(x+1)^4+(x+1)^2-6=0 решите пж

Question

(x+1)^4+(x+1)^2-6=0 решите пж

Answer 1

(x+1)^2=a   (x+1)^4=a^2
a^2+a-6=0
a1=2               a2=-3   (x+1)^2=-3  корней нет
(x+1)^2=2
x+1=+-√2
x1=√2-1  x2=-√2-1

Answer 2

( Х + 1 )^4 + ( Х + 1 )^2 - 6 = 0
( Х + 1 )^2 = а ; а > 0
а^2 + а - 6 = 0
D = 1 + 24 = 25 = 5^2
a1 = ( - 1 + 5 ) : 2 = 2
a2 = ( - 1 - 5 ) : 2 = - 3 ( < 0 )
( x + 1 )^2 = 2
X^2 + 2x + 1 - 2 = 0
X^2 + 2x - 1 = 0
D = 4 + 4 = 8 = ( 2 V 2 )^2
X1 = ( - 2 + ( 2 V 2 )) : 2 = - 1 + V 2
X2 = ( - 2 - ( 2 V 2 )) : 2 = - 1 - V 2

Comments

V корень квадратный