[MOVED] Rationalize denominator: sqrt[y]/2sqrt[x] + 3sqrt[y]

[Jade]

Square root y / 2 square root x + 3 square root y

Sorry I don't know the online square root symbol (if there is one)

I believe you would multiply by square root y on both top and bottom?

 

[skeeter]

if you meant ...

\(\displaystyle \L \frac{\sqrt{y}}{2\sqrt{x}+3\sqrt{y}}\)

multiply numerator and denominator by \(\displaystyle \L 2\sqrt{x}-3\sqrt{y}\)

 

[Jade]

This is a hard one

After multiplying 2 square root x - 3 square root y to both the top and bottom. I came up with the answer 1 - square root y over 1 square root x - 3. Does that look right?

 

[Denis]

Re: This is a hard one

After multiplying 2 square root x - 3 square root y to both the top and bottom. I came up with the answer 1 - square root y over 1 square root x - 3. Does that look right?

Jade, next time, post "2 square root x - 3 square root y"
like this: 2 sqrt(x) - 3 sqrt(y) : that's standard...plus less typing :idea:

Also, your original equation: Square root y / 2 square root x + 3 square root y
needed brackets: Square root y / (2 square root x + 3 square root y);
or: sqrt(y) / (2 sqrt(x) + 3 sqrt(y))

OK; your multiplications are not correct:

numerator: sqrt(y) * (2 sqrt(x) - 3 sqrt(y)) = 2 sqrt(x) sqrt(y) - 3y

denominator: (2 sqrt(x) + 3 sqrt(y)) * (2 sqrt(x) - 3 sqrt(y)) = 4x - 9y

Remember that sqrt(x) * sqrt (x) = x

 

[Jade]

Thanks

:lol: