Rationalize the denominator. (Photo of problem in comments)

Question

anonymous · Answer

attachment

mathstudent55 · Answer

Multiply the numerator and the denominator by the conjugate of the denominator.

anonymous · Answer

Also if anyone knows how to do this problem as well!

mathstudent55 · Answer

$ \huge x^{\frac{a}{b}} = \sqrt[b]{x^a} = (\sqrt[b]{x})^a $

anonymous · Answer

$$x^2/3$$

anonymous · Answer

how do I write that? like that^?

mathstudent55 · Answer

Use the equation editor.

primeralph · Answer

How about a screenshot?

anonymous · Answer

Oh right forgot about screenshots haha.

anonymous · Answer

I don't understand what to multiply the first question by

triciaal · Answer

do you have the answer 
not sure if I made an error
(28 + 19sqrt6)/15

triciaal · Answer

when you conjugate to rationalize the denominator  use the opposite of the second term in the expression. in this problem, you have 3rt 3 - 2 rt 2 so use 3rt3 +2 rt 2. multiply the top and bottom by the same (keep the equality) the purpose is to get rid of the rt   (i.e. to get a rational number)

mathstudent55 · Answer

This is your problem:

$\dfrac{2  \sqrt{3} + 5 \sqrt{2}}{{3  \sqrt{3} - 2 \sqrt{2}}}  $

The denominator is $3  \sqrt{3} \color{red}{-} 2 \sqrt{2}$.

The conjugate of the denominator is $3  \sqrt{3} \color{red}{+} 2 \sqrt{2}$.

Notice what @triciaal wrote above. You change the second part to the opposite of what it was, so $-2\sqrt{2} $ became $ + 2\sqrt{2} $.

Now you multiply the fraction you were given by a fraction whose numerator and denominator are the conjugate of the denominator.

$\dfrac{2  \sqrt{3} + 5 \sqrt{2}}{{3  \sqrt{3} - 2 \sqrt{2}}} \times \dfrac{ 3  \sqrt{3} + 2 \sqrt{2}   }{  3  \sqrt{3} + 2 \sqrt{2}  }$

Now you need to do the multiplication. Multiply the numerators together and multiply the denominators together. Use FOIL for the numerator. You can also use FOIL for the denominator, but a short cut is to use the pattern below.

Notice that the product of the denominators is the product of a sum and a difference and follows the pattern

$(a + b)(a - b) = a^2 - b^2$

Since you end up with $a^2$ and $b^2$, this eliminates the square roots from the denominator which is the purpose of rationalizing a denominator.