Question

If one of the roots of the equation \(6x^3 - 25x^2 + 2x + 8 = 0\) is an integer and \(\alpha>0\), \(\beta<0\) are the other two roots, then \( \frac{4}{\alpha} + \frac{1}{\beta} = \)

• A. 0
• B. 1
• C. -2
• D. 4

Hint:

When dealing with cubic polynomials where one root is known, use Vieta's formulas to find the sum and product of the other two roots. This avoids polynomial long division and quickly sets up a quadratic equation for the remaining roots.

Answer 1

The Correct Option is D

Let \(P(x) = 6x^3 - 25x^2 + 2x + 8 = 0\).
According to the Rational Root Theorem, any integer root of this polynomial must be a divisor of the constant term, 8.
The divisors of 8 are \(\pm 1, \pm 2, \pm 4, \pm 8\).
Let's test these values. Let's try \(x=4\).
\(P(4) = 6(4^3) - 25(4^2) + 2(4) + 8 = 6(64) - 25(16) + 8 + 8\).
\(P(4) = 384 - 400 + 16 = -16 + 16 = 0\).
So, \(x=4\) is the integer root.
Let the three roots be \(r_1=4\), \(r_2=\alpha\), and \(r_3=\beta\).
Using Vieta's formulas for a cubic equation:
Product of the roots: \(r_1 r_2 r_3 = 4\alpha\beta = -d/a = -8/6 = -4/3\).
This gives \(\alpha\beta = (-4/3) / 4 = -1/3\).
Sum of the roots taken two at a time: \(r_1r_2 + r_1r_3 + r_2r_3 = 4\alpha + 4\beta + \alpha\beta = c/a = 2/6 = 1/3\).
\(4(\alpha+\beta) + \alpha\beta = 1/3\).
Substitute \(\alpha\beta = -1/3\): \(4(\alpha+\beta) - 1/3 = 1/3\).
\(4(\alpha+\beta) = 2/3 \implies \alpha+\beta = (2/3)/4 = 1/6\).
Now we need to find \(\alpha\) and \(\beta\) from \(\alpha+\beta=1/6\) and \(\alpha\beta=-1/3\).
The quadratic equation with these roots is \(t^2 - (\alpha+\beta)t + \alpha\beta = 0\), which is \(t^2 - (1/6)t - 1/3 = 0\).
Multiplying by 6 gives \(6t^2 - t - 2 = 0\).
Factoring this: \((3t-2)(2t+1) = 0\). The roots are \(t=2/3\) and \(t=-1/2\).
Given \(\alpha>0\) and \(\beta<0\), we have \(\alpha = 2/3\) and \(\beta = -1/2\).
Finally, we calculate the required expression: \(\frac{4}{\alpha} + \frac{1}{\beta}\).
\( \frac{4}{2/3} + \frac{1}{-1/2} = 4 \cdot \frac{3}{2} - 2 = 6 - 2 = 4 \).