Why is the solution to this algebraic fraction simplification given like this and how was it reached?

Question

I have a question in my text book that asks me to simplify the above expression.

I reached the solution:

However, the textbook gives the answer as:

I appreciate that the value of these two expressions is the same. However, I am unsure as to how and why the answer above was reached instead of my answer.

Answer 1

Multiply your solution by $\dfrac{\sqrt a + \sqrt b}{\sqrt a + \sqrt b}$ and simplify.

Answer 2

Multiply your solution by $\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}}$, you can get the textbook answer.

It is because we usually like the denominator to be rational.

Answer 3

Hint:

$$\frac{a\sqrt{ab}}{\sqrt{a}-\sqrt{b}} = \frac{a\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\cdot\frac{\sqrt a + \sqrt b}{\sqrt a + \sqrt b}$$

Now, simplify the expression. Remember, $(x-y)(x+y)=x^2-y^2$

Answer 4

multiplying denominator and numerator by $$\sqrt{a}+\sqrt{b}$$ we get $$\frac{a\sqrt{ab}(\sqrt{a}+\sqrt{b})}{a-b}$$

Answer 5

Typically, the denominator does not carry a root. Therefore, we apply$(x-y)(x+y)=x^2-y^2$to the denominator. Then, numerator and denominator multiply $\sqrt{a}+\sqrt{b}$. We get the answer $$\frac{a^2\sqrt{b}+ab\sqrt{a}}{a-b}$$. P.S. The answer is a bit ugly.

Answer 6

What your textbook did is called rationalization. In which if we want to remove radicals($\sqrt.$) then whole expression is multiplied and divided by a conjugate of either numerator or denominator.

If you don't want radicals in numerator then divide whole expression with conjugate of numerator and if you don't want radicals in denominator then multiply and divide whole with conjugate of denominator.

Here you can learn more about it.

$$\frac{a\sqrt{ab}}{\sqrt{a}-\sqrt{b}} = \frac{a\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\cdot\frac{\sqrt a + \sqrt b}{\sqrt a + \sqrt b}$$