Question

If, x=3-2√2 then find the value of√x+1/√x

Answer 1

Answer:

√x + [1/√x] = 2√2  is the correct answer.

Step-by-step explanation:

Substituting the value of x as 3 - 2√2  in the equation √x + [1/√x].

⇒ √x + [1/√x] = √(3-2√2)  +  [1/√(3-2√2)]

⇒ √x + [1/√x] = √(1-√2)²  +  [1/√(1-√2)²]

⇒ √x + [1/√x] = √(√2-1)²  +  [1/√(√2-1)²]

⇒ √x + [1/√x] = √2-1  +  [1/√2-1]

⇒ √x + [1/√x] = √2-1  +  [(√2+1)/(2-1)]                [ Rationalising ]

⇒ √x + [1/√x] = √2 - 1 + √2 + 1

⇒ √x + [1/√x] = 2√2                  ←ANSWER

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Answer 2

Solution :-

x = 3 - 2√2

It can be written as

⇒ x = (√2)² + 1² - 2(√2)(1)

⇒ x = (√2 - 1)²

[ Because (a - b)² = a² + b² - 2ab ]

Taking square root on both sides

⇒ √x = ± √(√2 - 1)²

⇒ √x = ± (√2 - 1)

⇒ √x = (√2 - 1) or (1 - √2 )

Finding 1/√x value

when √x = √2 - 1

$\sf \dfrac{1}{ \sqrt{x} } = \dfrac{1}{ \sqrt{2} - 1 }$

Rationalizing the denominator

$\sf= \dfrac{1( \sqrt{2} + 1) }{( \sqrt{2} - 1)( \sqrt{2} + 1) }$

$\sf= \dfrac{ \sqrt{2} + 1 }{( \sqrt{2})^{2} - 1 ^{2} }$

$\sf= \dfrac{ \sqrt{2} + 1 }{2 - 1}$

$\sf= \dfrac{ \sqrt{2} + 1 }{1} = \sqrt{2} + 1$

When √x = 1 - √2

$\sf \dfrac{1}{ \sqrt{x} } = \dfrac{1}{1 - \sqrt{2} }$

$\sf= \dfrac{1(1 + \sqrt{2} ) }{( 1 - \sqrt{2} )(1 + \sqrt{2}) }$

$\sf= \dfrac{1 + \sqrt{2} }{ {1}^{2} - ( \sqrt{2} )^{2} }$

$\sf= \dfrac{1 + \sqrt{2} }{ 1 - 2}$

$\sf= \dfrac{ - ( - 1 - \sqrt{2}) }{ - 1} = - 1 - \sqrt{2}$

Now, √x + 1/√x

When √x = √2 - 1 and 1/√x = √2 + 1

√x + 1/√x = √2 - 1 + (√2 + 1)

= √2 - 1 + √2 + 1 = 2√2

When √x = 1 - √2 and 1/√x = - 1 - √2

√x + 1/√x = 1 - √2 + (-1 - √2) = 1 - √2 - 1 - √2 = - 2√2

Therefore the value of √x + 1/√x is 2√2 or - 2√2.