Let p, q and r be real numbers (p ≠ q, r ≠ 0), such that the roots of the equation (1/x + ρ) + (1/x + q) = (1/r) are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :

Question

Let p, q and r be real numbers (p ≠ q, r ≠ 0), such that the roots of the equation (1/x + ρ) + (1/x + q) = (1/r) are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to : (A) (p2 + q2/2) (B) p2 + q2 (C) 2(p2 + q2) (D) p2 + q2 + r2

Tardigrade Admin · Accepted Answer

(2x + p + q) r = (x + p) (x + q) x2+ (p + q - 2r) x + pq - pr - qr = 0 p + q = 2r ..............(i) α2 + β2= (α + β)2 - 2 α; β = 0 - 2 [pq - pr - qr] = - 2pq + 2r (p + q) = - 2pq + (p + q)2 = p2 + q2

Q. Let p, q and r be real numbers $(p \neq = q, r \neq = 0)$ , such that the roots of the equation $\frac{1}{x + ρ} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :

$\frac{p ^{2} + q ^{2}}{2}$

$p^{2} + q^{2}$

$2 (p^{2} + q^{2})$

$p^{2} + q^{2} + r^{2}$

$(2 x + p + q) r = (x + p) (x + q)$
$x_{2} + (p + q - 2 r) x + pq - p r - q r = 0$
$p + q = 2 r$ ..............(i)
$α_{2} + β_{2} = (α + β)_{2} - 2 α; β$
$= 0 - 2 [pq - p r - q r] = - 2 pq + 2 r (p + q) = - 2 pq + (p + q)_{2} = p_{2} + q_{2}$

Q. Let p, q and r be real numbers (p=q,r=0), such that the roots of the equation x+ρ1​+x+q1​=r1​ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :

2p2+q2​

p2+q2

2(p2+q2)

p2+q2+r2