How do you solve 2x4128=0?

1 Answer
Jan 10, 2017

The roots of this quartic equation are:

±22 and ±22i

Explanation:

The difference of squares identity can be written:

a2b2=(ab)(a+b)

So we find:

0=2x4128

0=2(x464)

0=2((x2)282)

0=2(x28)(x2+8)

0=2(x2(22)2)(x2(22i)2)

0=2(x22)(x+22)(x22i)(x+22i)

So the roots of the quartic equation are:

±22 and ±22i