Is this right MK2

**Joel** · June 3rd 2009, 03:58 AM

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?

**Prove It** · June 3rd 2009, 04:26 AM

Originally Posted by Joel

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?

$\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$ .

If $x + \frac{1}{x} = 3$ , then $\left(x + \frac{1}{x}\right)^2 = 3^2 = 9$ .

Thus $x^2 + 2 + \frac{1}{x^2} = 9$

And so $x^2 + \frac{1}{x^2} = 7$ .

**masters** · June 3rd 2009, 04:31 AM

Originally Posted by Joel

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?

Hi Joel,

(a) Expand $\left(x+\frac{1}{x}\right)^2$

$\left(x+\frac{1}{x}\right)^2=x^2+2+\frac{1}{x^2}$

(b) Suppose $x+\frac{1}{x}=3$

Use part (a) to evaluate $x^2+\frac{1}{x^2}$ without attempting to find the value of x.

$x + \frac{1}{x}=3$

$\left(x+\frac{1}{x}\right)^2=9$

$x^2+\frac{1}{x^2}+2=9$

$x^2+\frac{1}{x^2}=7$

Edit: Too slow......Prove it got there first. Rats!

**Joel** · June 3rd 2009, 04:38 AM

Ok Thanks heaps guys,

I was on the right track but looking at your solutions, I was going about the last step by solving it rather than representing it via x.

Cheers

Math Help - Is this right MK2

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