Question

Match the following:

• A. 1
• B. $2-\sqrt{3}$
• C. $3+2\sqrt{2}$
• D. $\frac{3-\sqrt{5}}{2}$

Answer 1

Using the identity,
$(a+b{)}^2$ = $(a{)}^2$ + $(b{)}^2$ + $2ab$,
$(1+\sqrt{2}{)}^2=1^2+{\sqrt{2}}^2+2\times 1\times \sqrt{2}$
$=1+2+2\sqrt{2}=3+2\sqrt{2}$

Using the identity,

(a+b)(a-b) = $(a{)}^2$ - $(b{)}^2$,
$(2+\sqrt{3})(2-\sqrt{3})=2^2-{\sqrt{3}}^2$
=$4-3=1$

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

$\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{(\sqrt{5}-1{)}^2}{(\sqrt{5}+1)\times (\sqrt{5}-1)}$
$=\frac{5-2\sqrt{5}+1}{4}=\frac{6-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}$