Question

b) (5x + 3y + 2x) = [(5x + 3y)+22
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Answer 1

Respuesta:

$x=\sqrt{\frac{1-3y-2z^2}{5}},\:x=-\sqrt{\frac{1-3y-2z^2}{5}},\:x=i\sqrt{\frac{3y+2z^2}{5}},\:x=-i\sqrt{\frac{3y+2z^2}{5}}$

Explicación paso a paso:

$\left(5x^2+3y+2z^2\right)=\left[\left(5x^2+3y\right)+2z^2\right]^2$

$5x^2+3y+2z^2=25x^4+30x^2y+20x^2z^2+9y^2+12yz^2+4z^4$

$25x^4+30x^2y+20x^2z^2+9y^2+12yz^2+4z^4=5x^2+3y+2z^2$

$25x^4+30x^2y+20x^2z^2+9y^2+12yz^2+4z^4-\left(3y+2z^2\right)=5x^2+3y+2z^2-\left(3y+2z^2\right)$

$25x^4+30x^2y+20x^2z^2+9y^2+12yz^2+4z^4-3y-2z^2=5x^2$

$25x^4+30x^2y+20x^2z^2+9y^2+12yz^2+4z^4-3y-2z^2-5x^2=5x^2-5x^2$

$25x^4+\left(30y+20z^2-5\right)x^2+9y^2+12yz^2+4z^4-3y-2z^2=0$

$25u^2+\left(30y+20z^2-5\right)u+9y^2+12yz^2+4z^4-3y-2z^2=0$

$u=\frac{1-3y-2z^2}{5},\:u=-\frac{3y+2z^2}{5}$

$x=\sqrt{\frac{1-3y-2z^2}{5}},\:x=-\sqrt{\frac{1-3y-2z^2}{5}},\:x=i\sqrt{\frac{3y+2z^2}{5}},\:x=-i\sqrt{\frac{3y+2z^2}{5}}$