Question

Last problem for the night! Help? Please and thank you :)

Answer 1

\[y ^{-2}-13y ^{-1}+40=0\]

Answer 2

I will start understanding this stuff sooner or later :/

Answer 3

a = 1/y

y = 13/2 +- sqrt(13^2 -4(40))/2

Answer 4

a = ..... not y lol

Answer 5

a = 13/2 +- 3/2

Answer 6

Eh.

Let \(u = y^{-1} \implies u^2 - y^{-2} \)
Then solve the quadratic the way you usually do
\[u^2-13u +40 = 0\]

Answer 7

a = 16/2 = 8

a = 10/2 = 5

Answer 8

8 = 1/y ; y=1/8

5 = 1/y ; y = 1/5

Answer 9

1/8 and 1/5 it is. I will try it out! Thanks