Question

If a and b are the roots of the equation $x^2+px+1=0$ and ; c and d are the roots of the equation $x^2+qx+1=0$, then (a-c)(b-c)(a+d)(b+d)=?

• A. p²-q²
• B. a²+b²
• C. q²-p²
• D. q²+p²

Answer 1

The correct option is C q²-p²

Shortcut: Assumption

Product of roots = 1 is the only constraint.

Assume values that satisfy the constraint and substitute, take a= -1 and b= -1, then p= 2

Assume c= $\frac{-1}{2}$and d = - 2; q = $\frac{5}{2}$

Question is = (a-c)(b-c)(a+d)(b+d) = $\frac{9}{4}$

Only option (c) satisfies this $\Rightarrow q^2-p^2=\frac{25}{4}-4=\frac{9}{4}$