SSC CGL CHSL CPO GD MTS Tier-II &Tier -I 2021 Mathematics Full Solution

यदि $3\sqrt\frac{1-a}{a}+9=19-3\sqrt\frac{a}{1-a}$ है तो a का मान क्या है ?

Solve-
$3\sqrt\frac{1-a}{a}+3\sqrt\frac{a}{1-a}= 19-9$

$3(\sqrt\frac{1-a}{a}+\sqrt\frac{a}{1-a})=10$

$\sqrt\frac{1-a}{a}+\sqrt\frac{a}{1-a}=\frac{10}{3}$ ........................(1)
Now we let

$\sqrt\frac{1-a}{a}=y$

Then

$\sqrt\frac{a}{1-a}= \frac{1}{y}$

Put the above value in Equcation No. (1) we have

$y+\frac{1}{y}=\frac{10}{3}$

$\frac{y^2+1}{y}= \frac{10}{3}$

$3y^2+3=10y$

$3y^2-10y+3=0$

$3y^2-9y-y+3=0$

$3y(y-3)-( y-3 )=0$

$(y-3)(3y-1)=0$

Here $(y-3)= 0$

Then $y=3$

& $(3y-1)=0$

$3y=1$

$y=\frac{1}{3}$

Putting the Real Value of $y$ then we have----

$y=3$ & $y = \frac{1}{3}$

$\sqrt\frac{1-a}{a}=3$ & $\sqrt\frac{1-a}{a}=\frac{1}{3}$

On Squaring bothsides we have

$\frac{1-a}{a}=9$ & $\frac{1-a}{a}=\frac{1}{9}$

$1-a= 9a$ & $9-9a= a$

$10a=1$ & $10a=9$

$a=\frac{1}{10}$ & $a=\frac{9}{10}$

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