How do you find all the zeros of $f (x) = 2 x^{4} - 2 x^{2} - 40$ $f(x)=2x^4-2x^2-40$ ?

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Answer 1 · 2016-03-01T00:45:51

$x = - \sqrt{5}, \sqrt{5}, - 2 i, 2 i$ $x=-sqrt5,sqrt5,-2i,2i$

Set $f (x) = 0$ $f(x)=0$ .

$2 x^{4} - 2 x^{2} - 40 = 0$ $2x^4-2x^2-40=0$

Divide both sides by $2$ $2$ .

$x^{4} - x^{2} - 20 = 0$ $x^4-x^2-20=0$

We can make this resemble a quadratic if we let $u = x^{2}$ $u=x^2$ :

$u^{2} - u - 20 = 0$ $u^2-u-20=0$

Factor to see that:

$(u - 5) (u + 4) = 0$ $(u-5)(u+4)=0$

So:

$u = 5 or u = - 4$ $u=5" "" ""or"" "" "u=-4$

Since $u = x^{2}$ $u=x^2$ ,

$x^{2} = 5 or x^{2} = - 4$ $x^2=5" "" ""or"" "" "x^2=-4$

These give

$x = \pm \sqrt{5} x = \pm 2 i$ $x=+-sqrt5" "" "" "" "x=+-2i$

How do you find all the zeros of f(x)=2x4−2x2−40f(x)=2x^4-2x^2-40?