Question

1
If x +1/x=2
Then find
$\sqrt{x + 1 \div \sqrt{x} }$


​

Answer 1

${\underline{\underline{\red{\sf{Given:}}}}}$

• $\sf{x+\dfrac{1}{x}=2}$

${\underline{\underline{\red{\sf{To\:Find:}}}}}$

• The value of $\sf{\sqrt{\dfrac{x+1}{\sqrt{x}}}}$

${\underline{\underline{\red{\sf{Answer:}}}}}$

We are here given that $\sf{x+\dfrac{1}{x}=2}$.

Now ,let $\sf{\sqrt{\dfrac{x+1}{\sqrt{x}}}}$ = k.

$\sf{\implies \sqrt{\dfrac{x+1}{\sqrt{x}}}=k}$

Squaring both sides ;

$\sf{\implies (\sqrt{\dfrac{x+1}{\sqrt{x}}})^{2}=k^{2}}$

$\sf{\implies \dfrac{x+1}{\sqrt{x}}=k^{2}}$

$\sf{\implies \dfrac{x}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}=k^{2}}$

$\sf{\implies\dfrac{1}{\sqrt{x}}+\sqrt{x}=k^{2}}$

Again squaring both sides;

$\sf{\implies (\dfrac{1}{\sqrt{x}}+\sqrt{x})^{2}=(k^{2})^{2}}$

$\sf{\implies (\sqrt{x})^{2}+(\dfrac{1}{\sqrt{x}})^{2}+2\times\sqrt{x}\times\dfrac{1}{\sqrt{x}}=k^{4}}$

$\sf{\implies x+\dfrac{1}{x}+2=k^{4}}$

$\sf{\implies 2+2=k^{4}}$

$\sf{\implies k^{4}=4}$

$\sf{\implies k=\sqrt[4]{4}}$

$\sf{\implies k=2^{2\times\frac{1}{4}}}$

$\sf{\implies k=2^{\frac{1}{2}}}$

${\underline{\boxed{\red{\sf{\leadsto k=\sqrt{2}}}}}}$

Hence we found k as √2 .

Earlier we had assumed that the value of $\sf{\sqrt{\dfrac{x+1}{\sqrt{x}}}}$ =k .

Hence $\sf{\sqrt{\dfrac{x+1}{\sqrt{x}}}}$ =√2.

Answer 2

Answer:

Given:

\sf{x+\dfrac{1}{x}=2}x+

x

1

=2

{\underline{\underline{\red{\sf{To\:Find:}}}}}

ToFind:

The value of \sf{\sqrt{\dfrac{x+1}{\sqrt{x}}}}

x

x+1

{\underline{\underline{\red{\sf{Answer:}}}}}

Answer:

We are here given that \sf{x+\dfrac{1}{x}=2}x+

x

1

=2 .

Now ,let \sf{\sqrt{\dfrac{x+1}{\sqrt{x}}}}

x

x+1

= k.

\sf{\implies \sqrt{\dfrac{x+1}{\sqrt{x}}}=k}⟹

x

x+1

=k

Squaring both sides ;

\sf{\implies (\sqrt{\dfrac{x+1}{\sqrt{x}}})^{2}=k^{2}}⟹(

x

x+1

)

2

=k

2

\sf{\implies \dfrac{x+1}{\sqrt{x}}=k^{2}}⟹

x

x+1

=k

2

\sf{\implies \dfrac{x}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}=k^{2}}⟹

x

x

+

x

1

=k

2

\sf{\implies\dfrac{1}{\sqrt{x}}+\sqrt{x}=k^{2}}⟹

x

1

+

x

=k

2

Again squaring both sides;

\sf{\implies (\dfrac{1}{\sqrt{x}}+\sqrt{x})^{2}=(k^{2})^{2}}⟹(

x

1

+

x

)

2

=(k

2

)

2

\sf{\implies (\sqrt{x})^{2}+(\dfrac{1}{\sqrt{x}})^{2}+2\times\sqrt{x}\times\dfrac{1}{\sqrt{x}}=k^{4}}⟹(

x

)

2

+(

x

1

)

2

+2×

x

×

x

1

=k

4

\sf{\implies x+\dfrac{1}{x}+2=k^{4}}⟹x+

x

1

+2=k

4

\sf{\implies 2+2=k^{4}}⟹2+2=k

4

\sf{\implies k^{4}=4}⟹k

4

=4

\sf{\implies k=\sqrt[4]{4}}⟹k=

4

4

\sf{\implies k=2^{2\times\frac{1}{4}}}⟹k=2

2×

4

1

\sf{\implies k=2^{\frac{1}{2}}}⟹k=2

2

1

{\underline{\boxed{\red{\sf{\leadsto k=\sqrt{2}}}}}}

⇝k=

2

Hence we found k as √2 .

Earlier we had assumed that the value of \sf{\sqrt{\dfrac{x+1}{\sqrt{x}}}}

x

x+1

=k .

Hence \sf{\sqrt{\dfrac{x+1}{\sqrt{x}}}}

x

x+1

=√2.