Question

rationalise denominator of 1+root2/2-root2

Answer 1

$\frac{1 + \sqrt{2} }{2 - \sqrt{2} } = \frac{(1 + \sqrt{2})(2 + \sqrt{2} ) }{(2 - \sqrt{2} )(2 + \sqrt{2}) } \\ = \frac{2 + \sqrt{2 } + 2 \sqrt{2} + 2 }{ {2}^{2} - { \sqrt{2} }^{2} } = \frac{4 + 3 \sqrt{2} }{4 - 2} \\ = \frac{4 + 3 \sqrt{2} }{2}$

Answer 2

Given:

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

To rationalize denominator of $\dfrac{1+\sqrt{2}}{2-\sqrt{2}}$.

Solution:

∴ $\dfrac{1+\sqrt{2}}{2-\sqrt{2}}$

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

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

Rationalising numerator and denominator, we get

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

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

Using the algebraic identity,

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

$(a + b)^2$ = $a^{2} +b^{2}$ + 2ab

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

= $\dfrac{3+2\sqrt{2}}{\sqrt{2}}$

Thus, the rationalize denominator of $\dfrac{1+\sqrt{2}}{2-\sqrt{2}}$ = $\dfrac{3+2\sqrt{2}}{\sqrt{2}}$