Simplifying an Expression: $\sqrt{3} + \sqrt{2} / \sqrt{3} - \sqrt{2}$

[prasadini]

\sqrt{3} + \sqrt{2} / \sqrt{3} - \sqrt{2}

 

[MarkFL]

Hello and welcome to MHB, prasadini! (Wave)

I've moved your thread to a more fitting area. :D

So, we are given the expressions (I assume):

$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

I assume you are to rationalize the denominator...what form of $1$ do we need to multiply this expression by to accomplish this?

 

[prasadini]

Hello and welcome to MHB, prasadini! (Wave)

I've moved your thread to a more fitting area. :D

So, we are given the expressions (I assume):

$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

I assume you are to rationalize the denominator...what form of $1$ do we need to multiply this expression by to accomplish this?

$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

is equal to



5+2√6 and 1 /5−2√6 How can i get this answer

 

[MarkFL]

$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

is equal to



5+2√6 and 1 /5−2√6 How can i get this answer

Well, suppose we are given:

$$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$$

a) To rationalize the denominator, we would do the following:

$$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\cdot\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{(\sqrt{a}+\sqrt{b})^2}{a-b}=\frac{a+2\sqrt{ab}+b}{a-b}=\frac{a+b}{a-b}+\frac{2}{a-b}\sqrt{ab}$$

b) To rationalize the numerator, we would do the following:

$$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\cdot\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{a-b}{a-2\sqrt{ab}+b}=\frac{a-b}{a+b-2\sqrt{ab}}$$

Can you use these techniques to rationalize the denominator and numerator of the given expression?