Question

Please help!

If we write $\sqrt{2} + \frac{1}{\sqrt{2}} + \sqrt{3} + \frac{1}{\sqrt{3}}$ in the form $\frac{a \sqrt{2} + b \sqrt{3}}{c}$, such that $a$, $b$, and $c$ are positive integers and $c$ is as small as possible, then what is $a+b+c$?

Answer 1

Answer:

$a+b+c=9+8+6=23$

Step-by-step explanation:

Let’s first simplify our expression. We have:

$\displaystyle {\sqrt2+\frac{1}{\sqrt2}+\sqrt3+\frac{1}{\sqrt{3}}$

For the second term, multiply both the numerator and denominator by √2. This yields:

$\displaystyle \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

Similarly, for the fourth term, multiply both the numerator and denominator by √3. This yields:

$\displaystyle\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3}$

Hence, our expression is now:

$\displaystyle =\sqrt{2}+\frac{\sqrt{2}}{2}+\sqrt{3}+\frac{\sqrt{3}}{3}$

Let’s combine them. First, we will need common denominators.

Our denominators are 2 and 3. So, our common denominator will be its LCM.

The LCM of 2 and 3 is 6.

Hence, let’s make each term’s denominator 6.

For the first term, we can multiply both layers by 6. Hence:

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

For the second term, we can multiply both layers by 3. Hence:

$\displaystyle \frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{6}$

For the third term, we can multiply both layers by 6. Hence:

$\displaystyle \sqrt{3}=\frac{6\sqrt{3}}{6}$

And for the last term, we can multiply both layers by 2. Hence:

$\displaystyle \frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{6}$

So, our expression is:

$\displaystyle =\frac{6\sqrt{2}}{6}+\frac{3\sqrt{2}}{6}+\frac{6\sqrt{3}}{6}+\frac{2\sqrt{3}}{6}$

Add:

$\displaystyle =\frac{6\sqrt{2}+3\sqrt{2}+6\sqrt{3}+2\sqrt{3}}{6}$

Combine like terms:

$\displaystyle=\frac{9\sqrt{2}+8\sqrt{3}}{6}$

This cannot be simplified. So, c is as small as possible.

Hence: a=9, b=8, and c=6.

Therefore:

$a+b+c=9+8+6=23$