Question

Let m, n be the roots of the equation x2 + qx + r = 0 and let *, t be the roots of the equation x2 + bx + c = 0.

*1. If m/n = /t, then**

(A) r2c = qb2

(B) r2b = qc2

(C) rb2 = cq2

(D) rc2 = bq2

2.If mn = st, then q2 − b2 is equal to

(A) [(m + t) + (n + *)][(m + *) − (n + t)]

(B) [(m + t) + (n + *)][(m + *) + (n + t)]

(C) [(m + t) − (n + *)][(m + *) + (n + t)]

(D) [(m + t) − (n + *)][(m + *) − (n + t)]

*3. If m = and rq = bc, then n and t are the roots of the equation**

(A) x2 − (b + q)x + bq = 0

(B) x2 − (b + r)x + rb = 0

(C) x2 − (c + q)x + cq = 0

(D) x2 − (c + r)x + rc = 0

Answer 1

Correct option 1. (C), 2. (D) 3. (A)

1.We have

2. For mn = st,

q2 − b2 = (m + n)2 − ( + t) = (m − n) 2 − ( − t)2

= [(m + t) − (n + *)][(m + *) − (n + t)]

3.With m = , rq = bc, + n = − q, sn = r and * + t = − b, st = c, we have

n − t = b − q and n/t = r/c

⇒ rt/c - t = b - q

⇒ t(r − c) = cb - cq - q(r - c)

⇒ t = q ⇒ n = b

Hence, n and t are the roots of the equation x2 − (b + q)x + bq = 0.