Question 501819: x^-6 - 11x^-3 + 24 = 0
I'm confused as to where to start!
Thanks,
Tracy
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! x^-6 - 11x^-3 + 24 = 0
.
Let y = x^-3
then
y^2 = x^-6
.
So, instead of:
x^-6 - 11x^-3 + 24 = 0
we can write:
y^2 - 11y + 24 = 0
now, we can factor the left:
(y-3)(y-8) = 0
y = {3, 8}
But, we're looking for 'x' NOT 'y', so substitute back into:
y = x^-3
3 = x^-3
3 = 1/x^3
3x^3 = 1
x^3 = 1/3
x = cuberoot(1/3)
or, rationalizing
x = cuberoot(3)/3
and, similarly
x = cuberoot(8)/8 = 2/8 = 1/4
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