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(D) Given the reciprocal equation $12x^4-56x^3+89x^2-56x+12=0$. Dividing by $x^2$,we get $12(x^2+\frac{1}{x^2})-56(x+\frac{1}{x})+89=0$. Let $u = x+\f

Vedclass · Accepted Answer

$\frac{15}{13}$

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(D) Given the reciprocal equation $12x^4-56x^3+89x^2-56x+12=0$. Dividing by $x^2$,we get $12(x^2+\frac{1}{x^2})-56(x+\frac{1}{x})+89=0$. Let $u = x+\frac{1}{x}$. Then $x^2+\frac{1}{x^2} = u^2-2$. Substituting this,we get $12(u^2-2)-56u+89=0$,which simplifies to $12u^2-56u+65=0$. Solving for $u$ using the quadratic formula: $u = \frac{56 \pm \sqrt{56^2-4(12)(65)}}{2(12)} = \frac{56 \pm \sqrt{3136-3120}}{24} = \frac{56 \pm 4}{24}$. Thus,$u_1 = \frac{60}{24} = \frac{5}{2}$ and $u_2 = \frac{52}{24} = \frac{13}{6}$. Since $\alpha\beta=1$,$\alpha+\beta$ is one of the values of $u$. Let $\alpha+\beta = u_1 = \frac{5}{2}$ and $\gamma+\delta = u_2 = \frac{13}{6}$. Then $\frac{\alpha+\beta}{\gamma+\delta} = \frac{5/2}{13/6} = \frac{5}{2} 	imes \frac{6}{13} = \frac{15}{13}$. Since $\frac{15}{13} > 1$,this satisfies the condition.

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $12x^4-56x^3+89x^2-56x+12=0$ such that $\alpha\beta=\gamma\delta=1$ and $\frac{\alpha+\beta}{\gamma+\delta}>1$,then $\frac{\alpha+\beta}{\gamma+\delta}=$

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