Question

If x=3+2√2 find the value of (√x -1/√x)​

Answer 1

Answer:

$\large{\underline{\boxed{\sf \sqrt{x}+\dfrac{1}{\sqrt{x}}=2 \sqrt{2}}}}$

Step-by-step explanation:

x = 3 + 2√2

It can be written as;

⇒ x = 2 + 1 + 2√2

⇒ x = (√2)² + 1² + 2 * 1 * √2

⇒ x = (√2 + 1)²

Also,

√x = √(√2 + 1)²

⇒ √x = √2 + 1

Now, find the value of 1/√x ;

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

So, rationalise the denominator of RHS.

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

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

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

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

Now, find the value of √x + (1/√x) ;

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

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