How do you solve $x^4 – 8x^2 – 9 = 0$ ?

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Answer 1 · 2016-02-11T02:31:25

x can be 3, -3, i, or -i.

If you can't find the factoring by looking at this, simplify the equation.

let $x^2 = y$ to make things easier to see

Now we have $y^2 - 8y - 9 = 0$

See it now?

We can factor into $(y-9)(y+1)$

Now substitute back in $x^2$ for $y$ .

$(x^2-9)(x^2+1)$

Since $(x^2-9)$ is a difference of two squares,

$(x-3)(x+3)$

Now we have $(x-3)(x+3)(x^2+1)$
x can be 3, -3 for the first two parts

$x^2+1=0$ can become $x^2=-1$
Taking the positive and negative root means $x = +-sqrt(-1)$

Thus $x = +-i$ in addition to 3 and -3.

How do you solve x^4 – 8x^2 – 9 = 0x^4 – 8x^2 – 9 = 0?