Question

4^2 over 8x^2-8x = 2 over x

Answer 1

\[\frac{4^2}{8x^2-8x}=\frac{2}{x}\]?

Answer 2

yes but the 4^2 is 4x^2 , sorry typo
and i cross mult and got 2(8x^2-8x)=4x^2 times x and then i got 16x^2-16x=4x^3

Answer 3

okay so:
\[\frac{4x^2}{8x^2-8x}=\frac{2}{x}\]

Answer 4

am i right so far? and yes thats it (:

Answer 5

am i right so far? and yes thats it (:

Answer 6

\[4x^3=16x^2-16x\]\[4x^3-16x^2+16x=0\]\[4x(x^2-4x+4)=0\]

Answer 7

Can you continue from here? All I did was move everything to one side and pull out a 4x.

Answer 8

why isnt the 4x^3 not negative?

Answer 9

because it was already on that side. For example if we wanted to move 3 over in this case:
x+3=-7
We would do:
(x + 3) - 3 = (-7) - 3
x = -7 - 3

Answer 10

oh okay i see so now i can solve that equation using the quad. formula right?

Answer 11

yup.

Answer 12

okay thank you (:

Answer 13

And you know that x is either 0, or whatever the solutions to:
\[x^2-4x+4=0\]are

Answer 14

when i put it in the quad form i dont use their exponents right?

Answer 15

well actually that equation is fairly straightforward to solve:
\[x^2-4x+4=x^2-2*2*x+2^2=(x-2)^2\]

Answer 16

so your answer would be x=0, x=2 (with x=2 being a repeated root)

Answer 17

thats not one of my options ): @bahrom7893

Answer 18

thats not one of my options ): @bahrom7893

Answer 19

0 and 2 are not there?

Answer 20

no there is x=2 and x=-2 , x=2, x=16 and x=-16, x=-16

Answer 21

ohh x=2
x cannot be 0 because of 2/x

Answer 22

okay thank you (: