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(B) Given the equation $x^2-ax-b=0$,the roots $\alpha, \beta$ satisfy $\alpha+\beta=a$ and $\alpha\beta=-b$.Since $P_n = \alpha^n - \beta^n$,we have t

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(B) Given the equation $x^2-ax-b=0$,the roots $\alpha, \beta$ satisfy $\alpha+\beta=a$ and $\alpha\beta=-b$.Since $P_n = \alpha^n - \beta^n$,we have the recurrence relation $P_n = aP_{n-1} + bP_{n-2}$.Using $P_6 = aP_5 + bP_4$,we get $45\sqrt{7}i = a(11\sqrt{7}i) + b(-3\sqrt{7}i)$,which simplifies to $11a - 3b = 45$.Using $P_5 = aP_4 + bP_3$,we get $11\sqrt{7}i = a(-3\sqrt{7}i) + b(-5\sqrt{7}i)$,which simplifies to $-3a - 5b = 11$.Solving these equations,we find $a=3$ and $b=-4$.We need to find $|\alpha^4+\beta^4|$. Note that $(\alpha^4+\beta^4)^2 = (\alpha^4-\beta^4)^2 + 4\alpha^4\beta^4 = P_4^2 + 4(\alpha\beta)^4$.Substituting the values,$P_4^2 = (-3\sqrt{7}i)^2 = 9 	imes 7 	imes (-1) = -63$.Also,$4(\alpha\beta)^4 = 4(-b)^4 = 4(-(-4))^4 = 4(256) = 1024$.Thus,$(\alpha^4+\beta^4)^2 = -63 + 1024 = 961$.Therefore,$|\alpha^4+\beta^4| = \sqrt{961} = 31$.

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

Let $\alpha, \beta$ be the roots of the equation $x^2-ax-b=0$ with $\operatorname{Im}(\alpha) < \operatorname{Im}(\beta)$. Let $P_n=\alpha^n-\beta^n$. If $P_3=-5 \sqrt{7} i, P_4=-3 \sqrt{7} i, P_5=11 \sqrt{7} i$ and $P_6=45 \sqrt{7} i$,then $|\alpha^4+\beta^4|$ is equal to . . . . . .

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