Question

If x+1/x=2 Find √x+1/√x. I will give the first answer as brainliest.

Answer 1

Answer:

$\sf{The \ value \ of \ \sqrt{x}+\dfrac{1}{\sqrt{x}} \ is \ 2. }$

Given:

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

To find:

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

Solution:

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

$\sf{\leadsto{\dfrac{x^{2}+1}{x}=2}}$

$\sf{\leadsto{x^{2}+1=2x}}$

$\sf{\leadsto{x^{2}-2x+1=0}}$

$\sf{\leadsto{x^{2}-x-x+1=0}}$

$\sf{\leadsto{x(x-1)-1(x-1)=0}}$

$\sf{\leadsto{(x-1)(x-1)=0}}$

$\sf{\leadsto{x=1 \ or \ 1}}$

$\sf{Hence, \ the \ value \ of \ x \ is \ 1}$

$\sf{\leadsto{\sqrt{x}+\dfrac{1}{\sqrt{x}}}}$

$\sf{Substitute \ x=1, \ we \ get}$

$\sf{\leadsto{\sqrt1+\dfrac{1}{\sqrt1}}}$

$\sf{\leadsto{1+\dfrac{1}{1}}}$

$\sf{\leadsto{1+1}}$

$\sf{\leadsto{2}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ \sqrt{x}+\dfrac{1}{\sqrt{x}} \ is \ 2. }}}$

Answer 2

Question:

If $x + \frac{1}{x} = 2$ , find $\sqrt{x} + \frac{1}{ \sqrt{x} }$.

To find:

$\bigstar \: \sqrt{x} + \frac{1}{ \sqrt{x} }$

Answer:

$The \: value \: of \: \underline{ \: \sf \pink{\bf \sqrt{x} + \frac{1} {\sqrt{x} }}\: is \: \sf \red{2 \: }}$

Given:

$\bigstar \: x + \frac{1}{x} = 2$

Step-by-step explanation:

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

Now we have to change to ( a + b )²

$Here \: \: \underline{ \: a = \sqrt{x} , \: b = \frac{1}{ \sqrt{x} } \: }$

Squaring on both sides,

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

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

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

$\therefore The \: value \: of \: \underline{ \: \bf \sqrt{x} + \frac{1} {\sqrt{x} }\: is \: 2 \: }$

Formula used:

$\bigstar \: {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$