Question

If $x+y\sqrt{2}=\frac{6+3\sqrt{2}}{8+5\sqrt{2}}$, then what will be the values of $x$ and $y$?

• A. $x=\frac{9}{7},y=-\frac{3\sqrt{2}}{7}$
• B. $x=-\frac{9}{7},y=\frac{3\sqrt{2}}{7}$
• C. $x=-\frac{3\sqrt{2}}{7},y=\frac{9}{7}$
• D. $x=\frac{3\sqrt{2}}{7},y=-\frac{9}{7}$

Answer 1

The correct option is A $x=\frac{9}{7},y=-\frac{3\sqrt{2}}{7}$

Consider the given fraction $\frac{6+3\sqrt{2}}{8+5\sqrt{2}}$.

To simplify it, rationalize its denominator by multiplying and dividing the given fraction by the rationalizing factor of the denominator, i.e, $8-5\sqrt{2}$.

$\begin{array}{cc}\frac{6+3\sqrt{2}}{8+5\sqrt{2}}& =\frac{6+3\sqrt{2}}{8+5\sqrt{2}}\times \frac{8-5\sqrt{2}}{8-5\sqrt{2}}\\ & =\frac{6(8-5\sqrt{2})+3\sqrt{2}(8-5\sqrt{2})}{(8+5\sqrt{2})(8-5\sqrt{2})}\\ & =\frac{48-30\sqrt{2}+24\sqrt{2}-15\left(2\right)}{(8{)}^2-(5\sqrt{2}{)}^2}\\ & =\frac{48-6\sqrt{2}-30}{64-50}\\ & =\frac{18-6\sqrt{2}}{14}\\ & =\frac{18}{14}-\frac{6\sqrt{2}}{14}\\ & =\frac{9}{7}-\frac{3\sqrt{2}}{7}\end{array}$

It is given that $x+y\sqrt{2}=\frac{6+3\sqrt{2}}{8+5\sqrt{2}}$, therefore we get

$x+y\sqrt{2}=\frac{9}{7}-\frac{3\sqrt{2}}{7}$

On comparing the coefficients of both sides, we get

$x=\frac{9}{7}$, and $y=-\frac{3\sqrt{2}}{7}$