Question

If √2 = 1.4142…, then find the value of: √
√2−1
√2+1

Answer 1

Step-by-step explanation:

Let's solve the problem,

we have,

$\sqrt{ \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1} }$

The denominator is √2+1. Multiplying the numerator and denomination by √2-1, we get

$⟹ \sqrt{ \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1} \times \frac{ \sqrt{2} - 1 }{ \sqrt{2} - 1 } }$

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

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

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

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

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

$⟹ \sqrt{2 } - 1$

$⟹1.4142 - 1$

$⟹0.4142 \: \: Ans.$

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Answer 2

Step-by-step explanation:

$\huge \underline{ \sf{ \red{Answer:-}}}$

Given :

• √2 = 1.4142

Solution:

$\green{\huge \longrightarrow \: \sqrt{ \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1 } }}$

$\green{\huge \longrightarrow \: \sqrt{\frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1} \times \frac{ \sqrt{2} - 1 }{ \sqrt{2} - 1 } }}$

$\green{\huge \longrightarrow \: \sqrt{\frac{ {( \sqrt{2} - 1) }^{2} }{ {( \sqrt{2} )}^{2} - {(1)}^{2} } }}$

$\green{\huge \longrightarrow \: \frac{ \sqrt{2} - 1 }{1} }$

$\green{\huge \longrightarrow \:1.4142 - 1}$

$\green{\huge \longrightarrow \:0.4142}$

Formula Used:

• (a-b)²=(a+b)(a-b)