Question

Find the value of $a+b$

if $\frac{5+3\sqrt{2}}{3+2\sqrt{2}}=a+b\sqrt{2}$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A

$1$

Given $\frac{5+3\sqrt{2}}{3+2\sqrt{2}}=a+b\sqrt{2}$

We will have to rationalize $\frac{5+3\sqrt{2}}{3+2\sqrt{2}}$.

Rationalizing factor is $3-2\sqrt{2}$.

So, multiplying and dividing the expression by $3-2\sqrt{2}$, we get,

$\frac{5+3\sqrt{2}}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}$

$=\frac{(5+3\sqrt{2})\times (3-2\sqrt{2})}{(3+2\sqrt{2})\times (3-2\sqrt{2})}$

$=\frac{15-10\sqrt{2}+9\sqrt{2}-12}{3^2-(2\sqrt{2}{)}^2}$

$=\frac{3-\sqrt{2}}{1}$

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

$\Rightarrow a=3,b=-2$

$\Rightarrow a+b=3-2=1$