Question

If √x =1+√2,find the value of x+1/x​

Answer 1

Hey friends here is your answer :-

The value of the expression is 8√2+8

The value of x is given to be : x = √2 + 1

The value of the required expression can be found easily by substituting the value of x = √2 + 1 in the expression

So, The value is :

(√2+1)^2+1/√2+1

2+1+4+1/√2+1

8/√2+1 ×√2-1/√2-1

8√2+8/2-1

8√2+8

HOPE IT HELPED YOU.........

Answer 2

Answer:

Required value = 6

Step-by-step explanation:

Given that,

√x = 1 +√2

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

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

          =$\frac{\sqrt{2}-{1}}{2-1}$        [∵(a + b)(a - b) = a² - b²]

          = $\sqrt{2}-{1}$

∴$\sqrt{x}+\frac{1}{\sqrt{x}} = \sqrt{2}+{1}+\sqrt{2}-{1}$

            = √2 + √2

            = 2√2

Now,

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

            = $(\sqrt{x}+\frac{1}{\sqrt{x}})^{2}-2.\sqrt{x}.\frac{1}{\sqrt{x} }$                 [∵ a² + b² = (a + b)² - 2.a.b]

            =  (2√2)² - 2

            = 8 - 2

            = 6

∴Required value = 6

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