radius $\sqrt{|b{|}^2-c}$. Equality gives just the point $-b$, and $|b{|}^2<c$ gives the empty locus. In particular, the nondegenerate answer is

Consider the minimization problemThe objective function is a strictly convex function with its unconstrained optimization minimum at $x=2$. If $b<2$, it decreases throughout the feasible set until the bound $x=b$ is reached. If $b\ge 2$, the feasible set already contains $2$. HenceWith $g(x)=x-b$, the optimization Lagrangian and KKT conditions areSubstituting $x_{\ast }(b)$ gives the unique inequality Lagrange multiplierFor $b<2$, the active constraint has ${\lambda }_{\ast }>0$. For $b>2$, there is strict slack $b-x_{\ast }=b-2$ and ${\lambda }_{\ast }=0$. At $b=2$, the active constraint has zero Lagrange multiplier: the boundary happens to pass through the unconstrained optimum. Feasibility and complementary slackness must both be checked; solving only $2(x-2)+\lambda =0$ would allow infeasible candidates.

An independent certificate follows by completing the square:The Lagrange dual function is a concave function of $\lambda$. Its maximum over $\lambda \ge 0$ occurs at ${\lambda }_{\ast }(b)$ and equals $V(b)$. Thus every case has an explicit lower bound attained by a feasible point; the KKT conditions provide a certificate that can be checked without searching all real numbers.

Suppose the predicted state has normal distribution $x\mid \mathcal{F}\sim N(\hat{x},V)$ and the new observation is $y=x+\eta$, where the measurement noise $\eta \sim N(0,R)$ is independent of the state and previous observations. Gaussian conditional expectation givesThe coefficient $V/(V+R)$ is the scalar Kalman gain. The posterior mean is a weighted mean of the predicted state and the new observation, with the more precise quantity receiving more weight. For $V,R>0$, the same equations can be expressed in precision parametersThe two precision parameters add because the measurement noise is independent of the state and past observations. Completing the square in the product of the prior density and the likelihood function gives this form directly. This is a Gaussian conditional expectation for a fixed unknown state; a moving state additionally requires a Kalman prediction step.

Let $M=\left(\begin{array}{cc}A& B\\ B^T& D\end{array}\right)$ be a real symmetric matrix with $A$ a positive-definite matrix. Its quadratic form satisfies the completing the square identityFor each fixed vector $y$, the unique minimizing vector is $x=-A^{-1}By$, so the minimized value is $y^{T}Sy$. Also, $M$ is a positive-definite matrix exactly when $S$ is a positive-definite matrix, under the stated assumption on $A$.^[ref]

An original scalar example isHere $A=2$, $B=1$, $D=3$, and the Schur complement is $5/2$. Holding $x=0$ would give $3y^2$, but letting $x$ relax to $-y/2$ reduces the quadratic form to $(5/2)y^2$. At $y=2$ the respective values are $12$ and $10$. Eliminating a variable therefore changes the effective cost of the retained variable.

This interpretation supports quadratic optimization and the positive-definite symmetric matrix tests used in Cholesky decomposition. In a multivariate normal distribution, a covariance matrix also gives a conditional covariance through a Schur complement; the required probabilistic hypotheses are supplied by conditional multivariate normal distribution, rather than by the algebra alone.

For the static energy $E=\int [{\varphi }^{\prime 2}+W^{\prime }(\varphi {)}^2]dx/2$, completing the square givesThe Bogomolny bound is the absolute boundary difference. Equality requires the Bogomolny equation ${\varphi }^{\prime }=\pm W^{\prime }(\varphi )$ with the sign selected by that difference. Differentiating it gives the Euler-Lagrange field equation ${\varphi }^{\prime \prime }=W^{\prime }{W}^{\prime \prime }$. A stationary solution of that second-order equation need not saturate the bound in a general field theory.

Suppose $f,g:X\to \mathbb{R}$ are bounded functions on the same nonempty set, and the supremum norm of their difference is at most $\delta$. Then their optimal values differ by at most the same amount:Indeed, $f(x)\le g(x)+\delta \le {\sup }_{X}g+\delta$ for every $x$, so ${\sup }_{X}f\le {\sup }_{X}g+\delta$; exchange $f$ and $g$ for the other direction. Adding a constant shifts a supremum by that constant, a basic compatibility between upper bounds and addition.^[ref]

For a chosen example on $[-1,1]$, take $f(x)=-x^2$ and $g(x)=-x^2+0.2x$. Their supremum error is $0.2$, since $|g(x)-f(x)|=0.2|x|$. Completing the square givesThus $f$ has maximum zero at $x=0$, while $g$ has maximum $0.01$ at $x=0.1$. The change in optimal value, $0.01$, is smaller than the certified bound $0.2$. For the vertical translation $h(x)=f(x)+0.2$, the bound is exact: its supremum error and optimal-value change are both $0.2$.

A stable optimal value does not guarantee a stable maximizing input. On $[-1,1]$, the two linear functions $f_{\epsilon }(x)=\epsilon x$ and $g_{\epsilon }(x)=-\epsilon x$ satisfyFor every positive $\epsilon$, their unique maximizing inputs remain two units apart, at $1$ and $-1$. Both functions approach the flat zero function, which has many maximizing inputs. Consequently an accurate objective function value is not by itself a certificate that the optimizing decision is accurate.

The estimate also gives a convenient consequence of uniform convergence. If $f_n$ converges uniformly to a bounded function $f$ on a nonempty $X$, then ${\sup }_X{f}_n\to {\sup }_{X}f$ for the bounded tail of the sequence. Uniform control handles the changing input that defines the best value. Pointwise convergence alone supplies no such control; the existing examples under uniform convergence illustrate that distinction.

