Question

In each of the following determine rational number a and b:
$\begin{array}{cc}& i)\frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b\sqrt{3}\\ & ii)\frac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b}\end{array}$

Answer 1

$\left(1\right)\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$ {Rationalizing}

$=\frac{{(\sqrt{3}-1)}^2}{{\left(\sqrt{3}\right)}^2-{\left(1\right)}^2}$

$=\frac{4-2\sqrt{3}}{2}$

$=2-\sqrt{3}$

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

$a=2,b=1$

$\left(2\right)\frac{4+\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}$ {Rationalizing}

$=\frac{8-4\sqrt{2}+2\sqrt{2}-2}{{\left(2\right)}^2-{\left(\sqrt{2}\right)}^2}$

$=\frac{6-2\sqrt{2}}{2}$

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

$a=3,b=2$