Question

If p=1/3(x+x^-1) and q=1/3(x-x^-1), then what is p^2 - pq

Answer 1

\[\frac{ 1 }{3 }(x+x ^{-1} = p \]

Answer 2

similarly for q

Answer 3

@ganeshie8

Answer 4

mult out and see... p and q are conjugates

Answer 5

i'm thinking 2/9

Answer 6

\[p ^{2}-pq=\left[ \frac{ 1 }{ 3 }\left( x+\frac{ 1 }{ x } \right) \right]^{2}-\left[ \frac{ 1 }{ 3 }\left( x+\frac{ 1 }{ x } \right) \right]\left[ \frac{ 1 }{ 3 }\left( x-\frac{ 1 }{ x } \right) \right]\]

Answer 7

\(x\ne 0\)

Answer 8

\[1-q/p=1-\frac{ \left( x+\frac{ 1 }{ x } \right) }{ \left( x-\frac{ 1 }{ x } \right) }\]

Answer 9

\[q=\frac{ x+\frac{ 1 }{ x } }{ x-\frac{ 1 }{ x } }p-2p\]

Answer 10

reverse plus and minus signs of p and q in last and second to last equations.

Answer 11

but that doesn't help find the value of p^2-pq does it?

Answer 12

@hartnn can u explain

Answer 13

whats
\((x+x^{-1})^2 = ... ?\)

Answer 14

x^2 + 2 +1/x^2

Answer 15

and
\((x+x^{-1})(x-x^{-1})=.. ?\)

Answer 16

x^2 + a/x^2

Answer 17

you mean x^2 -1/x^2 ?

Answer 18

yes

Answer 19

\(\Large p^2-pq = (1/9)(x+x^{-1})^2 -(1/9)(x+x^{-1})(x-x^{-1})\)
plug in values ...

Answer 20

did you get that bdw ?

Answer 21

uh how should I start to simplify

Answer 22

@hartnn

Answer 23

i just asked whether u got it ?

Answer 24

not the answer but yes I understand

Answer 25

you already calculated the terms
\(p^2-pq = (1/9) [(x^2+2+x^{-2}) - (x^2-x^{-2})]\)

Answer 26

oh yea

Answer 27

simplify that

Answer 28

is it 2/9 (x^-2 -1)

Answer 29

no

Answer 30

\(p^2-pq = (1/9) [(x^2+2+x^{-2}) - (x^2-x^{-2})] \\ p^2-pq = (1/9) [2+x^{-2} +x^{-2})] = \huge 2/9 (1+x^{-2})\)

Answer 31

oops typo

Answer 32

ok :) np