Question

Решите уравнение:
√(3х+1)+√(х-1)=6

Answer 1

$sqrt{3x+1}+sqrt{x-1} = 6, \ left { {{3x+1 geq 0,} atop {x-1 geq 0;}} right. left { {{x geq -frac{1}{3},} atop {x geq 1;}} right. \ x geq 1;$
$sqrt{3x+1} = 6-sqrt{x-1}, \ (sqrt{3x+1})^2 = (6-sqrt{x-1})^2, \ 3x+1 = 36-12sqrt{x-1}+x-1, \ 12sqrt{x-1} = 34-2x, \ (12sqrt{x-1})^2 = (34-2x)^2, \ 144(x-1) = 1156-136x+4x^2, \ 4x^2-280x+1300 = 0, \ x^2-70x+325 = 0, \ D_{/4} = (-35)^2-1cdot325 = 1225-325 = 900=30^2, \ x_1=35-30=5, \ x_2=35+30=65.$