How do you find all the zeros of $f(x)=x^4+3x^2-2$ ?

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Answer 1 · 2016-08-05T16:23:22

$f(x)$ has zeros:

$+-sqrt((sqrt(17)-3)/2)$

$+-sqrt((sqrt(17)+3)/2)i$

$f(x) = x^4+3x^2-2$

$=(x^2+3/2)^2 - 9/4-2$

$=(x^2+3/2)^2-17/4$

$=(x^2+3/2)^2-(sqrt(17)/2)^2$

$=(x^2+3/2-sqrt(17)/2)(x^2+3/2+sqrt(17)/2)$

Hence zeros:

$x=+-sqrt((sqrt(17)-3)/2)$

$x=+-sqrt((sqrt(17)+3)/2)i$

How do you find all the zeros of f(x)=x^4+3x^2-2f(x)=x^4+3x^2-2?