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

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

How do you solve 64x^4-81>064x^4-81>0?