# Question c0b35

Sep 13, 2016

The Soln. is : $\left(1\right) : x = \frac{{p}^{2} + 3 p q + 4 {q}^{2}}{p + 3 q} , \mathmr{if} p + 3 q \ne 0$.

$\left(2\right) : I f p + 3 q = 0 , i . e . , p = - 3 q \Rightarrow q = 0 \text{, & the soln. is, } x = 0$

#### Explanation:

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

Dr. Cawas K.

Dr.'s soln. is $x = \frac{{p}^{2} + 3 p q + 4 {q}^{2}}{p + 3 q}$.

This soln. is based upon the assumption that $p + 3 q \ne 0$

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

$p + 3 q = 0 , \mathmr{and} , p = - 3 q$ And, here is the Result :

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

$\sqrt{x + 3 q} + \sqrt{x - q} = \frac{p + q}{\sqrt{x - q}} = \frac{- 2 q}{\sqrt{x - q}}$

$\therefore \sqrt{\left(x + 3 q\right) \left(x - q\right)} + x - q = - 2 q ,$ or,

$\sqrt{\left(x + 3 q\right) \left(x - q\right)} = - q - x$.

Squaring, $\left(x + 3 q\right) \left(x - q\right) = {\left(- q - x\right)}^{2}$

$\therefore {x}^{2} + 2 q x - 3 {q}^{2} = {q}^{2} + 2 q x + {x}^{2} , \mathmr{and} , - 4 {q}^{2} = 0 \Rightarrow q = 0$

Thus, $p = - 3 q \Rightarrow q = 0 \Rightarrow p = 0$.

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

$\sqrt{x} + \sqrt{x} = 0 \Rightarrow x = 0$.

Thus, the Soln. is : $\left(1\right) : x = \frac{{p}^{2} + 3 p q + 4 {q}^{2}}{p + 3 q} , \mathmr{if} p + 3 q \ne 0$.

$\left(2\right) : I f p + 3 q = 0 , i . e . , p = - 3 q \Rightarrow q = 0 \text{, & the soln. is, } x = 0$

Sep 14, 2016

With $x > p \mathmr{and} x > q$..
$p + 1$, 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

$\sqrt{x - p} + \sqrt{x - q} = \frac{p}{\sqrt{x - q}} + \frac{q}{\sqrt{x - p}}$, x> both p and q.

Rearranging,

$\sqrt{x - p} - \frac{q}{\sqrt{x - p}} = \frac{p}{\sqrt{x - q}} - \sqrt{x - q}$

Upon squaring.

$x - p + {q}^{2} / \left(x - p\right) - 2 q = {p}^{2} / \left(x - q\right) + x - q - 2 p$. Upon simplfiication,

$\left(p - q\right) \left(x - p\right) \left(x - q\right) = {p}^{2} \left(x - p\right) - {q}^{2} \left(x - q\right)$.

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

$p - q = 0$ and/or ${x}^{2} - 2 \left(p + q\right) x + {p}^{2} + {q}^{2} + 2 p q = 0$

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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