Question

Question #c0b35

Answer 1

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`

Explanation:

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`

Answer 2

With `x>p and x>q`..
`p+1`, when p=q.
`p+q`, including 2p, when p=q..

Explanation:

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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