Question

X^2+2x+√x^2+2x+8=12 ---------

Answer 1

$x^2+2x+\sqrt{x^2+2x+8}=12\; ,\quad ODZ:\; x^2+2x+8 \geq 0\\\\t=x^2+2x\; ,\; t+\sqrt{t+8}=12\\\\\sqrt{t+8}=12-t\; ,\; \; \to \quad 12-t \geq 0\; ,\; t \leq 12\\\\t+8=(12-t)^2\\\\t+8=144-24t+t^2\\\\t^2-25t+136=0\\\\D=625-544=81\\\\t_1= \frac{25-9}{2} =8\; ,\; \; t_2=\frac{25+9}{2}=17\; (ne\; podxodit\; ,\; t.k.\; t \leq 12)\\\\x^2+2x=8\\\\x^2+2x-8=0\\\\D/4=1+8=9\\\\x_1=-1-3=-4\; ,\; \; x_2=-1+3=2\\\\Proverka:\; \; a)\; \; x_1=-4:\\\\(-4)^2+2(-4)+\sqrt{(-4)^2+2(-4)+8}=12\\\\8+\sqrt{16}=12\\\\8+4=12\\\\12=12$

$b)\; \ ;x_2=2:\\\\2^2+2\cdot 2+\sqrt{2^2+2\cdot 2+8}=12\\\\8+\sqrt{16}=12\\\\8+4=12\\\\12=12\\\\Otvet:\; \; x_1=-4\; ,\; \; x_2=2\; .$