Question

√(2a+x)=x^(1/2)+a^(1/2)

Answer 1

Respuesta:

la respuesta es $x=\frac{a}{4}\space\left\{a\ge \:0\right\}$

Explicación paso a paso:

$\sqrt{\left\{2a+x\right\}}=x^{\left(\frac{1}{2}\right)}+a^{\left(\frac{1}{2}\right)}$

$\left(\sqrt{\left\{2a+x\right\}}\right)^2=\left(x^{\frac{1}{2}}+a^{\frac{1}{2}}\right)^2$

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

$\mathrm{Desarrollar\:}\left(x^{\frac{1}{2}}+a^{\frac{1}{2}}\right)^2:\quad x+2a^{\frac{1}{2}}x^{\frac{1}{2}}+a$

$2a+x=x+2a^{\frac{1}{2}}x^{\frac{1}{2}}+a$

$2a+x-\left(x+a\right)=x+2a^{\frac{1}{2}}x^{\frac{1}{2}}+a-\left(x+a\right)$

$a=2x^{\frac{1}{2}}a^{\frac{1}{2}}$

$a^2=\left(2x^{\frac{1}{2}}a^{\frac{1}{2}}\right)^2$

$\mathrm{Desarrollar\:}\left(2x^{\frac{1}{2}}a^{\frac{1}{2}}\right)^2:\quad 4ax$

$a^2=4ax$

$\mathrm{Resolver\:}\:a^2=4ax:\quad x=\frac{a}{4}$

$x=\frac{a}{4}$

$\mathrm{Verificando\:las\:soluciones}:\quad x=\frac{a}{4}\space\left\{a\ge \:0\right\}$

$x=\frac{a}{4}\space\left\{a\ge \:0\right\}$