Question

Решить уравнение
$\sqrt{2x^{2}+2x-3 } +x=0$

Answer 1

$\sqrt{2x^2+2x-3}+x=0 \\\sqrt{2x^2+2x-3}=-x\\(\sqrt{2x^2+2x-3})^2=(-x)^2\\2x^2+2x-3=x^2\\x^2+2x-3=0\\x_1=-3; x_2=1\\\\Proverka:\\x=-3\\\sqrt{2(-3)^2+2*(-3)-3}-3=0\\\sqrt{18-6-3}-3=0\\\sqrt{9}-3=0\\0=0\\\\x=1\\\sqrt{2*1^2+2*1-3}+1=0\\\sqrt{2+2-3}+1=0\\2\neq 0\\\\Otvet: x=-3$

Answer 2

$\sqrt{2x^2+2x-3}=-x\Leftrightarrow \left \{ {{2x^2+2x-3=(-x)^2} \atop {-x\ge 0}} \right. \Leftrightarrow \left \{ {{x^2+2x-3=0} \atop {x\le 0}} \right. \Leftrightarrow \left \{ {{(x+3)(x-1)=0} \atop {x\le 0}} \right. .$

Ответ:  - 3