Question

If the value of x is 8-√63 then find the value of √x-1/√x

Answer 1

If the value of x is 8-√63 then find the value of √x-1/√x

Answer 2

$\star\:\:\:\bf\large\underline\blue{Question:-}$

• If $x = 8 - \sqrt{63}$, find the value of $\sqrt{x} - \frac{1}{ \sqrt{x} }$

$\star\:\:\:\bf\large\underline\blue{Solution:-}$

$x = 8 - \sqrt{63}$

Therefore,

$\frac{1}{x} = \frac{1}{8 - \sqrt{63} }$

$= \frac{1}{8 - \sqrt{63} } \times \frac{8 + \sqrt{63} }{8 + \sqrt{63} }$

$= \frac{8 + \sqrt{63} }{ {(8)}^{2} - {( \sqrt{63} )}^{2} }$

$= \frac{8 + \sqrt{63} }{64 - 63}$

$= \frac{8 + \sqrt{63} }{1}$

$= 8 + \sqrt{63}$

So,

$x + \frac{1}{x} = 8 - \sqrt{63} + 8 + \sqrt{63}$

$= 16$

$\implies x + \frac{1}{x} + 2 = 18$

$\implies {( \sqrt{x} )}^{2} + \frac{1}{ \sqrt{ (x)}} + 2 \times \sqrt{x} \times \frac{1}{ \sqrt{x} } = 18$

$\implies {( \sqrt{x} + \frac{1}{ \sqrt{x} } )}^{2} = 20$

$\implies {( \sqrt{x} + \frac{1}{ \sqrt{x} } )} = \sqrt{20}$

$\implies \sqrt{x} + \frac{1}{ \sqrt{x} } = 2\sqrt{5}$

$\star\:\:\:\bf\large\underline\blue{Answer:-}$

• $\implies \sqrt{x} + \frac{1}{ \sqrt{x} } = 2\sqrt{5}$