Multiple choice

If $x$ and $y$ are the roots of the equation $x^2 + bx + 1= 0$, then the value of $\dfrac{1}{x+b} + \dfrac{1}{y+b}$ is

  1. $\dfrac{1}{b}$
  2. $b$
  3. $\dfrac{1}{2b}$
  4. $2b$
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B Correct answer
Explanation

Given x, y are roots of x^2 + bx + 1 = 0, we have x+y = -b and xy = 1. The expression is (x+b+y+b) / ((x+b)(y+b)) = (x+y+2b) / (xy + b(x+y) + b^2). Substituting values: (-b + 2b) / (1 + b(-b) + b^2) = b / (1 - b^2 + b^2) = b/1 = b.

AI explanation

For x^2 + bx + 1 = 0, Vieta's formulas state that x + y = -b and x*y = 1. The expression 1 / (x + b) can be rewritten by adding and subtracting y in the denominator to become 1 / (x + b + y - y), which equals 1 / ((x + y) + b - y). Substituting the sum of the roots yields 1 / (-b + b - y) = -1 / y. Applying this to the second fraction gives 1 / (y + b + x - x) = 1 / (-b + b - x) = -1 / x. Adding them together results in (-1/y) + (-1/x) = -(x + y) / (x*y). Plugging in the sum and product gives -(-b) / 1 = b.

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