Question

need to find the solution to this equation.

Answer 1

\[y ^{-1}-22y -_{2}^{1}+12-=0\]

Answer 2

I couldn't get this to look right. it is supposed to be y with the -1/2 as the square root

Answer 3

Is it
\[{1 \over y}-22y^{-1/2}+12=0?\]

Answer 4

let u=y^(-1/2) so we have
u^2=y^(-1)

Answer 5

u^2-22u+12=0

Answer 6

I got 12 and 10. is that correct?

Answer 7

or maybe they are negative?

Answer 8

one sec im scanning something

Answer 9

ok, thanks

Answer 10

wow, thanks for doing that! I can see where I maybe slipped up. I am a bit confused as to what the answer is though. Is there still one more step in solving those 2 solutions?

Answer 11

i didn't simplify

Answer 12

for some reason i'm not getting a whole number

Answer 13

i'm not getting a whole number

Answer 14

i just realized that you had 12 in the problem when it's supposed to be 120. Sorry if I typed it wrong.

Answer 15

yep, I did... ahh. sorry.

Answer 16

lol. okay lets take another look at it

Answer 17

Thanks! :)

Answer 18

so we get u=10 and u=12 like you been talking about this whole time but u=y^(-1/2)
y^(-1/2)=10 and y^(-1/2)=12
so y=10^(-2)=1/(10^2)=1/100
so y=12^(-2)=1/(12^2)=1/144
there you basically had it you just needed to solve for y

Answer 19

we got u=12 and 10 because u^2-22u+120=(u-12)*(u-10)=0
so very good

Answer 20

ok I just wasn't sure if they were supposed to stay positive.

Answer 21

Thanks for clarifying!

Answer 22

to get rid of that -1/2 exponent i just raised both sides to -2
because law of exponents tells me if we have (x^{-1/2)^{-2} to multiply the exponents but if i do this to one side i have to do it to the other

Answer 23

okk got it:)