Question

Find the real number solutions:
4w^4+40w^2-44=0
I need the steps! Thanks!

Answer 1

4w^4+40w^2-44=0
4(w^4+10w^2-11)=0
4(w^2-1)(w^2+11)=0
since the second set of parentheses give an imaginary answer, the only one to worry about it the first set, which yields
w={-1,1}

Answer 2

Thanks! :)