rationalizing a powered square root

[chutes]

Hi all,

I have to rationalize the following problem and show all work:

2-(sqrt) 5
_________

2 + 3(sqrt)5

read: 2 minus the square root of 5 divided by 2 plus the 3rd root of 5.


I took the opposite of the denominator (2 minus the 3rd root of 5), and when I do that, this is what I get:

((2- (sq rt 5)) * (2- 3(sqrt) 5)) / ((2+3(sqrt)5) (2- 3(sqrt)5))

simplified:

4- 6(sqrt)5- 2(sqrt)5+ 3(5) / (2-6 (sqrt) 5+ 6 (sqrt)5- 9(5))

then I cancelled the 6 sqrt 5:

4-4 (sqrt)5+15 / -43

The answer in the back of the book is: 19+ 8 (sqrt)5/ 41

My math went wrong somewhere, and I'm hoping somebody can help. Thanks.

 

[jwpaine]

\(\displaystyle \L \frac{2 - \sqrt{5}}{2 + 3\sqrt{5}}\)

Multiply by the denominator's conjugate:

\(\displaystyle \L \frac{2 - \sqrt{5}}{2 + 3\sqrt{5}} \cdot \frac{2-3\sqrt{5}}{2-3\sqrt{5}}\)

= \(\displaystyle \L \frac{(2-\sqrt{5})(2-3\sqrt{5})}{(2+3\sqrt{5})(2-3\sqrt{5})}\)

= \(\displaystyle \L \frac{19 - 8\sqrt{5}}{-41}\)

= \(\displaystyle \L \frac{8\sqrt{5} - 19}{41}\)

 

[chutes]

Ah, thank you.
Where did I go wrong in my foiling?

(The first line where you have the equal sign..)

 

[chutes]

duh!! the -6 (sqrt) 5 and the -2 (sqrt) 5 make the 8(sqrt)5. ahahaha

 

[chutes]

okay.. so do you distribute the negative in the denominator positive, and the numerator negative?

Where does the 41 come from? I kep getting -43.

 

[jwpaine]

okay.. so do you distribute the negative in the denominator positive, and the numerator negative?

Where does the 41 come from? I kep getting -43.

\(\displaystyle \L (2 + 3\sqrt{5})(2-3\sqrt{5}) = [2(2) + 2(-3\sqrt{5}) + 2(3\sqrt{5}) -(3\sqrt{5})^{2}] = -41\)

I'm not going to take the time to factor out like terms so that you can see; you should be able to do that yourself :wink:

John

 

[Deleted member 4993]

\(\displaystyle (2 + 3sqrt{5})(2 - 3sqrt{5}) = 2^2 - (3sqrt{5})^2 = 4 - 9\cdot5 = 4 - 45 = -41\)

 

[chutes]

Thanks.. After staring at it a while, I understand now. I appreciate all your help. Thanks, again.