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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