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The roots of $a{x}^2+bx+c=0$, where $a\ne 0$, are

For symmetric $A$, the quadratic form $x^{T}Ax$ has gradient $2Ax$.

A vector $v$ is isotropic for a quadratic form $q$ when $q(v)=0$.

A subspace $E$ is totally isotropic for $B$ when $B(u,v)=0$ for all $u,v\in E$. For a real nondegenerate bilinear form of inertia $(p,q)$, its dimension is at most $\min (p,q)$.

Every real quadratic form is congruent to a diagonal form with entries $+1,-1,0$, and the numbers of entries of each kind are invariant.

A real symmetric matrix $A$ is positive definite whenfor every nonzero real vector $x$. It therefore defines the inner product $⟨x,y{⟩}_A=x^{T}Ay$ and is invertible.

A real symmetric or complex Hermitian matrix $A$ is positive semidefinite when $x^{†}Ax\ge 0$ for every vector $x$, equivalently when all its eigenvalues are nonnegative.

A symmetric bilinear form is positive semidefinite when $B(v,v)\ge 0$ for every $v$, and positive definite when equality is possible only for $v=0$.

Over a field of characteristic other than two, a quadratic form recovers its symmetric bilinear form by

The signature of a real quadratic form is the number of positive squares minus the number of negative squares in its diagonal normal form.

The rank of a quadratic form is the rank of its associated symmetric bilinear form.

The factorizationshows that $X^3=-1$ always has $X=-1$ as a solution. For $p\ne 3$, it has two further solutions exactly when $-3$ is a quadratic residue; for $p=3$, all three factors coalesce at the unique root $-1$.

If $M/L$ is purely inseparable, an embedding of $L$ into an algebraic closure has at most one extension to $M$: if $x^{p^n}\in L$, the image of $x$ must be the unique $p^n$th root of the image of $x^{p^n}$. Existence in a normal overfield therefore implies uniqueness.

An algebraic element $x$ is purely inseparable over $K$ when $x^{p^n}\in K$ for some $n\ge 0$.

For a word consisting of three long constrained blocks, a pumpable window shorter than each block cannot alter both outer blocks; pumping then breaks a required three-way count relation.

A set of propositions has a model exactly when every finite subset has a model. Syntactically, a contradiction uses only finitely many assumptions, and every consistent set extends by Zorn lemma to a maximal consistent set in propositional logic.

A consistent set $T$ is maximal consistent when no proper extension is consistent. It is deductively closed and decides every formula: exactly one of $t$ and $\neg t$ belongs to $T$. Therefore $v(t)=1$ exactly when $t\in T$ defines a model of $T$.

Ifare deductively closed, then each $T_i$ is consistent and their union is consistent and deductively closed. The union is not finitary. Each conclusion follows from the finite character of formal proofs.

A set $S$ is finitary when it is deductively equivalent to some finite set $F$: both $S⊢F$ and $F⊢S$.

A set $S$ is syntactically consistent when $S/⊢\perp$. Equivalently in classical logic, there is no formula $\phi$ for which both $S⊢\phi$ and $S⊢\neg \phi$.