## anonymous 5 years ago need to find the solution to this equation.

1. anonymous

$y ^{-1}-22y -_{2}^{1}+12-=0$

2. anonymous

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

3. anonymous

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

4. myininaya

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

5. myininaya

u^2-22u+12=0

6. anonymous

I got 12 and 10. is that correct?

7. anonymous

or maybe they are negative?

8. myininaya

one sec im scanning something

9. anonymous

ok, thanks

10. myininaya

11. anonymous

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?

12. myininaya

i didn't simplify

13. anonymous

for some reason i'm not getting a whole number

14. myininaya

i'm not getting a whole number

15. anonymous

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

16. anonymous

yep, I did... ahh. sorry.

17. myininaya

lol. okay lets take another look at it

18. anonymous

Thanks! :)

19. myininaya

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

20. myininaya

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

21. anonymous

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

22. anonymous

Thanks for clarifying!

23. myininaya

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

24. anonymous

okk got it:)