Question

If \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + 3x - 4 = 0 \), then \( \frac{1}{\alpha} + \frac{1}{\beta} \) is equal to:

• A. \( \frac{-3}{4} \)
• B. \( \frac{3}{4} \)
• C. \( \frac{-4}{3} \)
• D. \( \frac{4}{3} \)
• E. \( \frac{3}{2} \)

Hint:

For a quadratic equation \(ax^2 + bx + c = 0\), the sum of reciprocals of roots is: \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = -\frac{b}{c} \] provided \( \alpha, \beta \neq 0 \).

Answer 1

The Correct Option is B

Concept: The sum of the reciprocals of the roots is given by: \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \]

Step 1: Find the sum and product of the roots.
Given quadratic equation: \[ x^2 + 3x - 4 = 0 \] Compare with standard form \(ax^2 + bx + c = 0\), where: \[ a = 1,\quad b = 3,\quad c = -4 \] Using Vieta’s formulas: Sum of roots: \[ \alpha + \beta = -\frac{b}{a} = -\frac{3}{1} = -3 \] Product of roots: \[ \alpha \beta = \frac{c}{a} = \frac{-4}{1} = -4 \] Thus, \[ \alpha + \beta = -3, \quad \alpha\beta = -4 \]

Step 2: Calculate the required expression.
We use the identity: \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \] Substituting values: \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{-3}{-4} \] Now simplify the fraction: \[ \frac{-3}{-4} = \frac{3}{4} \] Hence, \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4} \] Final Answer: \[ \boxed{\frac{3}{4}} \]