Consider an objective function that is a quadratic polynomial, with a singular optimization constraint:Its only feasible point is $x_{\ast }=0$, giving $p_{\ast }=9$. For $\lambda \ge 0$, completing the square yieldsEvery finite Lagrange multiplier gives $q(\lambda )<9$, but $q(\lambda )\to 9$ as $\lambda \to \infty$. Thus $d_{\ast }=p_{\ast }=9$: strong duality holds, and the primal optimum is attained, yet dual attainment fails. The inner minimizers $3/(1+\lambda )$ approach $0$ without ever becoming feasible at finite $\lambda$.

Both the objective function and the constraint function are convex functions. However, Slater's condition fails because $x^2<0$ is impossible. At the optimum, stationarity would require $-6+\lambda \cdot 0=0$, which no finite Lagrange multiplier satisfies. Compare KKT regularity at a local minimum: a zero gradient of an active constraint can prevent a multiplier certificate. This example has zero duality gap without the Strong Lagrangian property.

For a real symmetric matrix $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$, completing the square gives, when $a\ne 0$,Thus $A$ is a positive-definite matrix exactly when $a>0$ and $\det A=ac-b^2>0$. For necessity, test $(1,0)$ first, then $(-b/a,1)$; for sufficiency, the displayed two positive squares give the result. These are the first two leading principal minors in Sylvester's criterion. A positive determinant alone is insufficient: $-I_2$ has determinant $1$ but is a negative-definite matrix.

The eigenvalues provide an equivalent test under the same real-symmetric hypothesis:The spectral theorem for real symmetric matrices supplies coordinates in an orthonormal basis in which $q={\lambda }_{+}{u}^2+{\lambda }_{-}{v}^2$. More generally, a complex Hermitian matrix has an orthonormal eigenbasis and $q(z)={\sum }_j{\lambda }_j|w_j{|}^2$, so it is a Hermitian positive-definite matrix exactly when every eigenvalue is positive. This is a short proof outline: test the eigenvectors for necessity, and sum the positive squared absolute values for sufficiency.^[ref]

For a real positive semidefinite matrix, the condition is instead $q(x,y)\ge 0$ for all vectors, equivalent here to $a\ge 0$, $c\ge 0$, and $ac-b^2\ge 0$. If $a>0$, the same completing the square proves this; if $a=0$, nonnegativity forces $b=0$ and leaves $c{y}^2$ with $c\ge 0$. Merely replacing the strict inequalities on the leading principal minors by nonstrict ones fails: $\mathrm{diag}(0,-1)$ has both leading principal minors zero yet gives $q(0,1)=-1$. An indefinite quadratic form takes both positive and negative values; its real symmetric matrix has eigenvalues of both signs.

A translation-invariant connected generating functional with a constant external source is extensive in spacetime volume ${\mathcal{V}}_d$. Its total logarithm is dimensionless, while its density has mass dimension $d$. Consequently its dimensionless representation at fixed physical volume must retain ${\mu }^d{\mathcal{V}}_d$. For a free massive scalar with source coupling ${\mu }^{-ϵ/2}h\varphi$, completing the square gives $F={\mathcal{V}}_d{\mu }^{-ϵ}{h}^2/(2m^2)$. This already disproves a literal rewriting of the full uniform-source logarithm using only $\lambda ,m^2/{\mu }^2,h/{\mu }^3$ at fixed volume. In infinite volume the total functional is generally infinite; use the density, a source of compact support, or an explicit volume regulator. After writing the density as ${\mu }^{d}f$, its renormalization group equation has the engineering term $+df$. Ultraviolet subtraction alone does not establish infrared finiteness for a massless constant source.

Let $A$ be a real positive-definite matrix and minimize the displayed quadratic function over all of ${\mathbb{R}}^n$. Its gradient is $Ax-b$, so its only stationary point solves the linear system $A{x}_{\ast }=b$. The matrix is an invertible matrix, making this solution exist and be unique. Completing the square around that solution proves the stronger global statement:The quadratic function is strictly convex, and its level sets above the minimum are translated ellipsoids. Positivity therefore turns solving a linear system into locating the bottom of a bowl.^[ref]

For the opening matrix $A=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)$, take $b=(1,1{)}^T$ and $c=0$. Then $A^{-1}=\frac{1}{8}\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)$, $x_{\ast }=(1/4,1/4{)}^T$, and $f(x_{\ast })=-1/4$. Every displacement $h$ raises the value by $\frac{1}{2}{h}^{T}Ah$; along the unit eigenvectors $e_{+}$ and $e_{-}$ the increases are respectively $2s^2$ and $s^2$ for displacement $s{e}_{+}$ or $s{e}_{-}$.

For a singular positive semidefinite matrix, a solution of $Ax=b$ is still a global minimizer, but adding any kernel vector gives another. If $b$ is outside the image of a linear map defined by $A$, some kernel vector $v$ has $b^{T}v\ne 0$, and $f(sv)=c-s{b}^{T}v$ is unbounded below. For an indefinite quadratic form, a negative eigenvalue gives a direction along which the quadratic term tends to $-\infty$ regardless of $b$. Thus a stationary point alone does not establish a minimum. The same energy-difference identity extends to the quadratic variational principle for a symmetric positive operator, with additional existence hypotheses in infinite dimension.

Completing the square in $S[q]=\frac{1}{2}qKq+Jq$ with $K=-{\partial }_t^2-m^2+iϵ$ gives the displayed source factor, where $G=K^{-1}$. In the frequency convention $e^{i\omega t}$, $\tilde{G}=({\omega }^2-m^2+iϵ{)}^{-1}$ and $G(t)=-i{e}^{-im|t|}/(2m)$ for $m>0$. The physical time-ordered oscillator quantum field theory propagator is $iG$. Sources are first taken as test functions so the pairings are defined.