Question

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

Answer 1

Answer:

$\boxed{\bf \sqrt{x} + \dfrac{1}{\sqrt{x}} = 2\sqrt{2}}$

Step-by-step explanation:

Given :

• x = 3 + 2√2

Step 1 : Find the value of √x.

⇒ x = 3 + 2√2

⇒ x = 2 + 1 + 2√2

⇒ x = (√2)² + (1)² + 2. 1. √2

⇒ x = ( √2 + 1)²

⇒ √x = √2 + 1

Step 2 : Find the value of 1/√x.

⇒ √x = √2 + 1

⇒ $\tt \dfrac{1}{\sqrt{x}} = \dfrac{1}{\sqrt{2}+1}$

⇒ $\tt \dfrac{1}{\sqrt{x}} = \dfrac{1}{\sqrt{2}+1} \times \dfrac{\sqrt{2}-1}{\sqrt{2}-1}$

⇒ $\tt \dfrac{1}{\sqrt{x}} = \dfrac{\sqrt{2}-1}{(\sqrt{2})^2 - (1)^2}$

⇒ $\tt \dfrac{1}{\sqrt{x}} = \dfrac{\sqrt{2}-1}{2 - 1}$

⇒ $\tt \dfrac{1}{\sqrt{x}} = \sqrt{2} - 1$

Step 3 : Find the value of √x + 1 /√x.

⇒ $\tt \sqrt{x} + \dfrac{1}{\sqrt{x}} = \sqrt{2} + 1 + \sqrt{2} - 1$

⇒ $\tt \sqrt{x} + \dfrac{1}{\sqrt{x}} = \sqrt{2} +\sqrt{2}$

⇒ $\tt \sqrt{x} + \dfrac{1}{\sqrt{x}} = 2\sqrt{2}$

Hence, the answer is 2√2.

Answer 2

$\huge \mathfrak \red{answer}$

$\rm \blue{ \: answer \: is \: 2}$

✤Question:✤

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

✥step to step explanation✥

➹$\sf{x = 3 + 2 \sqrt{2}}$

➹$\sf{1 + 2 \sqrt{2} + 2}$

➹$\sf {1}^{2} + 2.1. \sqrt{2} + \sqrt{2 {}^{2} )}$

➹$\sf{(1 + \sqrt{2 }^{2}) }$

➹$\sf{ \sqrt{x} = 1 + \sqrt{2}}$

________________________________

then

◇$\tt{ \sqrt{x} - \frac{1}{ \sqrt{x} }}$

◇$\tt{ = 1 + \sqrt{2} - \frac{1}{1 + \sqrt{2} }}$

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

◇$\tt{ \frac{ = 1 + 2 \sqrt{2} + 2 - 1 }{1 + \sqrt{2} }}$

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

◇$\tt \red{ = 2}$

Answer is 2