How do you find the complex roots of $x^{4} - 10 x^{2} + 9 = 0$ $x^4-10x^2+9=0$ ?

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How do you find the complex roots of $x^{4} - 10 x^{2} + 9 = 0$ $x^4-10x^2+9=0$ ?

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Answer 1 · 2017-01-10T22:45:46

The roots are $\pm 3$ $+-3$ and $\pm 1$ $+-1$

Explanation:

Note that $9 + 1 = 10$ $9+1 = 10$ and $9 \cdot 1 = 9$ $9*1 = 9$ .

Hence we find:

$0 = x^{4} - 10 x^{2} + 9$ $0 = x^4-10x^2+9$

$0 = (x^{2} - 9) (x^{2} - 1)$ $color(white)(0) = (x^2-9)(x^2-1)$

$0 = (x^{2} - 3^{2}) (x^{2} - 1^{2})$ $color(white)(0) = (x^2-3^2)(x^2-1^2)$

$0 = (x - 3) (x + 3) (x - 1) (x + 1)$ $color(white)(0) = (x-3)(x+3)(x-1)(x+1)$

So the four roots of this quartic equation are $\pm 3$ $+-3$ and $\pm 1$ $+-1$

These are Real numbers, but any Real number is also a Complex number (of the form $a + 0 i$ $a+0i$ if you wish).

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How do you find the complex roots of $x^{4} - 10 x^{2} + 9 = 0$ $x^4-10x^2+9=0$ ?

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How do you find the complex roots of x^4-10x^2+9=0x4−10x2+9=0x^4-10x^2+9=0?

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How do you find the complex roots of $x^{4} - 10 x^{2} + 9 = 0$ $x^4-10x^2+9=0$ ?