Let p, q and r be real numbers (p [ ] q, r [ ] 0), such that the roots of the equation [1 …

Question

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

Accepted Answer

Given,

$${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$$

$$ \Rightarrow $$ $\,\,\,$${{x + p + x + q} \over {\left( {x + p} \right)\left( {x + q} \right)}} = {1 \over r}$$ \Rightarrow $$ $\,\,\,$$ (2x + p + q) r = x2 + px + qx + pq

$$ \Rightarrow $$ $\,\,\,$x2 + (p + q$-$2r) x + pq$-$pr$-$qr = 0

Let$\alpha $and$\beta $are the roots,$\therefore\,\,\,$$\alpha $+$\beta $=$-$(p + q$-$2r)

and$\alpha $$\beta $= pq$-$pr$-$qr

Given that,$\alpha $=$-$$\beta $$ \Rightarrow $$\alpha $+$\beta $= 0$\therefore\,\,\,$$-$(p + q$-$2r) = 0

Now,$\alpha $2 +$\beta $2

= ($\alpha $+$\beta $)2$-$2$\alpha $$\beta $= ($-$(p + q$-$2r))2$-$2 (pq$-$pr$-$qr)

= p2 +q2 + 4r2 + 2pq$-$4pr$-$4qr$-$2pq + 2pr + 2qr

= p2 + q2 + 4r2$-$2pr$-$2qr

= p2 + q2$-$2r (p + q$-$2r)

= p2 + q2$-$

Solutions · 1