>>7983908
Not even gonna bother to TeX pre-school maths

1 = 16x2/(x4 + 64)
16x2 = x4 + 64
16x2 - x4 - 64 = 0

find the roots of that polynomial, then plug in numbers into the x4 + 64 part because if it equals 0 then the function isn't defined for that number, so you can disregard those points in your answer set.

After that you are done and can then proceed to kill yourself, faggot. Learn 2 fucking math.

You just replace f(x) with 1, and solve.

sqrt(8)

>>7983908
[math]f(x) = \frac{ 16x^{2} }{ x^{4} + 64 } = 1[\math]

[math]16x^{2} = x^{4} + 64[\math]


[math]x^{4} - 16x^{2} + 64 = 0[\math]


[math](x^{2} - 8)^{2} = 0[\math]


[math]x^{2} = 8[\math]

>>7983915
>x^4 + 64
>ever being 0 when x is real
wew laddy, sounds like you need to learn to math

1 = 16x^2/(x^4 + 64)

x^4 + 64 = 16x^2

x^4 - 16x^2 + 64 = 0

(x^2 - 8)^2 = 0

x^2 - 8 = 0

x^2 = 8

x = +- sqrt(8)

x = +- 2sqrt(2)

If you can't guess the factorization, another way to do it is:

x^4 - 16x^2 + 64 = 0

(x^2)x^2 - (16x)x + 64 = 0

Use quadratic equation:

x = (16x +- sqrt(256x^2 - 256x^2)) / (2x^2)

x = 8/x

x^2 = 8

x = sqrt(8)

x = 2sqrt(2)

>>7983926
Damn you are right. However, I said IF that is the case.

I'm not going to bother to understand the problem, there is a standard procedure. You'd usually have to check.

Actually, I don't know why the fuck wouldn't the problem involve checking becasue that would already be easy, but having to only solve the polynomial is poor.

To what community college are you going, OP?

>>7983936
I forgot +- on the last two lines.

>>7983936
>My life is so meaningless and boring that I sit all day just pressing refresh on /sci/ until someone posts a homework question so that I can do it for them so that they can go back to having sex with their girlfriend, something I will never do.

Ever heard of sudoku?

>>7983957
I think you're on the wrong board, friendo.