Question

(z-a)^2 (z-b)^2=(a+b)^2

Answer 1

Respuesta:

$a=\frac{z^3-2bz^2+b^2z+b+b^2-z^2}{z^2-2bz+b^2-1}$

Explicación paso a paso:

$\left(z-a\right)^2\left(z-b\right)^2=\left(a+b\right)^2$

$\mathrm{Desarrollar\:}\left(z-a\right)^2\left(z-b\right)^2:\quad z^4-2bz^3+b^2z^2-2az^3+4abz^2-2ab^2z+a^2z^2-2a^2bz+a^2b^2$

$\mathrm{Desarrollar\:}\left(a+b\right)^2:\quad a^2+2ab+b^2$

$\mathrm{Restar\:}b^2\mathrm{\:de\:ambos\:lados}$

$z^4-2bz^3+b^2z^2-2az^3+4abz^2-2ab^2z+a^2z^2-2a^2bz+a^2b^2-b^2=a^2+2ab+b^2-b^2$

$\mathrm{Simplificar}$

$z^4-2bz^3+b^2z^2-2az^3+4abz^2-2ab^2z+a^2z^2-2a^2bz+a^2b^2-b^2=a^2+2ab$

$\mathrm{Restar\:}2ab\mathrm{\:de\:ambos\:lados}$

$z^4-2bz^3+b^2z^2-2az^3+4abz^2-2ab^2z+a^2z^2-2a^2bz+a^2b^2-b^2-2ab=a^2+2ab-2ab$

$\mathrm{Simplificar}$

$\left(z^2-2bz+b^2-1\right)a^2+\left(-2z^3+4bz^2-2b^2z-2b\right)a+z^4-2bz^3+b^2z^2-b^2=0$

$\mathrm{La\:solucion\:a\:la\:ecuacion\:de\:segundo\:grado\:es:\:}$

$a=\frac{z^3-2bz^2+b^2z+b+b^2-z^2}{z^2-2bz+b^2-1}$