Question

(Root 2 +1/Root2-1)whole square

Answer 1

• Given:- (Root 2 +1/Root2-1)whole square=?

• To find:- The value

• Solution:- According to the question, we have to determine the value of (Root 2 +1/Root2-1)whole square

and

(Root 2 +1/Root2-1)whole square= [{(√2 +1)(√2+1)}/{(√2 -1)(√2+1)}]²

= [(√2+1)²/1]²= 2+1+2√2 (according to the rule (a+b)²= a²+b²+2ab)

= 3+2√2

So the final answer

(Root 2 +1/Root2-1)whole square= 3+2√3

Answer 2

Given:

(i) The expression is $\sqrt{2} + \frac{1}{\sqrt{2} - 1}$.

To find:

(i) The whole squared simplification of the above expression.

Solution:

Given,

The expression = $\sqrt{2} + \frac{1}{\sqrt{2} - 1}$

Multiplying by (√2 + 1) with $\frac{1}{\sqrt{2} - 1}$, we get

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

Squaring according to (a+b)² = a² + b² + 2ab, we get

(2√2)² + 1² + 2(2√2)(1)

= 8 + 1 + 4√2

= 9 + 4√2

So, 9 + 4√2 is the simplified answer.