Question

If a, b are the roots of x² + px + 1 = 0 and g, d are the roots of x² + qx + 1 = 0, then q² – p² is equal to –

• A. (a – g) (b – g) (a + d) (b + d)
• B. (a + g) (b + g) (a – d) (b + d)
• C. (a + g) (b + g) (a + d) (b + d)
• D. None of the above

Answer 1

Answer: (A)

(a – g) (b – g) (a + d) (b + d)

Answer 2

Since, a and b are the roots of equation

x² + px + 1 = 0

Ž a + b = – p, ab = 1

and g, d are the roots of the equation x² + qx + 1 = 0

Ž g + d = – q, gd = 1

Now, (a – g) (b – g) (a + d) (b + d)

= {(ab – g (a + b) + g²}{ab + d (a + b) + d²}

= (1 + pg + g²) (1 – pd + d²)

= (pg – qg) (– pd – qd)

$\left\{ \begin{array} { l } \text { Since, } \gamma \text { is a root of } \mathrm { x } ^ { 2 } + \mathrm { qx } + 1 = 0 \\ \Rightarrow \quad \gamma ^ { 2 } + \mathrm { q } \gamma + 1 = 0 \Rightarrow \gamma ^ { 2 } + 1 = - \mathrm { q } \gamma \\ \text { and similarly, } \delta ^ { 2 } + 1 = - \mathrm { q } \delta \end{array} \right\}$

= – gd (p – q) (p + q) = q² – p² [Q gd = 1]