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Question
If α, β are the roots of the equation \[x^2 + px + 1 = 0; \gamma, \delta\] the roots of the equation \[x^2 + qx + 1 = 0, \text { then } (\alpha - \gamma)(\alpha + \delta)(\beta - \gamma)(\beta + \delta) =\]
Options
\[q^2 - p^2\]
\[p^2 - q^2\]
\[p^2 + q^2\]
none of these.
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Solution
\[q^2 - p^2\]
Given:
\[\alpha \text { and } \beta\] are the roots of the equation \[x^2 + px + 1 = 0\].
Also,
\[\gamma \text { and } \delta\] are the roots of the equation \[x^2 + qx + 1 = 0\].
Then, the sum and the product of the roots of the given equation are as follows:
\[\alpha + \beta = - \frac{p}{1} = - p\]
\[\alpha\beta = \frac{1}{1} = 1\]
\[\gamma + \delta = - \frac{q}{1} = - q\]
\[\gamma\delta = \frac{1}{1} = 1\]
\[\text { Moreover,} (\gamma + \delta )^2 = \gamma^2 + \delta^2 + 2\gamma\delta\]
\[ \Rightarrow \gamma^2 + \delta^2 = q^2 - 2\]
\[\therefore (\alpha - \gamma) (\alpha + \delta) (\beta - \gamma) (\beta + \delta) = (\alpha - \gamma) (\beta - \gamma) (\alpha + \delta) (\beta + \delta)\]
\[ = \left( \alpha\beta - \alpha\gamma - \beta\gamma + \gamma^2 \right)\left( \alpha\beta + \alpha\delta + \beta\delta + \delta^2 \right)\]
\[ = \left[ \alpha\beta - \gamma\left( \alpha + \beta \right) + \gamma^2 \right] \left[ \alpha\beta + \delta \left( \alpha + \beta \right) + \delta^2 \right]\]
\[ = (1 - \gamma( - p) + \gamma^2 ) (1 + \delta( - p) + \delta^2 )\]
\[ = (1 + \gamma p + \gamma^2 ) (1 - \delta p + \delta^2 )\]
\[ = 1 - p\delta + \delta^2 + p\gamma - p^2 \gamma\delta + p\gamma \delta^2 + \gamma^2 - p\delta \gamma^2 + \gamma^2 \delta^2 \]
\[ = 1 - p\delta + p\gamma + \delta^2 - p^2 \gamma\delta + p\gamma \delta^2 + \gamma^2 - p\delta \gamma^2 + \gamma^2 \delta^2 \]
\[ = 1 - p(\delta - \gamma) - p^2 \gamma\delta + p\gamma\delta (\delta - \gamma) + ( \gamma^2 + \delta^2 ) + 1\]
\[ = 1 - p^2 \gamma\delta + p\gamma\delta (\delta - \gamma) - p(\delta - \gamma) + ( \gamma^2 + \delta^2 ) + 1\]
\[ = 1 - p^2 + (\delta - \gamma) p (\gamma\delta - 1) + q^2 - 2 + 1\]
\[ = - p^2 + (\delta - \gamma) p (1 - 1) + q^2 \]
\[ = q^2 - p^2\]
