Question

How do I add 2x^2 + (16/x^2)?

Answer 1

Any ideas?

Answer 2

[(2x^2)(x^2) + 16] / x^2 ?

Answer 3

Correct :)

Answer 4

It all equals 18, if that makes a difference.

Answer 5

So I would end up with (2x^4 + 16)/x^2 = 18, then
[(2x^4 + 16)/ x^2] - 18 = 0

Answer 6

Keep going...

Answer 7

..try to simplify further :-)

Answer 8

Okay, so here's what I've done:
[2x^2(x^2) + 16 - 18(x^2)] / x^2 = 0
(2x^4 - 18x^2 + 16)/x^2 = 0
[2(x-1)(x+1)(x^2-8)]/x^2 = 0

I know I'm really close, but how do I get x from all this mess? @_@

Answer 9

Can I make the x in the denominator be 1 and solve for the top?

Answer 10

Nope...go back to the beginning and try to solve the entire equation:
$$\Huge 2x^2 +\frac{16}{x^2}=18$$

Answer 11

But I don't know how to get anything other than this:
[2(x-1)(x+1)(x^2-8)]/x^2 = 0

Answer 12

You can use the idea of balance to transform the equation by doing the same thing on both sides, right?

Answer 13

So multiply both sides by x^2?

Answer 14

Sounds like a reasonable idea :-)

Answer 15

Okay, so I tried that and then I only had to work with the numerator!
2(x-1)(x+1)(x^2-8) = 0
2=0 (not a solution) x=1 x=-1 x=+/-2sqrt(2)

I checked in the back and those are right :) Thank you!

Answer 16

Nice work :-)

Answer 17

Actually you should get $$\Huge 2x^2 +\frac{16}{x^2}=18$$ $$\Huge 2x^4 +16=18x^2 $$ $$\Huge 2x^4 -18x^2+16 = 0$$ $$\Huge x^4 -9x^2+8 = 0$$ $$\Huge \text{Let } y= x^2$$ $$\Huge \text{Then }y^2-9y+8=0$$ $$\Huge(y-8)(y-1)=0$$ $$\Huge x^2-8=0 \text{ or }x^2-1=0$$