Question 14

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

Solution

The wrong equation had product of roots $$(-2)(-15) = 30$$, so $$q = 30$$ is correct. The original equation is $$x^2 + 13x + 30 = 0$$ with roots $$\alpha = -3$$ and $$\beta = -10$$. Then $$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{9 + 100}{30} = \frac{109}{30}$$ and the product is 1, so the required equation is $$30x^2 - 109x + 30 = 0$$.

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