Question 228346
Here's a couple of ways:<ol><li>Direct. Try raising numbers to the fourth power and see if you can get 256.<ul><li>{{{2^4 = 2*2*2*2 = 16}}}</li><li>{{{3^4 = 3*3*3*3 = 81}}}</li><li>{{{4^4 = 4*4*4*4 = 256}}} Bingo!</li></ul>The problem with approach is that you tend to forget the "other" number. -4 works, too!</li><li>Solve by factoring:<ol><li>Subtract 256 from each side:
{{{x^4-256=0}}}</li><li>Factor. This is a difference of squares:
{{{(x^2+16)(x^2-16) = 0}}}
We still have a difference of squares:
{{{(x^2+16)(x+4)(x-4) = 0}}}</li><li>Solve. In order for the product to be zero one of the factors must be zero. So:
{{{x^2+16 = 0}}} or {{{x +4 = 0}}} or {{{x-4 =0}}}
The first equation is impossible (i.e. no solution) because you can't square something, add 16 and end up with zero. (Think about it.) But the other two equation can be solved:
{{{x = -4}}} or {{{x = 4}}}
This way takes a little more work but you are less likely to overlook the -4.</li></ol></li></ol>