Question

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 $\frac{\alpha+\beta}{\gamma+\delta} =$

• A. $65/6$
• B. $13/2$
• C. $17/15$
• D. $15/13$

Hint:

Recognize reciprocal equations by their symmetric coefficients. The standard technique is to divide by the middle power of $x$ (here $x^2$) and use the substitution $y=x+1/x$ to reduce it to a simpler equation.

Answer 1

The Correct Option is D

The given equation is a reciprocal equation of type I, since the coefficients are symmetric.
$12x^4-56x^3+89x^2-56x+12=0$.
Divide the equation by $x^2$ (since $x=0$ is not a root):
$12x^2 - 56x + 89 - \frac{56}{x} + \frac{12}{x^2} = 0$.
Group the terms: $12(x^2 + \frac{1}{x^2}) - 56(x + \frac{1}{x}) + 89 = 0$.
Let $y = x + \frac{1}{x}$. Then $y^2 = x^2 + 2 + \frac{1}{x^2}$, which means $x^2 + \frac{1}{x^2} = y^2 - 2$.
Substitute this into the equation:
$12(y^2 - 2) - 56y + 89 = 0$.
$12y^2 - 24 - 56y + 89 = 0$.
$12y^2 - 56y + 65 = 0$.
We solve this quadratic equation for $y$:
$y = \frac{-(-56) \pm \sqrt{(-56)^2 - 4(12)(65)}}{2(12)} = \frac{56 \pm \sqrt{3136 - 3120}}{24} = \frac{56 \pm \sqrt{16}}{24} = \frac{56 \pm 4}{24}$.
The two possible values for $y$ are:
$y_1 = \frac{56+4}{24} = \frac{60}{24} = \frac{5}{2}$.
$y_2 = \frac{56-4}{24} = \frac{52}{24} = \frac{13}{6}$.
The roots of the original equation come in reciprocal pairs. Let the pairs be $(\alpha, \beta)$ and $(\gamma, \delta)$. Given $\alpha\beta = 1$ and $\gamma\delta = 1$. So $\beta=1/\alpha$ and $\delta=1/\gamma$. The sums are $\alpha+\beta = \alpha + 1/\alpha$ and $\gamma+\delta = \gamma+1/\gamma$. These sums are the values of $y$ we found. So, $\{\alpha+\beta, \gamma+\delta\} = \{\frac{5}{2}, \frac{13}{6}\}$.
We are given the condition $\frac{\alpha+\beta}{\gamma+\delta}>1$, which implies $\alpha+\beta>\gamma+\delta$.
Let's compare the two values: $\frac{5}{2} = \frac{15}{6}$. Since $\frac{15}{6}>\frac{13}{6}$, we must have $\alpha+\beta = \frac{15}{6} = \frac{5}{2}$ and $\gamma+\delta = \frac{13}{6}$.
The required ratio is $\frac{\alpha+\beta}{\gamma+\delta} = \frac{5/2}{13/6} = \frac{5}{2} \times \frac{6}{13} = \frac{30}{26} = \frac{15}{13}$.