# Thread: Solving radicals in the denominator AND flipping the sign(?)

1. ## Solving radicals in the denominator AND flipping the sign(?)

Hello Math Community,

I am a 34 year old who hasn't practiced math in a while. I am usually able to help my kiddo with assignments, but I am simply stumped on this problem. I have tried several different ways to research why it's necessary to flip the sign when solving radicals in the denominator, but I am struggling to find answers: I have, 1.) looked in the algebra book, 2.) looked at my kiddo's notes 3.) googled it.

I am not looking for a solution to this problem. I am looking to find WHY its necessary to flip the sign to be able to remove the radical from the denominator. Can you help me?

Solve:

(5) / (-5 - 3 * sqrt(3) )

Solution as determined by my kiddo's notes & photomath's app:

The step that confuses me is the bolded portion below. Why is it appropriate to multiply by a positive 3*sqrt(3) and not the negative 3*sqrt(3) like the denominator in the original expression above?

[ (5) * (-5 + 3 * sqrt(3) ) ] / [ (-5 - 3 * sqrt(3) ) * (-5 + 3 * sqrt(3) ) ]

Secondly, is there a resource that someone can recommend for me to use on how to solve problems like this? I am a fan of Khan Academy, but I am not sure why topic to search for.

Thank you for your help community.

2. ## Re: Solving radicals in the denominator AND flipping the sign(?)

It's all got to do with the fact that $\displaystyle (a+b)(a-b) = a^2 -ab +ab -b^2 =a^2 - b^2$ which is commonly called the difference of two squares.

You can see how the middle two terms in the second expression will always cancel each other out. Squaring a square root will get rid of the square root sign.

In your case: $\displaystyle (-5 - 3\sqrt{3})(-5+3\sqrt{3}) = (-5)^2 -15\sqrt{3} +15\sqrt{3} -(3\sqrt{3})^2 = 25 -27 =-2$

See how the middle terms cancel out. You are left with a rational number (hence the term "rationalise the denominator").

Try multiplying $\displaystyle (-5 -3\sqrt{3})(-5-3\sqrt{3})$ , (that is, without changing the sign to +).

You will see that the middle terms don't cancel and you will not have a rational result.

3. ## Re: Solving radicals in the denominator AND flipping the sign(?)

Thank you very much! That made COMPLETE sense. And I absolutely understand everything you said.