If a and b are the roots of the equation $x^{2} + p x + 1 = 0$ and c and d are the roots of the equation $x^{2} + q x + 1 = 0$ ,then $(a - c) (b - c) (a + d) (b + d) = ?$

Open in App

Solution

$(a - c) (b - c) (a + d) (b + d) = (a b - a c - b c + c^{2}) (a b + a d + b d + d^{2}) = {a b - c (a + b) + c^{2}} {a b + d (a + b) + d^{2}} = (1 + p c + c^{2}) (1 - p d + d^{2}) [a b = c d = 1;] = (c d + p c + c^{2}) (c d - p d + d^{2}) [a + b = - p; c + d = - q] = (d + p + c) (c^{2} d - p c d + c d^{2}) = (d + p + c) (c - p + d) = (- q + p) (- q - p) = (p + q) (q - p) = (q + p) (q - p) = q^{2} - p^{2}$ .

Suggest Corrections

Similar questions

a, b

x^{2} + p x + 1 = 0

c, d

x^{2} + q x + 1 = 0,

(a - c) (b - c) (a + d) (b + d) =

a

b

x^{2} + px + 1 = 0 and c,

x^{2} + qx + 1 = 0

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

a, b

x^{2} + p x + 1 = 0

c, d

x^{2} + q x + 1 = 0

\frac{(a - c) (b - c) (a + d) (b + d)}{(q^{2} - p^{2})} =

If a,b are roots of equation

x²+px-q=0 and

c,d are roots of equation

x²+px+r=0

Prove that

(a-c)(a-d)=(b-c)(b-d)=q+r

Join BYJU'S Learning Program