# Math Help - Quadratic equations????? PART-2

well, here's another tough one, please explain the steps ...
4(x – 1/x)² + 8(x + 1/x) – 29 = 0

[Ans: x=1/2, (- 9 ± √65)/ 4]

2. Originally Posted by ice_syncer
well, here's another tough one, please explain the steps ...
4(x – 1/x)² + 8(x + 1/x) – 29 = 0

[Ans: x=1/2, (- 9 ± √65)/ 4]

To see that more clearly First write $\left(x - \frac1{x}\right)^2 = \left(x + \frac1{x}\right)^2 - 4$

So

4(x – 1/x)² + 8(x + 1/x) – 29 = 4((x + 1/x)² - 4) + 8(x + 1/x) – 29 = 4(x + 1/x)² + 8(x + 1/x) – 45 = 0

Now the last equation is a quadratic in (x+1/x). So solve it. Once you get a couple of answers for that solve the equation "x+1/x = previous answer", which is another quadratic.

3. Originally Posted by Isomorphism
To see that more clearly First write $\left(x - \frac1{x}\right)^2 = \left(x + \frac1{x}\right)^2 - 4$