Question

$\large{\bf{ If\: x + \frac{1}{x} = 2}}$

Then find the value of :-

$\large{\bf{ \sqrt{x} + \frac{1}{ \sqrt{x} } }}$

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Answer 1

Answer:

the answer is æ=1æ.

Step-by-step explanation:

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Answer 2

Answer:

t = 2

Step-by-step explanation:

$x + \frac{1}{x} = 2\\\\\sqrt{x}+ \frac{1}{\sqrt{x}} = t$

Now we have to find the value of t

Squaring both sides :

$(\sqrt{x})^{2} + (\frac{1}{\sqrt{x}})^{2} + 2(\sqrt{x})(\frac{1}{\sqrt{x}}) = t^{2}\\\\x + \frac{1}{x} + 2 = t^{}2\\\\x + \frac{1}{x} = 2\\\\2 + 2 = t^{2}\\\\t^{2} = 4\\\\t = \pm2$

Now please note that $t \neq -2$ because $t = \sqrt{x} + \frac{1}{\sqrt{x}}$ and square root function is always defined for positive numbers only.

For future reference:

Now You may also note that if $x^2 = a$ then $x = \pm{\sqrt{a}}$ But if x = $\sqrt{a}$ then x is positive because $\sqrt{a}$ can never be negative and  -$\sqrt{a}$ is negative.

So t > 0 for all positive values of x.

Thus t = 2

Might Help !! Thanks!