Solving Quadratic Equations without CD: Better Direction?

[karush]

ok this was posted on LinkedIn and sure it has already be answered
but usually these types of problems are resolved by way too many steps
so just wanted to proceed with this without looking at previous attempts

my first reaction was to get a CD but would introduce a bigger problem
however with the exp i presume you could do this by Quadratics and grouping

or is there a better direction?

 

[pasmith]

Set $u = x + x^{-1}$. Then $x^2 + x^{-2} = u^2 - 2$ and $x^3 + x^{-3} = u^3 - 3u$ so that $$u^3 + u^2 - 2u - 30 = (u- 3)(u^2 + 4u - 10) = 0.$$ Then completing the square in $x$ gives $$(2x - u)^2 = u^2 - 4$$ and the choice $u = 3$ leads directly to $$(2x - 3)^2 = 5.$$ The other roots $u = -2 \pm \sqrt{14}$ lead to $$x^2 + 2x \mp \sqrt{14}x + 1 = 0.$$