Question

If√(1+27/169)=1+x/13,find thw value of x?

Answer 1

Hey dear!


Here is yr answer.....


$= > \sqrt{( \frac{1 + 27}{169} )} = \frac{1 + x}{13} \\ \\ = > \frac{1 + \sqrt{27} }{13} = \frac{1 + x}{13} \\ \\ = > (13)(1 + \sqrt{27} ) = (1 + x)(13) \\ \\ = > 13 + 13 \sqrt{27} = 13 + 13x \\ \\ = > 13 \sqrt{27} - 13x = 13 - 13 \\ \\ = > 13 \sqrt{27} - 13x = 0 \\ \\ = > - 13x = - 13 \sqrt{27} \\ \\ = > x = \frac{ - 13 \sqrt{27} }{ - 13} \\ \\ = > x = \sqrt{27}$

Hope it hlpz....

Answer 2

Given,

$\sf{\sqrt{1+\dfrac{27}{169} }=1+\dfrac{x}{13} }$

Objective :-

To find the value of x.

Solution :-

$\sf{\sqrt{1+\dfrac{27}{169} }=1+\dfrac{x}{13} }\\\\\displaystyle{\sf{\longrightarrow \sqrt{\frac{169+27}{169} } =1+\frac{x}{13} }}\\\\\displaystyle{\sf{\longrightarrow \sqrt{\frac{196}{169}}=\frac{13+x}{13} }}\\\\\displaystyle{\sf{\longrightarrow \frac{14}{13}=\frac{13+x}{13} }}\\\\\displaystyle{\sf{\longrightarrow 14\times13=13(13+x)}}\\\\\displaystyle{\sf{\longrightarrow 182=169+13x}}\\\\\displaystyle{\sf{\longrightarrow 196-182=13x}}\\\\\displaystyle{\sf{\longrightarrow 13=13x}}$

$\large\displaystyle{\sf{\color{lime}{\longrightarrow x=1 }}}$