Question

x^2 = 4+корень из (2x^2-5)

Answer 1

$\x^2=4+sqrt{2x^2-5}\ x^2-4=sqrt{2x^2-5}\ 2x^2-5geq0wedge x^2-4geq0\ 2x^2geq5wedge x^2geq4\ x^2geqfrac{5}{2}wedge xin(-infty,-2ranglecuplangle,2,infty)\ xin((-infty,-sqrt{frac{5}{2}}ranglecuplanglesqrt{frac{5}{2}},infty))cap((-infty,-2ranglecuplangle,2,infty))\ D:xin(-infty,-2ranglecuplangle,2,infty)\\ (x^2-4)^2=2x^2-5\ x^4-8x^2+16=2x^2-5\ x^4-10x^2+21=0\ x^4-3x^2-7x^2+21=0\ x^2(x^2-3)-7(x^2-3)=0\ (x^2-7)(x^2-3)=0$

$\x=-sqrt 7 vee x=sqrt 7vee x=-sqrt 3vee x=sqrt 3\ {-sqrt 3,sqrt 3}not in D\\ underline{x=-sqrt7 vee x=sqrt 7}$