Question

$\frac{ \sqrt{2} - 1}{ \sqrt{2} + 1 } = a + b \sqrt{2}$
find a and b​

Answer 1

Answer:

a= 3 ,b= -2

Step-by-step explanation:

Formulas used (a+b)^2 = a^2 -2ab+ b^2

Answer 2

$\textbf {\underline {\underline {Step-By-Step\:Explanation}}}$

$\sf\pink {Given:}$

• $\frac {\sqrt {2}-1}{\sqrt {2}+1} = a + b \sqrt {2}$

$\sf\pink {To\:Find:}$

• Value of a and b

$\huge\underline\green{\tt Solution:}$

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

( Rationalise it )

= $\dfrac {\sqrt {2}-1}{\sqrt {2}+1}$ × $\dfrac {\sqrt {2}-1}{\sqrt {2}-1}$

➡ Use the following 2 identities

• ${( a + b )}^2 = a^2 + b^2 + 2ab$

• ${a}^2 - {b}^2 = ( a - b ) ( a + b )$

Now,

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

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

= 3 - 2 $\sqrt {2}$

➡ On comparing both sides

3 - 2 $\sqrt {2}$ = a + b $\sqrt {2}$

a = 3

b = - 2

Hence,

Value of a is 3 and b is - 2