Question 142906
Anne, I bet you can solve this equation
{{{b^2 - 5b + 4=0}}}

You would factor and get
{{{(b - 4)(b -1) = 0 }}} and then solve for b


So how is the given problem different than the one above? If you set {{{b = a^2}}}, they are the same.

Thus
{{{a^4-5a^2+4=0 }}}
{{{(a^2 -4)(a^2 -1) = 0 }}}

Solving yields a^2 = 4  and a^2 = 1
Take the square roots and you'll get 4 solutions.

Make sense?