The coefficient of $$x$$ in the equation $$x^2 + px + q = 0$$ was taken as 17 , in place of 13 and its roots were found to be -2 and -15 . If $$\alpha, \beta$$ are the roots of the original equation, then the equation whose roots are $$\frac{\alpha}{\beta}$$ and $$\frac{\beta}{\alpha}$$ is

The constant $$q$$ was taken correctly which can be found from the prouct of the wrong roots obtained. 

$$(-2)(-15) = 30$$, so $$q = 30$$.

The original equation is 

$$x^2 + 13x + 30 = 0$$ 

Factorising we get

$$ (x+3)(x+10) = 0$$

with roots $$\alpha = -3$$ and $$\beta = -10$$. 

Then

$$\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{9 + 100}{30} = \dfrac{109}{30}$$ and the product is 1.

Hence, the required equation is 

$$x^2 - (\text{sum of roots})x + \text{(product of roots)}= 0$$.

$$x^2 - \left(\dfrac{109}{30}\right)x + 1= 0$$.

$$30x^2 - 109x + 30 = 0$$.