Verified Answer :

konrad509

\(x^4-81=0 \\ (x^2)^2-9^2=0\\ (x^2-9)(x^2+9)=0\\ (x-3)(x+3)(x^2+9)=0\\ x=3 \vee x=-3\)

apologiabiology

so this is a difference of 2 perfect squares
so if
x^2-y^2 then it is equal to (x-y)(x+y)

so
x^4-81=[(x^2)^2]-[(9)^2]=(x^2-9)(x^2+9)

so we know if xy=0 then x or/and y=0 so assume that they are equal to zero

x^2+9=0
we can't do anything about this because
x^2=-9
and that is not possible in the real realm


x^2-9=0
yet another difference of 2 perfect squares
x^2-3^2=(x-3)(x+3)=0
set them to zero
x-3=0
x=3
x+3=0
x=-3

so x=3 or -3

Related Questions