Question

If a, b are zeros of f(x) = x2+ px+1 and c, d are the zeros of g(x) = x2 +qx + 1 then 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) none of these​

Answer 1

Answer:

Option (B), E = q² - p² is correct.

Step-by-step explanation:

The given polynomials are

f(x) = x² + px + 1

g(x) = x² + qx + 1

Since a, b are the zeroes of f(x),

a + b = - p ..... (1)

ab = 1 ..... (2)

Since c, d are the zeroes of g(x),

c + d = - q ..... (3)

cd = 1 ..... (4)

Now, E = (a - c) (b - c) (a + d) (b + d)

= {ab - (a + b)c + c²} {ab + (a + b)d + d²}

= (1 + pc + c²) (1 - pd + d²) [by (1) & (2)]

= 1 - pd + d² + pc - p²cd + pcd² + c² - pc²d + c²d²

= 1 - pd + d² + pc - p²(cd) + p(cd)d + c² - pc(cd) + (cd)²

= 1 - pd + d² + pc - p² + pd + c² - pc + 1 [by (4)]

= 2 + c² + d² - p²

= 2 + (c + d)² - 2(cd) - p²

= 2 + (- q)² - 2 - p² [by (3) & (4)]

= q² - p²

∴ option (B), E = q² - p² is correct.