Question

If α, β are the roots of the equation \[x^2 + px + q = 0 \text { then } - \frac{1}{\alpha} + \frac{1}{\beta}\] are the roots of the equation

Options

• \[x^2 - px + q = 0\]

• \[x^2 + px + q = 0\]

• \[q x^2 + px + 1 = 0\]

• \[q x^2 - px + 1 = 0\]

Answer 1

\[q x^2 - px + 1 = 0\]

Given equation: 

\[x^2 + px + q = 0\]

Also, 

\[\alpha\] and \[\beta\] are the roots of the given equation.
Then, sum of the roots = \[\alpha + \beta = - p\]

Product of the roots = \[\alpha\beta = q\]

Now, for roots 

\[- \frac{1}{\alpha} , - \frac{1}{\beta}\] , we have:

Sum of the roots = \[- \frac{1}{\alpha} - \frac{1}{\beta} = - \frac{\alpha + \beta}{\alpha\beta} = - \left( \frac{- p}{q} \right) = \frac{p}{q}\]

Product of the roots = \[\frac{1}{\alpha\beta} = \frac{1}{q}\]

Hence, the equation involving the roots \[- \frac{1}{\alpha}, - \frac{1}{\beta}\] is as follows:

\[x^2 - \left( \alpha + \beta \right)x + \alpha\beta = 0\]

\[\Rightarrow x^2 - \frac{p}{q}x + \frac{1}{q} = 0\]

\[ \Rightarrow q x^2 - px + 1 = 0\]