Question

find rational numbers A and B such that root 2 minus one upon root 2 + 1 equals to a + b root 2​

Answer 1

$\mathfrak{\large{\underline{\underline{Answer :-} }}}$

a = 3, b = -2

$\mathfrak{\large{\underline{\underline{Explanation :-} }}}$

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

Consider Left Hand Side :

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

The Rationalising factor of  √2 + 1 is √2 - 1. So, multiply the numerator and denominator of the given fraction with Rationalising factor.

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

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

Since (x + y)(x - y) ≡ x² - y²

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

Since (x - y)² ≡ x² - 2xy + y²

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

$=3-2\sqrt{2}$

Now consider

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

$i.e,3-2\sqrt{2}=a+\sqrt{b}$

Equating corresponding rational and irrational factors, we have

a = 3

b√2 = -2√2

b = -2

So, a = 3, b = -2