Question

If \(a,b\) are the roots of \(x^2+px+1=0\) and \(c,d\) are the roots of \(x^2+qx+1=0\), the value of \(E=(a-c)(b-c)(a+d)(b+d)\) is

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

Hint:

For quadratic root questions, use: \[ \alpha+\beta=-\frac{b}{a},\qquad \alpha\beta=\frac{c}{a} \] and avoid full expansion unless necessary.

Answer 1

The Correct Option is A

Concept:
For a quadratic equation: \[ x^2+px+1=0 \] if roots are \(a,b\), then: \[ a+b=-p \] and \[ ab=1 \] Similarly, for: \[ x^2+qx+1=0 \] if roots are \(c,d\), then: \[ c+d=-q \] and \[ cd=1 \]

Step 1: Use roots of first equation.
Since \(a,b\) are roots of: \[ x^2+px+1=0 \] For any value \(x=c\): \[ (c-a)(c-b)=c^2+pc+1 \] Now: \[ (a-c)(b-c)=(c-a)(c-b) \] So: \[ (a-c)(b-c)=c^2+pc+1 \]

Step 2: Use relation for \(c\).
Since \(c\) is a root of: \[ x^2+qx+1=0 \] we have: \[ c^2+qc+1=0 \] So: \[ c^2+1=-qc \] Then: \[ c^2+pc+1=(p-q)c \] Therefore: \[ (a-c)(b-c)=(p-q)c \]

Step 3: Evaluate \((a+d)(b+d)\).
Using roots \(a,b\): \[ (a+d)(b+d)=ab+d(a+b)+d^2 \] Now: \[ ab=1,\quad a+b=-p \] So: \[ (a+d)(b+d)=1-pd+d^2 \] Since \(d\) is a root of: \[ x^2+qx+1=0 \] \[ d^2+qd+1=0 \] So: \[ d^2+1=-qd \] Hence: \[ 1-pd+d^2=(-q-p)d \] \[ (a+d)(b+d)=-(p+q)d \]

Step 4: Multiply both parts.
\[ E=(a-c)(b-c)(a+d)(b+d) \] \[ E=(p-q)c\cdot [-(p+q)d] \] \[ E=-(p-q)(p+q)cd \] Since: \[ cd=1 \] \[ E=-(p^2-q^2) \] \[ E=q^2-p^2 \]

Step 5: Compare with the given answer key.
The direct algebraic simplification gives: \[ q^2-p^2 \] However, the provided answer key marks: \[ p^2-q^2 \] So there may be a sign issue in the printed expression or option key. If the expression is exactly: \[ (a-c)(b-c)(a+d)(b+d) \] then the calculated value is: \[ q^2-p^2 \] Hence, mathematically the answer is: \[ \boxed{(B)\ q^2-p^2} \]