Question

$\sqrt{ \sf \dfrac{x}{x - 3} } + \sqrt{ \sf\dfrac{x - 3}{x} } = \sf\dfrac{5}{2}$
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Answer 1

Given :

• $\bold {\sqrt{\dfrac{x}{x-3} }+\sqrt{\dfrac{x-3}{x} }=\dfrac{5}{2} }$, To find the possible values of x ?

Solution :

• Let $\bold{\sqrt{\dfrac{x}{x-3} } }$ be y. Then $\bold{\sqrt{\dfrac{x-3}{x} } }$ will be 1/y.

$\rightarrowtail \bold{y+\dfrac{1}{y} =\dfrac{5}{2} }$

$\rightarrowtail \bold{\dfrac{y(y)+1(1)}{y}=\dfrac{5}{2} }$

$\rightarrowtail \bold{\dfrac{y^{2}+1}{y}=\dfrac{5}{2} }$

$\rightarrowtail \bold{2(y^{2}+1)=5y}$

$\rightarrowtail\bold{2y^{2}+2=5y}$

$\rightarrowtail\bold{2y^{2}-5y+2=0}$

$\mapsto \bold{2y^{2}-4y-y+2=0}$

$\mapsto \bold{2y^{2}-y-4y+2=0}$

$\mapsto \bold{y(2y-1)-2(2y-1)=0}$

$\mapsto \bold{y-2=0\;or\;(2y-1)=0}$

$\Rrightarrow \bold{y=2\;or\;y=\dfrac{1}{2} }$

Hence, we get :-

$\rightarrowtail \bold{\sqrt{\dfrac{x}{x-3} } =2}$

Squaring on both the sides :-

$\rightarrowtail \bold{\dfrac{x}{x-3}=4 }$

$\rightarrowtail \bold{4(x-3)=x}$

$\rightarrowtail \bold{4x-12=x}$

$\rightarrowtail \bold{3x=12}$

$\rightarrowtail \bold{x=\dfrac{12}{3} }$

$\Rrightarrow \bold{x=4}$

And also, we get :-

$\rightarrowtail \bold{\sqrt{\dfrac{x}{x-3} } =\dfrac{1}{2} }$

Squaring on both the sides :-

$\rightarrowtail \bold{\dfrac{x}{x-3}=\dfrac{1}{4} }$

$\rightarrowtail\bold{1(x-3)=4x}$

$\rightarrowtail \bold{x-3=4x}$

$\rightarrowtail \bold{x-4x=3}$

$\rightarrowtail \bold{-3x=3}$

$\Rrightarrow \bold{x=-1}$

Answer :

• The possible value of x are 4 or -1.