How do you solve #64x^4-81>0#?

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Answer 1 · 2017-02-06T11:05:34

#-(3sqrt2)/2 < x < (3sqrt2)/2#

and

#-(3isqrt2)/2 < x < (3isqrt2)/2#

Let's first see that all of the terms are perfect squares:

#64x^2=(8x^2)^2; 81=9^2#

If we assign #A=8x, B=9#, then we'll have a left hand side that is in the form of:

#(A^2-B^2)#

which can be factored as

#(A-B)(A+B)#

Substituting back in:

#(8x^2-9)(8x^2+9)>0#

And now let's solve the inequality. First we change the #># for an #=#, we get

#(8x^2-9)(8x^2+9)=0#

we get:

#8x^2-9=0 => x^2=9/8=>x=pmsqrt(9/8)=pm3/sqrt2=pm(3sqrt2)/2#

#8x^2=0=>x^2=-9/8=>x=pmsqrt(-9/8)=pm(3i)/sqrt2=pm(3isqrt2)/2#

as points of significance.

Working with the roots, if we set #x=0#, we can see that it isn't valid for the original equation:

#64(0)^4-81>0=>-81>0 color(white)(00)color(red)("X")#

and so our solution lies outside the points (as opposed to inside), and so the final answer is:

#-(3sqrt2)/2 < x < (3sqrt2)/2#

and

#-(3isqrt2)/2 < x < (3isqrt2)/2#