Question

If -1+p+q=1-r, where p,q,and r are constants, solve the given relation.
$\frac{\sqrt{x-1}-q-r}{p} +\frac{\sqrt{x-1}-r-p}{q}+\frac{\sqrt{x-1}-p-q }{r}=3$

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: - 1 + p + q = 1 - r$

$\rm\implies \:\boxed{ \rm{ \:p + q + r = 2 \: \: }} - - - (1)$

Now, Consider

$\rm \: \frac{\sqrt{x-1}-q-r}{p} +\frac{\sqrt{x-1}-r-p}{q}+\frac{\sqrt{x-1}-p-q }{r}=3$

can be re-arranged as

$\rm \: \frac{\sqrt{x-1}-q-r}{p} +\frac{\sqrt{x-1}-r-p}{q}+\frac{\sqrt{x-1}-p-q }{r}=1 + 1 + 1$

$\rm \: \frac{\sqrt{x-1}-q-r}{p} +\frac{\sqrt{x-1}-r-p}{q}+\frac{\sqrt{x-1}-p-q }{r} - 1 - 1 - 1 = 0$

$\rm \: \frac{\sqrt{x-1}-q-r}{p} - 1 +\frac{\sqrt{x-1}-r-p}{q} - 1+\frac{\sqrt{x-1}-p-q }{r} - 1 = 0$

$\rm \: \frac{\sqrt{x-1}-q-r - p}{p} +\frac{\sqrt{x-1}-r-p - q}{q}+\frac{\sqrt{x-1}-p-q - r}{r}= 0$

$\rm \: \frac{\sqrt{x-1}-(q + r + p)}{p} +\frac{\sqrt{x-1}-(r + p + q)}{q}+\frac{\sqrt{x-1}-(p + q + r)}{r}= 0$

$\rm \: \frac{\sqrt{x-1}-2}{p} +\frac{\sqrt{x-1}-2}{q}+\frac{\sqrt{x-1}-2}{r}= 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{[ \: \because \: \rm \: p + q + r = 2 \: ]}$

$\rm \: ( \sqrt{x - 1} - 2)\bigg(\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} \bigg) = 0$

$\rm\implies \:\rm \: \sqrt{x - 1} - 2 = 0$

$\rm \: \sqrt{x - 1} =2$

On squaring both sides, we get

$\rm \: x - 1 = 4$

$\rm \: x = 4 + 1$

$\rm\implies \:\boxed{ \rm{ \:x = 5 \: }}$

Verification :-

Consider, LHS

$\rm \: \frac{\sqrt{x-1}-q-r}{p} +\frac{\sqrt{x-1}-r-p}{q}+\frac{\sqrt{x-1}-p-q }{r}$

On substituting x = 5, we get

$\rm \: = \frac{\sqrt{5-1}-q-r}{p} +\frac{\sqrt{5-1}-r-p}{q}+\frac{\sqrt{5-1}-p-q }{r}$

$\rm \: = \frac{\sqrt{4}-q-r}{p} +\frac{\sqrt{4}-r-p}{q}+\frac{\sqrt{4}-p-q }{r}$

$\rm \: = \frac{2-q-r}{p} +\frac{2-r-p}{q}+\frac{2-p-q }{r}$

$\rm \: = \frac{p}{p} +\frac{q}{q}+\frac{r}{r}$

$\rm \: = \: 1 + 1 + 1$

$\rm \: = \: 3$

Hence, Verified