How do you find the zeroes for $f (x) = 2 x^{4} - 2 x^{2} - 40$ $f(x)=2x^4-2x^2-40$ ?

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How do you find the zeroes for $f (x) = 2 x^{4} - 2 x^{2} - 40$ $f(x)=2x^4-2x^2-40$ ?

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Answer 1 · 2015-05-29T00:19:02

Simplify by replacing $x^{2}$ $x^2$ with $y$ $y$ ;
solve as a standard quadratic for expressions in terms of $y$ $y$ that result in zeros;
replace $y$ $y$ in those expressions with $x^{2}$ $x^2$ and solve for values of $x$ $x$

If $y = x^{2}$ $y=x^2$
$2 x^{4} - 2 x^{2} - 40 = 0$ $2x^4-2x^2-40 = 0$
is equivalent to
$2 y^{2} - 2 y - 40 = 0$ $2y^2-2y-40 = 0$

$(2 y + 8) (y - 5) = 0$ $(2y+8)(y-5) = 0$

For zeros either
Case 1: $(2 y + 8) = 0$ $(2y+8) = 0$
or
Case 2: $(y - 5) = 0$ $(y-5) = 0$

Case 1: $(2 y + 8) = 0$ $(2y+8)=0$
$\to y = - 4$ $rarr y=-4$
$\to x^{2} = - 4$ $rarr x^2 = -4$
There are no Real solutions, but there are Complex solutions $x = \pm 2 i$ $x=+-2i$

Case 2: $(y - 5) = 0$ $(y-5)=0$
$\to y = 5$ $rarr y = 5$
$\to x^{2} = 5$ $rarr x^2 = 5$
$\to x = \pm \sqrt{5}$ $rarr x = +-sqrt(5)$

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How do you find the zeroes for $f (x) = 2 x^{4} - 2 x^{2} - 40$ $f(x)=2x^4-2x^2-40$ ?

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How do you find the zeroes for f(x)=2x4−2x2−40f(x)=2x^4-2x^2-40?

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How do you find the zeroes for $f (x) = 2 x^{4} - 2 x^{2} - 40$ $f(x)=2x^4-2x^2-40$ ?