Question

If $\frac{3+2\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}$, where a and b are rationals. Find the values of a and b.

• A. $a=\frac{9}{7}$
• B. $b=\frac{13}{7}$
• C. $a=\frac{13}{7}$
• D. $b=\frac{9}{7}$

Answer 1

Answer: (C), (D)

$b=\frac{9}{7}$
$a=\frac{13}{7}$

L.H.S $=\frac{3+2\sqrt{2}}{3-\sqrt{2}}=\frac{(3+2\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=\frac{9+3\sqrt{2}+6\sqrt{2}+4}{9-2}=\frac{13}{7}+\frac{9}{7}\sqrt{2}$
$\therefore \frac{13}{7}+\frac{9}{7}\sqrt{2}=a+b\sqrt{2}$
equating the rational and irrational parts:
We get $a=\frac{13}{7},b=\frac{9}{7}$