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.