Question

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Answer 1

Given

The value of x is 5 - 2√6

TO FIND

The value of √x + 1/√x

CONCEPT USED

• The given value should be used only in adequate situation where no more steps left to be done or it couldn't be simplified further without that given value.
• Rationalising the denominator, it is the most wide spread method used to simplify the number by using the rules.

1. If there is one irrational term in the denominator then the same term is multiplied both at numerator as well as denominator to maintain the consistency of same value and to find a rational value of denominator.
2. If there is binomial term in denominator then it's conjucate (2nd term interchanged sign) is multiplied both numerator and denominator so as to obtain

a^2 - b2 form where both terms will be rationalised.

If there is more than 2 terms in denominator then we have to use the 2nd method more than one times to simplify into rationalise form.

ANSWER

2√3 (Steps of solution is given in the attachment)

Answer 2

$\huge \bold \green{question}$

$\mathsf{x = 5 - 2 \sqrt{6} \: find \: the \: value \: of \: \sqrt{x} + \frac{1}{ \sqrt{x} } }$

$\mathsf{ \underline {\pink{given : }}x = 5 - 2 \sqrt{6} } \\ \\ \implies \: \sqrt{x} = \mathsf{ \sqrt{5 - 2 \sqrt{6} } }$

$\mathsf{ \implies \: \sqrt{ \sqrt{ {(3)}^{2} } + \sqrt{(2) {}^{2} } - 2 \sqrt{3} \times \sqrt{2} } }$

$\mathsf{ \implies \: \boxed{ \red{ \sqrt{x} = \sqrt{3} - \sqrt{2} }}}$

$\mathsf{ \frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{3} - \sqrt{2} } } \\ \\ \mathsf{ \implies \: \frac{1}{ \sqrt{x} } = \frac{1}{( \sqrt{3} - \sqrt{2)} } }$

$\mathsf{ \implies \: \frac{ 1}{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{3} } } \\ \\ \mathsf{ \implies \: \frac{ \sqrt{3} + \sqrt{2} }{( \sqrt{3} {)}^{2} - ( \sqrt{2 {)}^{2} } } }$

$\mathsf{ \implies \: \frac{ \sqrt{3} + \sqrt{2} }{3 - 2} } = \frac{ \sqrt{3} + \sqrt{2} }{1}$

$\mathsf{ \implies \: \frac{1}{ \sqrt{x} } = \sqrt{3} + \sqrt{2} }$

$\mathsf{ \implies \: \sqrt{x} + \frac{1}{ \sqrt{x} } = \sqrt{3} - \sqrt{2} + \sqrt{3} + \sqrt{2} }$

$\mathsf{ \boxed{ \sqrt{x} + \frac{1}{x} = 2 \sqrt{3} }}$