In the quadratic equation
\( x^2 + \left(k-\frac{1}{k}\right)x - 1 = 0,\)
solve for x in terms of k .
+36916 Use Quadratic Formula
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) a = 1 b = (k- 1/k) c = -1
[(1/k - k) +- sqrt ((k-1/k)^2 + 4 ) ] / 2
By the qudaratic formula
\(x = \dfrac{k - 1/k \pm \sqrt{(k - 1/k)^2 - 4}}{2} = \dfrac{k - 1/k \pm \sqrt{k^2 - 8 + 1/k^2}}{2}\)
+36916 Check the -b part of the Quad Form in your answer.......AND the -4ac part too 
