Factorizatio polynomii qvadratici et trinomii gradvs 2n

Svmmarivm

Haec lectio, qvadratvm complendo, formvlam radicvm polynomii qvadratici dedvcit, discriminans et relationes inter radices et coefficientes perscrvtatvr, et factorizationem $a(x - x_1)(x - x_2)$ obtinet. Deinde methodvm ad trinomia gradvs $2n$ extendit, formae $a x^{2n} + b x^n + c$ , per mvtationem variabilis $u = x^n$ , et ostendit trinomivm gradvs $4$ in factores resolvi posse etiam si discriminans negativvm sit.

Proposita discendi

Formvlam qvadraticam qvadrato complendo dedvcere eamqve ad factores inveniendos adhibere.
Discriminantem interpretari et relationibvs inter radices et coefficientes vti.
Trinomia gradvs $2n$ per mvtationem variabilis $u = x^n$ et prodvcta notabilia in factores resolvere.
Agnoscere qvando polynomivm qvadraticvm in realibvs irredvcibile sit, et trinomivm gradvs $4$ cvivs discriminans negativvm est in factores resolvere.
Problemata applicata constitvere qvae ad aeqvationem qvadraticam dvcvnt, vt volatvm pilae vel designationem horti, eivsqve radices et discriminans in contextv interpretari.

Polynomivm qvadraticvm

Polynomivm qvadraticvm est polynomivm gradvs $2$ ; coefficientibvs realibvs $a$ , $b$ et $c$ , et cvm $a \neq 0$ , sic scribitvr:

P(x) = ax^2 + bx + c

Id in factores resolvere est id vt prodvctvm dvorvm polynomiorvm gradvs $1$ scribere. Per theorema factoris, in lectione de algebra polynomiorvm tractatvm, qvaeqve radix realis $r$ factorem $x - r$ affert; itaqve polynomivm qvadraticvm in factores resolvere idem est ac radices eivs invenire. Methodvs ad eas inveniendas est qvadratvm complere.

TheoremaFormvla qvadratica

Sit $ax^2 + bx + c$ polynomivm qvadraticvm, cvm $a \neq 0$ . Pro omni reali $x$ valet aeqvalitas seqvens, qvae forma completa vocatvr; et si $b^2 - 4ac \geq 0$ , radices polynomii svnt exacte nvmeri $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .

ax^2 + bx + c = a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2}\right]

Idea geometrica est: $x^2 + \frac{b}{a}x$ est area qvadrati lateris $x$ addita area rectangvli latervm $x$ et $\frac{b}{a}$ . Si rectangvlvm in dvo dimidia secatvr et haec ad dvo latera qvadrati applicantvr, solvs angvlvs deest ad qvadratvm maivs complendvm, lateris $x + \frac{b}{2a}$ .

Demonstratio

$ax^2 + bx + c = a\left(x^2 + \dato{s}{\frac{b}{a}x} + \frac{c}{a}\right)$
Cvm $a \neq 0$ , $a$ vt factorem commvnem extrahimvs.
$x^2 + \resaltar{\frac{b}{a}x} = \dato{c}{\left(x + \frac{b}{2a}\right)^2} - \resaltar{\frac{b^2}{4a^2}}$
Qvadratvm ipsivs $x + \frac{b}{2a}$ continet $x^2$ , terminvm $\frac{b}{a}x$ et angvlvm $\frac{b^2}{4a^2}$ ; angvlo detracto svmma qvaesita obtinetvr.
$ax^2 + bx + c = a\left[\left(x + \frac{b}{2a}\right)^2 - \dato{d}{\frac{b^2 - 4ac}{4a^2}}\right]$
In parenthesi svbstitvimvs et terminos constantes in vnam fractionem colligimvs, qvia $\frac{c}{a} = \frac{4ac}{4a^2}$ .
$ax^2 + bx + c = 0 \leftrightarrow \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$
Cvm $a \neq 0$ , polynomivm evanescit exacte cvm vncvs evanescit.
$\left(x + \frac{b}{2a}\right)^2 - \resaltar{\left(\frac{\sqrt{b^2 - 4ac}}{2a}\right)^2} = 0 \leftrightarrow x + \frac{b}{2a} = \resaltar{\pm \frac{\sqrt{b^2 - 4ac}}{2a}}$
Si $b^2 - 4ac \geq 0$ , nvmervs $w = \frac{\sqrt{b^2 - 4ac}}{2a}$ aeqvationi $w^2 = \frac{b^2 - 4ac}{4a^2}$ satisfacit, qvodcvmqve signvm ipsivs $a$ sit. Si $u$ vocamvs expressionem $x + \frac{b}{2a}$ , per differentiam qvadratorvm aeqvatio est $(u - w)(u + w) = 0$ ; et prodvctvm realivm nvllvm est solvm si aliqvis factor nvllvs est, per lemma de signo prodvcti ex lectione de fvnctionibvs algebraicis. Vnde aeqvatio exacte aeqvivalet ad $u = w$ vel $u = -w$ : nvlla radix amittitvr neqve additvr.
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Ipsvm $x$ expedimvs: haec est formvla qvadratica.