How do you solve #x ^ { 4} + x ^ { 2} - 2= 0#?

1 Answer
mason m
Dec 20, 2016

#x=+-isqrt2, +-1#.

Explanation:

Make the substitution #u=x^2#.

So #x^4+x^2-2=u^2+u-2#.

This is a quadratic in #u#, so solve it as you would any other quadratic.

#u^2+u-2=(u+2)(u-1)=0#, so #u=-2# or #u=1#.

So #x^2=-2# or #x^2=1#.

So #x=+-isqrt2, +-1#.

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