Дано x1+x2=7 x1*x2=2 найти x1^4-x2^4 (^4 - степень)

X1+x2=7,x1*x2=2
x1=7-x2
x2*(7-x2)=2
x2²-7x2+2=0
D=49-8=41
1)x2=(7-√41)/2⇒x1=7-(7-√41)/2=(7+√41)/2
2)x2=(7+√41)/2⇒x1=(4-√41)/2

x1^4-x2^4=(x1²-x2²)(x1²+x2²)=(x1+x2)(x1-x2)*[(x1+x2)²-2x1*x2]=
=7*(49-4)(х1-х2)=315*(х1-х2)
1)315*[(7+√41)/2-(7-√41)/2]=315*√41
2)315*[(7-√41)/2-(7+√41)/2]=-315√41