How do you solve for all solutions, including complex solutions, to $x^{4} + 4 x^{2} = 45$ $x^4 + 4x^2 = 45$ ?

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Answer 1 · 2016-05-02T14:57:23

$x = \pm \sqrt{5}$ $x=+-sqrt(5)$ or $x = \pm 3 i$ $x=+-3i$

OK, let's complete the square to start, adding $4$ $4$ to both sides to get:

$x^{4} + 4 x^{2} + 4 = 49$ $x^4+4x^2+4 = 49$

That is:

${(x^{2} + 2)}^{2} = 7^{2}$ $(x^2+2)^2 = 7^2$

Hence:

$x^{2} + 2 = \pm 7$ $x^2+2 = +-7$

Subtract $2$ $2$ from both sides to get:

$x^{2} = - 2 \pm 7$ $x^2 = -2+-7$

That is:

$x^{2} = 5$ $x^2 = 5$ or $x^{2} = - 9$ $x^2 = -9$

If $x^{2} = 5$ $x^2 = 5$ then $x = \pm \sqrt{5}$ $x = +-sqrt(5)$

If $x^{2} = - 9$ $x^2 = -9$ then $x = \pm 3 i$ $x = +-3i$

How do you solve for all solutions, including complex solutions, to x4+4x2=45x^4 + 4x^2 = 45?