Question

The roots of the given equation \( (p - q)x^2 + (q - r)x + (r - p) = 0 \) are:

• A. \( \frac{p - q}{r - p}, 1 \)
• B. \( \frac{q - r}{p - q}, 1 \)
• C. \( \frac{r - p}{p - q}, 1 \)
• D. None of these

Hint:

For solving quadratic equations, always compare the given equation with the standard form and apply the quadratic formula correctly.

Answer 1

The Correct Option is C

The given quadratic equation is: \[ (p - q)x^2 + (q - r)x + (r - p) = 0 \] Step 1: Comparing with the standard quadratic equation. A quadratic equation is generally given as: \[ ax^2 + bx + c = 0 \] From the given equation: - \( a = (p - q) \) - \( b = (q - r) \) - \( c = (r - p) \) Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Step 2: Substituting values. \[ x = \frac{-(q - r) \pm \sqrt{(q - r)^2 - 4(p - q)(r - p)}}{2(p - q)} \] Expanding the discriminant: \[ (q - r)^2 - 4(p - q)(r - p) = (q - r)^2 - 4(p - q)(r - p) \] Solving the quadratic equation, one root simplifies to: \[ x = \frac{r - p}{p - q} \] The other root is: \[ x = 1 \] Thus, the roots of the equation are \( \frac{r - p}{p - q} \) and 1.