Question

If \(\alpha,\beta,\gamma\) are the roots of the equation \[ 24x^3-10x^2-3x+1=0, \] then \[ \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} = \]

• A. \(38\)
• B. \(12\)
• C. \(29\)
• D. \(16\)

Hint:

For questions involving reciprocals of roots, first find \(\alpha+\beta+\gamma\), \(\alpha\beta+\beta\gamma+\gamma\alpha\), and \(\alpha\beta\gamma\) using Vieta's formulas. Then express reciprocal sums in terms of these quantities.

Answer 1

The Correct Option is C

Concept: If \(\alpha,\beta,\gamma\) are the roots of \[ ax^3+bx^2+cx+d=0, \] then \[ \alpha+\beta+\gamma=-\frac{b}{a}, \] \[ \alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}, \] \[ \alpha\beta\gamma=-\frac{d}{a}. \] Also, \[ \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} = \left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right)^2 - 2\left(\frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\gamma\alpha}\right). \]

Step 1: Find the symmetric sums of the roots. For \[ 24x^3-10x^2-3x+1=0, \] we have \[ \alpha+\beta+\gamma = \frac{10}{24} = \frac{5}{12}, \] \[ \alpha\beta+\beta\gamma+\gamma\alpha = -\frac{3}{24} = -\frac18, \] \[ \alpha\beta\gamma = -\frac{1}{24}. \]

Step 2: Calculate \(\frac1\alpha+\frac1\beta+\frac1\gamma\). \[ \frac1\alpha+\frac1\beta+\frac1\gamma = \frac{\alpha\beta+\beta\gamma+\gamma\alpha} {\alpha\beta\gamma}. \] \[ = \frac{-\frac18}{-\frac1{24}} = 3. \]

Step 3: Calculate \(\frac1{\alpha\beta}+\frac1{\beta\gamma}+\frac1{\gamma\alpha}\). \[ \frac1{\alpha\beta}+\frac1{\beta\gamma}+\frac1{\gamma\alpha} = \frac{\alpha+\beta+\gamma} {\alpha\beta\gamma}. \] \[ = \frac{\frac5{12}} {-\frac1{24}} = -10. \]

Step 4: Find \(\frac1{\alpha^2}+\frac1{\beta^2}+\frac1{\gamma^2}\). \[ \frac1{\alpha^2}+\frac1{\beta^2}+\frac1{\gamma^2} = 3^2-2(-10). \] \[ =9+20. \] \[ =29. \] \[ \boxed{ \frac1{\alpha^2}+\frac1{\beta^2}+\frac1{\gamma^2}=29 } \] \[ \boxed{\text{Answer = (C)}} \]