Question

x^2+2x+√x^2+2x+8=12 ---------

Answer 1

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

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