An algebra problem by Yeldo Pailo

Let $r_1,r_2$ be the two (nonzero) roots of the general quadratic equation $ax^2+bx+c=0$ . Start by using the quadratic formula to derive the following formula: $\frac{r_1}{r_2}+\frac{r_2}{r_1}=\frac{b^2}{ac}-2$ . Apply this formula to the given equation $\lambda x^2+(1-\lambda)x+5=0$ to find $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{(1-\lambda)^2}{5\lambda}-2=4\Rightarrow \lambda^2-32\lambda+1=0$ . Applying the formula once more, one finds $\frac{\lambda_1}{\lambda_2}+\frac{\lambda_2}{\lambda_1}=\frac{(-32)^2}{(1)(1)}-2=1022$ .

The above quadratic equation is expressible as:

$(x-\alpha)(x-\beta) = x^2 + \frac{1-\lambda}{\lambda}x + \frac{5}{\lambda} = 0 \Rightarrow \alpha + \beta = \frac{\lambda-1}{\lambda}, \alpha\beta = \frac{5}{\lambda}$ .

If $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = 4$ holds true, then:

$\frac{\alpha^{2} + \beta^{2}}{\alpha\beta} = \frac{(\frac{\lambda-1}{\lambda})^{2} - 2\alpha\beta}{\alpha\beta} = \frac{(\frac{\lambda-1}{\lambda})^{2} - 2\frac{5}{\lambda}}{\frac{5}{\lambda}} = 4$ ,

or $\lambda^{2} - 32\lambda + 1 = 0 \Rightarrow \lambda = 16 \pm \sqrt{255}$ .

So we finally compute:

$\frac{\lambda_{1}^{2} + \lambda_{2}^{2}}{\lambda_1\lambda_2} = \frac{(16+\sqrt{255})^{2} + (16-\sqrt{255})^{2}}{(16+\sqrt{255})(16-\sqrt{255})} = \frac{2 \cdot (256 + 255)}{256 - 255} = \boxed{1022}.$