Hi guys,
Got this bit right.......... thanks for the conformation
(a) Expand ( x + ⅟x )² = x² + 2x⅟x + ⅟x₂ = x² + 2 + ⅟x₂
(b) Suppose that x + ⅟x = 3. Use part (a) to evaluate x² + ⅟x² without attempting to find the value of x.
ok my thoughts are..... ( x + ⅟x )² = (3)² from above.
so i think i should use the solution x² + 2 + ⅟x₂ which would be 9 + 2.
The question asks me to evaluate, but I think I may be on the wrong track. as I feel I am solving for x.
Am I doing the wrong thing here?
$\displaystyle \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$.Originally Posted by Joel
If $\displaystyle x + \frac{1}{x} = 3$, then $\displaystyle \left(x + \frac{1}{x}\right)^2 = 3^2 = 9$.
Thus $\displaystyle x^2 + 2 + \frac{1}{x^2} = 9$
And so $\displaystyle x^2 + \frac{1}{x^2} = 7$.
Hi Joel,
(a) Expand $\displaystyle \left(x+\frac{1}{x}\right)^2$
$\displaystyle \left(x+\frac{1}{x}\right)^2=x^2+2+\frac{1}{x^2}$
(b) Suppose $\displaystyle x+\frac{1}{x}=3$
Use part (a) to evaluate $\displaystyle x^2+\frac{1}{x^2}$ without attempting to find the value of x.
$\displaystyle x + \frac{1}{x}=3$
$\displaystyle \left(x+\frac{1}{x}\right)^2=9$
$\displaystyle x^2+\frac{1}{x^2}+2=9$
$\displaystyle x^2+\frac{1}{x^2}=7$
Edit: Too slow......Prove it got there first. Rats!
  |
LinkBack URL
About LinkBacks