Question

Rationalize the denominator and simplify.

Answer 1

so this was solved before when posted it

Answer 2

Multiply the fraction by \(\dfrac{\sqrt x - \sqrt y}{\sqrt x - \sqrt y} \)

Answer 3

Multiplying a square root by a square root will get rid of the square root. Which is why you multiply the entire fraction by the denominator with one change: changing the sign (in this case from subtraction to addition). Works everytime

Answer 4

still confussed

Answer 5

to rationalise the denominator u should remove any kind of roots in it so u multiply both the numerator and the denominator by the above mentioned quantity to remove the square roots

Answer 6

Thomas,

Here is an explanation.

Just like if you have a problem whose solution is a fraction, you are expected to write the final answer as a reduced fraction, a similar thing exists with roots. It is the custom to leave answers with no roots on the denominator.
Rationalizing the denominator is a process to rid the denominator of roots.

If the root in the denominator is a simple square root, then you multiply the numerator and denominator by that same root.

For example, if you have:

\(\dfrac{1}{\sqrt 2}\)

you have a root in the denominator.
That means you need an equivalent fraction without a root in the denominator.
To find that fraction, you multiply the fraction by

\(\dfrac{\sqrt 2}{\sqrt 2} \)

The product of the square roots in the denominator will get rid of the sqare root, and you will have your answer.

\(\dfrac{1}{\sqrt 2} \times\dfrac{\sqrt 2}{\sqrt 2} = \dfrac{\sqrt 2}{2}\)

Now the answer has no root in the denominator, and you are done.

Answer 7

In your case, it is a little more complicated because you don;t have a simple root in the denominator. You have a sum of roots in the denominator. When you have a sum or a difference in the denominator, and it involves a root, then you use a different technique.

Let's look at the product of a sum and a difference.

\((a + b)(a - b) = a^2 - b^2\)

Notice that ion the left side, you have a and b.
There is a sum of a and b, and there is a difference of a and b.
The important thing is to see what the product is.
In the product, we get the difference of a^2 and b^2.
Both a and b end up squared.

Now imagine that instead of having a and b, you had the square roots of both a and b (like your problem has), or the square root of just one of them, but you still had a sum or a difference. By using the proper product of a sum and a difference, the two quantities will end up squared, and you will no longer have a square root.

Answer 8

Let's look at your denominator.

It is

\(\sqrt x + \sqrt y\)

It is a sum of two roots. If we multiply it by the proper difference of two roots, then by following the pattern above, we will end up squaring \(\sqrt x\) and \(\sqrt y\), and we will no longer have roots, but we will have x and y instead.

Let's do it. Since this denominator has a sum of the roots, we need to multiply it by the difference of the same roots, \(\sqrt x - \sqrt y\).

\((\sqrt x + \sqrt y)(\sqrt x - \sqrt y)\)

We follow the pattern abnove of the product of a sum and a difference, and we get:

\((\sqrt x)^2 - (\sqrt y)^2 = x - y\)

As you can see, we no longer have roots. All we have is x - y.

Answer 9

^

Answer 10

That is how we get rid of the roots of the denominator.
Now remember that your problem is a fraction, so to get an equivalent fraction, we need to multiply the numerator and denominator by the same quantity.

\(\dfrac{2 \sqrt x - 3 \sqrt y}{\sqrt x + \sqrt y} \times \dfrac{ \sqrt x - \sqrt y }{\sqrt x - \sqrt y } \)

We already know what the denominator is from above.
Now we need to work on the numerator.

\(=\dfrac{(2 \sqrt x - 3 \sqrt y)(\sqrt x - \sqrt y)}{x - y} \)

Now we multiply the two binomials in the numerator.

\(=\dfrac{2 (\sqrt x)^2 - 2 \sqrt x \sqrt y - 3 \sqrt x \sqrt y + 3 (\sqrt y)^2}{x - y} \)

\(=\dfrac{2 x - 2 \sqrt {xy} - 3 \sqrt {xy} + 3y}{x - y} \)

\(=\dfrac{2 x -5 \sqrt {xy} + 3y}{x - y} \)

Answer 11

I hope you understand the explanation.
If you have any questions, just ask.