If alpha, beta are the roots of the equation | vedclass

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Question

(C) Given that $\alpha + \beta = -p, \alpha \beta = 1$ and $\gamma + \delta = -q, \gamma \delta = 1$.Consider the expression $(\alpha - \gamma)(\beta

Vedclass · Accepted Answer

$(\alpha + \gamma)(\beta + \gamma)(\alpha + \delta)(\beta + \delta)$

Vedclass · Answer

(C) Given that $\alpha + \beta = -p, \alpha \beta = 1$ and $\gamma + \delta = -q, \gamma \delta = 1$.Consider the expression $(\alpha - \gamma)(\beta - \gamma)(\alpha + \delta)(\beta + \delta)$.$= \{\alpha \beta - \gamma(\alpha + \beta) + \gamma^2\} \{\alpha \beta + \delta(\alpha + \beta) + \delta^2\}$$= (1 + p\gamma + \gamma^2)(1 - p\delta + \delta^2)$Since $\gamma$ is a root of $x^2 + qx + 1 = 0$,we have $\gamma^2 + q\gamma + 1 = 0$,so $\gamma^2 + 1 = -q\gamma$.Similarly,$\delta^2 + 1 = -q\delta$.Substituting these into the expression:$= (p\gamma - q\gamma)(-p\delta - q\delta) = \gamma(p - q) \cdot \delta(-p - q)$$= -\gamma \delta (p - q)(p + q) = -(1)(p^2 - q^2) = q^2 - p^2$.

List-$I$	List-$II$
$(i) \alpha = \beta$	$(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii) \alpha = 2\beta$	$(B) 2b^2 = 9ac$
$(iii) \alpha = 3\beta$	$(C) b^2 = 6ac$
$(iv) \alpha = \beta^2$	$(D) 3b^2 = 16ac$
	$(E) b^2 = 4ac$
	$(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$

If $\alpha, \beta$ are the roots of the equation $x^2 + px + 1 = 0$ and $\gamma, \delta$ are the roots of the equation $x^2 + qx + 1 = 0$,then $q^2 - p^2 = \dots$

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