Question

If \( \alpha, \beta \) are the real roots of the equation \( x^2 + ax + b = 0 \). If \( \alpha + \beta = \frac{1}{2} \) and \( \alpha^3 + \beta^3 = \frac{37}{8} \), then \( {a}-{\frac{1}{b}} =\)

• A. \(-\frac{1}{6}\)
• B. \(\frac{3}{2}\)
• C. \(-\frac{3}{2}\)
• D. \(\frac{1}{6}\)

Hint:

When working with roots and their properties in polynomial equations, applying Vieta's formulas correctly and ensuring careful handling of algebraic manipulation is essential for accurate solutions.

Answer 1

The Correct Option is A

Step 1: Use Vieta's formulas to find \(a\) and \(b\). From Vieta's formulas for a quadratic equation \(x^2 + ax + b = 0\): \[ \alpha + \beta = -a \quad \text{and} \quad \alpha \beta = b \] Given \(\alpha + \beta = \frac{1}{2}\), we find: \[ a = -\left(\frac{1}{2}\right) = -\frac{1}{2} \] 

Step 2: Calculate \(b\) using the identity for the sum of cubes. Using the formula: \[ \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta) \] Substituting the known values: \[ \frac{37}{8} = \left(\frac{1}{2}\right)^3 - 3b\left(\frac{1}{2}\right) \] Simplify and solve for \(b\): \[ \frac{37}{8} = \frac{1}{8} - \frac{3b}{2} \Rightarrow \frac{36}{8} = -\frac{3b}{2} \Rightarrow b = -\frac{36 \times 2}{8 \times 3} = -3 \] 

Step 3: Calculate \( {a} - {\frac{1}{b}} \). Now, substitute \(a\) and \(b\) into the expression: \[ a - \frac{1}{b} = -\frac{1}{2} - \frac{1}{-3} = -\frac{1}{2} + \frac{1}{3} \] To combine the fractions: \[ = \frac{-3 + 2}{6} = -\frac{1}{6} \] 

Conclusion: Thus, \( {a} - {\frac{1}{b}} = -\frac{1}{6} \), which matches the correct answer option (1)