Multiple choice

If the roots of the equation $\dfrac{x^{2}-bx}{ax-c}=\dfrac{m-1}{m+1}$ are equal but opposite in sign, then the value of $m$ will be

  1. $\dfrac{a-b}{a+b}$
  2. $\dfrac{ b-a}{a+b}$
  3. $\dfrac{a+b}{a-b}$
  4. $\dfrac{b+a}{b-a}$
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A Correct answer
Explanation

For the roots to be equal but opposite in sign, the sum of the roots must be zero. Rearranging the equation to standard quadratic form ax^2 + bx + c = 0, the coefficient of the linear term must be zero, which leads to the ratio m = (a-b)/(a+b).

AI explanation

By cross-multiplying the given equation, we get (m + 1)x^2 - (m + 1)bx - (m - 1)ax + (m - 1)c = 0, which simplifies to (m + 1)x^2 - [b(m + 1) + a(m - 1)]x + c(m - 1) = 0. Since the roots are equal in magnitude but opposite in sign, the sum of the roots must be zero, meaning the coefficient of x must equal zero. Setting the coefficient of x to zero gives b(m + 1) + a(m - 1) = 0, which expands to bm + b + am - a = 0. Factoring out m yields m(a + b) = a - b, so the value of m is (a - b)/(a + b).

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