Question

if x+1/x =8, find the x^2 + x^-2

Answer 1

Is it
\[{x+1 \over x}=8?\]

Answer 2

no, it is x+(1/x)

Answer 3

(x + 1/x)^2 = x^2 + 2 + 1/x^2

Answer 4

It is further = x^2 + 2 + x^-2

Answer 5

finally
(x + 1/x)^2 - 2 = x^2 + x^-2 a

Answer 6

\[x+{1 \over x}=8 \implies x^2+2+x^{-2}=16 \implies x^2+x^{-2}=14\]

Answer 7

You hv x + 1/x = 8
Squaring both sides (x + 1/x)^2 = 8^2 = 16

S0, 16 - 2 = x^2 + x^-2

so 14 is the answer

Answer 8

Sorry not 16, it's 64. So the final answer is 62.

Answer 9

the answer I have are:
a) 60, b)58, c)62, d)0, e)non
So the answer must be e)

Answer 10

Correct Anwar!!!

Answer 11

So answer is 62

Answer 12

option c)

Answer 13

\[x+{1 \over x}=8 \implies (x+{1 \over x})^2=8^2 \implies x^2+2+x^{-2}=64 \]
\[\implies x^2+x^{-2}=62\]

Answer 14

Here is the corrected version

You hv x + 1/x = 8
Squaring both sides (x + 1/x)^2 = 8^2 = 64

S0, 64 - 2 = x^2 + x^-2

so 62 is the answer

Answer 15

You got the idea emunrradtvamg?

Answer 16

yes, thanks guys its clear now

Answer 17

You're welcome!