Question

if x=3 + 2 root 2 find the value of root x minus one by root x do it in step by step process​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\sqrt{x}+\frac{1}{\sqrt{x}}=2\sqrt{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies x = 3 + 2 \sqrt{2} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies \sqrt{x} + \frac{1}{ \sqrt{x} } = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt\circ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} = {( \sqrt{x}) }^{2} + ( \frac{1}{ \sqrt{x} } )^{2} + 2 \times \sqrt{x} \times \frac{1}{ \sqrt{x} } \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =x + \frac{1}{x} + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =3 + 2 \sqrt{2} + \frac{1}{3 + 2 \sqrt{2} } + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =3 + 2 \sqrt{2} + \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =3 + 2 \sqrt{2} + \frac{3 - 2 \sqrt{2} }{ {3}^{2} - (2 \sqrt{2})^{2} } + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =3 + 2 \sqrt{2} + \frac{3 - 2 \sqrt{2} }{9 - 8} + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =6 + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =8\\\\ \green{\tt:\implies \sqrt{x}+\frac{1}{\sqrt{x}}=2\sqrt{2}}$

Answer 2

$\tt{\red{\huge{Solution _{BQ} \:: - }}}$

$\begin{lgathered} \tt: \leadsto ( \sqrt{x} + \dfrac{1}{ \sqrt{x} } )^{2} = {( \sqrt{x}) }^{2} + ( \dfrac{1}{ \sqrt{x} } )^{2} + 2 \times \sqrt{x} \times \dfrac{1}{ \sqrt{x} } \\ \\ \tt: \mapsto ( \sqrt{x} + \dfrac{1}{ \sqrt{x} } )^{2} =x + \dfrac{1}{x} + 2 \\ \\ \tt: \leadsto ( \sqrt{x} + \dfrac{1}{ \sqrt{x} } )^{2} =3 + 2 \sqrt{2} + \frac{1}{3 + 2 \sqrt{2} } + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =3 + 2 \sqrt{2} + \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =3 + 2 \sqrt{2} + \frac{3 - 2 \sqrt{2} }{ {3}^{2} - (2 \sqrt{2})^{2} } + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =3 + 2 \sqrt{2} + \frac{3 - 2 \sqrt{2} }{9 - 8} + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =6 + 2 \\ \\ \tt: \implies ( \sqrt{x} + \frac{1}{ \sqrt{x} } )^{2} =8\\\\ \purple{\tt:\leadsto \sqrt{x}+\frac{1}{\sqrt{x}}=2\sqrt{2}}......(A)\end{lgathered}$