Question

$\text{If \: \alpha,\beta \: are \: roots of equation } \\ \text{ x^2+px+1;\gamma,\delta the roots of equation }\\\text{ x^2+qx+1=0 , then find (\alpha-\gamma)(\alpha+\delta)(\beta-\gamma)(\beta+\delta) }$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto \alpha,\beta \: are \: \: roots \: of \: x^2+px+1$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \:\alpha + \beta = - \dfrac{p}{1} = - \: p$

Also,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \:\alpha \beta = \dfrac{1}{1} =1$

Also, given that,

$\rm :\longmapsto \: \gamma,\delta \: are \: \: roots \: of \: x^2+qx+1$

So,

$\bf\implies \:\gamma + \delta = - \: \dfrac{q}{1} = - q$

and

$\bf\implies \:\gamma \delta = \: \dfrac{1}{1} = 1$

Now, Consider

$\rm :\longmapsto\:(\alpha-\gamma)(\alpha+\delta)(\beta-\gamma)(\beta+\delta)$

can be re-arranged as

$\rm :\longmapsto\:(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)$

$\rm \: = \:(\alpha\beta - \alpha\gamma - \beta\gamma + {\gamma}^{2})(\beta\alpha + \alpha\delta + \beta\delta + {\delta}^{2} )$

$\rm \: = \:(\alpha\beta -[ \alpha + \beta]\gamma + {\gamma}^{2})(\beta\alpha + [\alpha + \beta]\delta + {\delta}^{2} )$

On substituting the values, we get

$\rm \: = \:(1 + p\gamma + {\gamma}^{2})(1 - p\delta + {\delta}^{2})$

$\rm \: = \:( p\gamma - q\gamma)( - q\delta- p\delta )$

$\red{\bigg \{ \because \: {\gamma}^{2} + q\gamma + 1 = 0 \: and \: {\delta}^{2} + q\delta + 1 = 0 \bigg \}}$

$\rm \: = \: - \gamma(p - q)\delta(p + q)$

$\rm \: = \:\gamma\delta(q - p)(q + p)$

$\rm \: = \:1 \times ( {q}^{2} - {p}^{2} )$

$\rm \: = \: {q}^{2} - {p}^{2}$

$\red{\rm \implies\:\boxed{ \sf{ \: (\alpha-\gamma)(\alpha+\delta)(\beta-\gamma)(\beta+\delta) = {q}^{2} - {p}^{2} }}}$