# Math Help - Finding the roots of this equation: x^2 + (a - 1/a)x-1 = 0

1. ## Finding the roots of this equation: x^2 + (a - 1/a)x-1 = 0

Any help would be greatly appreciated.

2. Originally Posted by -DQ-
Any help would be greatly appreciated.

The roots of this equation are

$x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

where a = 1, b= a-1/a, c=-1

3. Correct, but it is not a good idea to use "a" to represent two different quantities!

Perhaps it would be better to write
"The roots to the equation $ix^2+ jx+ k= 0$ are $x= \frac{-j\pm\sqrt{j^2- 4ik}}{2i}$. For this problem, i= 1, j= (a-1)/a, k= -1".

(I am assuming that in the first post, "a- 1/a" meant "(a-1)/a", not "a- (1/a)".)

4. This quadratic factorises as $(x+a)(x-\tfrac1a)$. So the solutions of the equation are $x=-a$ and $x-\tfrac1a$.

(I am making the opposite assumption from HallsofIvy, namely that, "a- 1/a" meant "a- (1/a)", not "(a-1)/a".)