Question #c0b35

I have to add a little to the Answer submitted by Respected

Dr. Cawas K.

Dr.'s soln. is #x=(p^2+3pq+4q^2)/(p+3q)#.

This soln. is based upon the assumption that #p+3q!=0#

So, out of curiosity, I tried to see what happens to the Soln. if,

#p+3q=0, or, p=-3q# And, here is the Result :

Letting #p=-3q#, in the eqn., we get,

#sqrt(x+3q)+sqrt(x-q)=(p+q)/sqrt(x-q)=(-2q)/sqrt(x-q)#

#:. sqrt((x+3q)(x-q))+x-q=-2q,# or,

#sqrt((x+3q)(x-q))=-q-x#.

Squaring, #(x+3q)(x-q)=(-q-x)^2#

#:. x^2+2qx-3q^2=q^2+2qx+x^2, or, -4q^2=0 rArr q=0#

Thus, #p=-3q rArr q=0 rArr p=0#.

Hence, in case, #p=-3q=0,# the eqn. becomes,

#sqrtx+sqrtx=0 rArr x=0#.

Thus, the Soln. is : #(1) : x=(p^2+3pq+4q^2)/(p+3q), if p+3qne0#.

#(2) : If p+3q=0, i.e., p=-3q rArr q=0", & the soln. is, "x=0#

I have learnt there is a typographical error in the equation and the

correct equation is

#sqrt (x-p)+sqrt (x-q)=p/sqrt (x-q)+q/sqrt (x-p)#, x> both p and q.

Rearranging,

#sqrt (x-p)-q/sqrt (x-p)=p/sqrt (x-q)-sqrt (x-q)#

Upon squaring.

#x-p+q^2/(x-p)-2q=p^2/(x-q)+x-q-2p#. Upon simplfiication,

#(p-q)(x-p)(x-q)=p^2(x-p)-q^2(x-q)#.

#(p-q)(x^2-2(p+q)x+(p^2+q^2+2pq)=0#

#p-q=0# and/or #x^2-2(p+q)x+p^2 + q^2 +2pq=0#

The zeros of the quadratic are

#x = p+q, p+q#.,

So, the list is

#x=p+1#, when p=q

#x=p+q#, including 2p, when p=q.

I think there are no bugs now, and this is my final edition.
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