How do you solve the quadratic with complex numbers given $x^{4} + 16 x^{2} - 225 = 0$ $x^4+16x^2-225=0$ ?

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How do you solve the quadratic with complex numbers given $x^{4} + 16 x^{2} - 225 = 0$ $x^4+16x^2-225=0$ ?

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Answer 1 · 2016-11-05T06:21:21

This quartic equation has roots $\pm 3$ $+-3$ and $\pm 5 i$ $+-5i$

Explanation:

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

We will use this several times.

$color(white)()$
Given:

$x^{4} + 16 x^{2} - 225 = 0$ $x^4+16x^2-225 = 0$

This is a quartic in $x^{4}$ $x^4$ , but effectively a quadratic in $x^{2}$ $x^2$ . So we can make a start by treating it as such and using whatever method we like to solve the quadratic:

$0 = x^{4} + 16 x^{2} - 225$ $0 = x^4+16x^2-225$

$0 = {(x^{2})}^{2} + 16 (x^{2}) + 64 - 289$ $color(white)(0) = (x^2)^2+16(x^2)+64-289$

$0 = {(x^{2} + 8)}^{2} - 17^{2}$ $color(white)(0) = (x^2+8)^2-17^2$

$0 = ((x^{2} + 8) - 17) ((x^{2} + 8) + 17)$ $color(white)(0) = ((x^2+8)-17)((x^2+8)+17)$

$0 = (x^{2} - 9) (x^{2} + 25)$ $color(white)(0) = (x^2-9)(x^2+25)$

$0 = (x^{2} - 3^{2}) (x^{2} - {(5 i)}^{2})$ $color(white)(0) = (x^2-3^2)(x^2-(5i)^2)$

$0 = (x - 3) (x + 3) (x - 5 i) (x + 5 i)$ $color(white)(0) = (x-3)(x+3)(x-5i)(x+5i)$

Hence roots:

$x = \pm 3$ $x = +-3$ and $x = \pm 5 i$ $x = +-5i$

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How do you solve the quadratic with complex numbers given $x^{4} + 16 x^{2} - 225 = 0$ $x^4+16x^2-225=0$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you solve the quadratic with complex numbers given x^4+16x^2-225=0x4+16x2−225=0x^4+16x^2-225=0?

1 Answer

Explanation:

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Impact of this question

How do you solve the quadratic with complex numbers given $x^{4} + 16 x^{2} - 225 = 0$ $x^4+16x^2-225=0$ ?