Question

решите уравнение:
$sqrt{3+2x} = x-6$
$sqrt{1-5x} = 7+x$

Answer 1

$1) sqrt{3+2x} =x-6 \ 3+2x=(x-6)^2 \ OD3: \ 3+2x geq 0 \ 2x geq 3 \ x geq 1.5 \ 3+2x=x^2-12x+36 \ x^2-12x+36-2x-3=0 \ x^2-14x+33=0 \ D=196-132=64 \ x_1= frac{14+8}{2} =11 \ x_2= frac{14-8}{2} =3 \ 2) sqrt{1-5x} =7+x \ OD3: \ 1-5x geq 0 \ 5x leq 1 \ x leq frac{1}{5} \ 1-5x=49+17x+x^2 \ x^2+14x+49+5x-1=0 \ x^2+19x+48=0 \ D=361-192=169 \ x_1= frac{-19+13}{2} =-3 \ x_2= frac{-19-13}{2} =-16$