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

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

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Answer 1 · 2016-08-05T16:23:22

$f (x)$ $f(x)$ has zeros:

$\pm \sqrt{\frac{\sqrt{17} - 3}{2}}$ $+-sqrt((sqrt(17)-3)/2)$

$\pm \sqrt{\frac{\sqrt{17} + 3}{2}} i$ $+-sqrt((sqrt(17)+3)/2)i$

Explanation:

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

$= {(x^{2} + \frac{3}{2})}^{2} - \frac{9}{4} - 2$ $=(x^2+3/2)^2 - 9/4-2$

$= {(x^{2} + \frac{3}{2})}^{2} - \frac{17}{4}$ $=(x^2+3/2)^2-17/4$

$= {(x^{2} + \frac{3}{2})}^{2} - {(\frac{\sqrt{17}}{2})}^{2}$ $=(x^2+3/2)^2-(sqrt(17)/2)^2$

$= (x^{2} + \frac{3}{2} - \frac{\sqrt{17}}{2}) (x^{2} + \frac{3}{2} + \frac{\sqrt{17}}{2})$ $=(x^2+3/2-sqrt(17)/2)(x^2+3/2+sqrt(17)/2)$

Hence zeros:

$x = \pm \sqrt{\frac{\sqrt{17} - 3}{2}}$ $x=+-sqrt((sqrt(17)-3)/2)$

$x = \pm \sqrt{\frac{\sqrt{17} + 3}{2}} i$ $x=+-sqrt((sqrt(17)+3)/2)i$

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

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Explanation:

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