## anonymous 5 years ago if x+1/x =8, find the x^2 + x^-2

1. anonymous

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

2. anonymous

no, it is x+(1/x)

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

Correct Anwar!!!

11. anonymous

12. anonymous

option c)

13. anonymous

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

14. anonymous

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

15. anonymous

16. anonymous

yes, thanks guys its clear now

17. anonymous

You're welcome!