Question:
If \( x \) and \( y \) are the roots of the equation \( x^2 + bx + 1 = 0 \), then the value of \( \frac{1}{x+b} + \frac{1}{y+b} \) is:
If \( x \) and \( y \) are the roots of the equation \( x^2 + bx + 1 = 0 \), then the value of \( \frac{1}{x+b} + \frac{1}{y+b} \) is:
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When the variable from the denominator appears in the original equation (like \( x+b \)), try to isolate that term in the quadratic equation first. It often turns a fraction into a simple polynomial.
Updated On: May 1, 2026
- \( \frac{1}{b} \)
- \( b \)
- \( \frac{1}{2b} \)
- \( 2b \)
- \( 1 \)
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The Correct Option is B
Solution and Explanation
Concept: Since \( x \) and \( y \) are roots of \( x^2 + bx + 1 = 0 \), they satisfy the equation. This gives us the identities \( x^2 + bx + 1 = 0 \implies x(x+b) = -1 \) and \( y^2 + by + 1 = 0 \implies y(y+b) = -1 \). These substitutions simplify the denominators significantly.
Step 1: Simplifying the individual terms.
From the root properties: \[ x+b = -\frac{1}{x} \quad \text{and} \quad y+b = -\frac{1}{y} \] Substituting these into the given expression: \[ \frac{1}{x+b} + \frac{1}{y+b} = \frac{1}{-1/x} + \frac{1}{-1/y} = -x - y = -(x+y) \]
Step 2: Using the sum of roots.
For the quadratic \( x^2 + bx + 1 = 0 \), the sum of roots is: \[ x + y = -b \] Substituting this back into our simplified expression: \[ -(x+y) = -(-b) = b \]
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