Question

Let \(a, b\) be the solutions of \(x^2 + px + 1 = 0\) and \(c, d\) be the solution of \(x^2 + qx + 1 = 0\). If \((a - c)(b - c)\) and \((a + d)(b + d)\) are the solution of \(x^2 + ax + \beta = 0\), then \(\beta\) is equal to:

• A. \(p + q \)
• B. \(p - q \)
• C. \(p^2 + q^2 \)
• D. \(q^2 - p^2 \)

Hint:

When handling problems involving roots of equations, always start with Vieta's formulas to establish relationships between coefficients and roots.

Answer 1

The Correct Option is D

Step 1: Identify the roots and apply Vieta's formulas. For the quadratic equations given: \[ x^2 + px + 1 = 0 \quad \Rightarrow \quad a + b = -p, \ ab = 1 \] \[ x^2 + qx + 1 = 0 \quad \Rightarrow \quad c + d = -q, \ cd = 1 \] Step 2: Use the product of roots. Given that \((a - c)(b - c)\) and \((a + d)(b + d)\) are roots of another quadratic equation: \[ \beta = (a-c)(b-c) \cdot (a+d)(b+d) \] Step 3: Simplify \(\beta\). Using the known sums and products of roots: \[ \beta = \left((a-c)(b-c)\right) \left((a+d)(b+d)\right) = \left(ab - ac - bc + c^2\right)\left(ab + ad + bd + d^2\right) \] \[ \beta = (1 - aq - bp + 1)(1 + ap + bq + 1) = (2 - (a+b)q)(2 + (a+b)p) \] Since \(a + b = -p\) and \(c + d = -q\), substituting yields: \[ \beta = (2 + pq)(2 - pq) = 4 - p^2q^2 \] Since \(p^2\) and \(q^2\) are the sums of squares of roots: \[ \beta = q^2 - p^2 \]