Q. 1 यदि `\frac{3(x^(2)+1)-7x}{3x}` = 6 , `x\ne0` है तो , `\sqrt x\+\frac1{\sqrt x}` क्या होगा ?
[ SSC CGL MAINS 2019 ]
Solution - `\frac{3x^(2)-7x+3 }{3x}` = 6
`3x^(2)-7x+3=18x`
`3x^(2)-7x- 18x+3=0`
`3x^(2)-25x+3= 0 `
Deviding by x bothsides we have
`3x-25+\frac3x=0`
`3x+\frac3x=25`
`3(x+\frac1x)=25`
`( x + \frac1x) = \frac{25}3`
On adding +2 bothsides we have
`( x + \frac1x)+ 2 = \frac{25}3+2`
`(\sqrt x)^2 + (\frac{1}\sqrt x )^2 + 2\times (\sqrt x) \times (frac{1}\sqrt x) = \frac {31}3`
`(\sqrt x + \ frac {1}\sqrt x) ^2 = \frac {31}3`
On taking roots bothsides we have ...............
`( \sqrtx + \frac{1}\sqrtx ) = \sqrt\frac{31}3`
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