Multiple choice

If the equation $\dfrac { 1 }{ x } +\dfrac { 1 }{ x+\alpha } =\dfrac { 1 }{ \lambda } +\dfrac { 1 }{ \lambda +\alpha }$ has real roots that are equal in magnitude and opposite in sign, then

  1. ${\lambda}^{2}=3{a}^{2}$
  2. ${\lambda}^{2}=2{a}^{2}$
  3. ${\lambda}^{2}={a}^{2}$
  4. ${a}^{2}=2{\lambda}^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For roots equal in magnitude and opposite in sign, their sum must be zero. Simplifying the equation and setting the coefficient of x to zero gives the required relation a^2 = 2λ^2. Therefore, option D is correct.

AI explanation

The given equation has roots equal in magnitude but opposite in sign, meaning if x is a root, -x is also a root. To solve for the parameters, let the equal and opposite roots be r and -r. By substituting x with r and then -r into the equation 1/x + 1/(x + a) = 1/y + 1/(y + a), and applying the condition for equal and opposite real roots, the discriminant relationship of the resulting polynomial yields the equation a^2 = 2y^2.

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