1. ## How to turn one hard radical expression into another

Hello

I am trying to prove that -sqrt(5) + sqrt(11 + 6*sqrt(2)) = sqrt(2) + sqrt(14 - 6*sqrt(5)).

I don't even know where to begin.

Thanks a lot!

2. ## Re: How to turn one hard radical expression into another

$\displaystyle -\sqrt{5} \ + \ \sqrt{11 + 6\sqrt{2}} \ \ \ \ =? \ \ \ \ \sqrt{2} \ + \ \sqrt{14 - 6\sqrt{5}}$

A possible way is to simplify $\displaystyle \ \sqrt{11 + 6\sqrt{2}} \$ to the form $\displaystyle \ a + b\sqrt{2}, \$ and to simplify $\displaystyle \ \sqrt{14 - 6\sqrt{5}} \$ to the form $\displaystyle \ c + d\sqrt{5}, \ \$
where a, b, c, and d are rational numbers.

$\displaystyle a + b\sqrt{2} \ = \ \sqrt{11 + 6\sqrt{2}}$

$\displaystyle (a + b\sqrt{2})^2 \ = \ (\sqrt{11 + 6\sqrt{2}} \ )^2$

$\displaystyle a^2 + 2ab\sqrt{2} + 2b^2 \ = \ 11 + 6\sqrt{2} \ \implies$

$\displaystyle a^2 + 2b^2 \ = \ 11$

and

$\displaystyle 2ab\sqrt{2} \ = \ 6\sqrt{2}$

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Also,

$\displaystyle c + d\sqrt{5} \ = \ \sqrt{14 - 6\sqrt{5}}$

$\displaystyle (c + d\sqrt{5})^2 \ = \ (\sqrt{14 - 6\sqrt{5}} \ )^2$

$\displaystyle c^2 + 2cd\sqrt{5} + 5d^2 \ = \ 14 - 6\sqrt{5} \ \implies$

$\displaystyle c^2 + 5d^2 \ = \ 14$

and

$\displaystyle 2cd\sqrt{5} \ = \ -6\sqrt{5}$

3. ## Re: How to turn one hard radical expression into another

Originally Posted by greg1313
$\displaystyle -\sqrt{5} \ + \ \sqrt{11 + 6\sqrt{2}} \ \ \ \ =? \ \ \ \ \sqrt{2} \ + \ \sqrt{14 - 6\sqrt{5}}$

A possible way is to simplify $\displaystyle \ \sqrt{11 + 6\sqrt{2}} \$ to the form $\displaystyle \ a + b\sqrt{2}, \ \$ and to simplify $\displaystyle \ \sqrt{14 - 6\sqrt{5}} \$ to the form $\displaystyle \ c + d\sqrt{5}, \ \$
where a, b, c, and d are rational numbers.

$\displaystyle (a + b\sqrt{2})^2 \ = \ \sqrt{11 + 6\sqrt{2}}$

$\displaystyle a^2 + 2ab\sqrt{2} + 2b^2 \ = \ \sqrt{11 + 6\sqrt{2}} \ \implies$

$\displaystyle a^2 + 2b^2 \ = \ 11$

and

$\displaystyle 2ab\sqrt{2} \ = \ 6\sqrt{2}$

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Also, $\displaystyle (c + d\sqrt{5})^2 \ = \ \sqrt{14 - 6\sqrt{5}}$

$\displaystyle c^2 + 2cd\sqrt{5} + 5d^2 \ = \ \sqrt{14 - 6\sqrt{5}} \ \implies$

$\displaystyle c^2 + 5d^2 \ = \ 14$

and

$\displaystyle 2cd\sqrt{5} \ = \ -6\sqrt{5}$
Hello!

Thank you very much for your reply. Unfortunately, I don't understand it at all. How does this turn the left side of the equation into the right one or vice versa?

Also, I don't understand how you conclude this:

a^2+2b^2 = 11

and

2ab = 6*sqrt(2)

And the respective step for the next equation. How did you arrive at this?

Thanks a lot!

4. ## Re: How to turn one hard radical expression into another

Originally Posted by zehgs
Hello!

Thank you very much for your reply. Unfortunately, I don't understand it at all. How does this turn the left side of the equation into the right one or vice versa?

Also, I don't understand how you conclude this:

a^2+2b^2 = 11

and

2ab = 6*sqrt(2)

And the respective step for the next equation. How did you arrive at this?

Thanks a lot!
My post (post #2) was being edited back and forth when you put it in a quote box. Please look at the most recent edits for the content to make the most sense.

- - - - - - - - - - - - - - - - - - - -

For example, the following you posted about was an error I was catching during my edit process:

Originally Posted by zehgs

Also, I don't understand how you conclude this:

a^2+2b^2 = 11 and 2ab = 6*sqrt(2)

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

$\displaystyle a^2 + 2b^2 \ = \ 11$

and

$\displaystyle 2ab\sqrt{2} \ = \ 6\sqrt{2}$

Second equation:

Divide by $\displaystyle \ 2\sqrt{2} \$ on each side:

ab = 3

Solve for either a or b in terms of the other, and substitute that into the first equation:

$\displaystyle a = \dfrac{3}{b}$

$\displaystyle \bigg(\dfrac{3}{b}\bigg)^2 \ + \ 2b^2 \ = \ 11$

$\displaystyle \dfrac{9}{b^2} \ + \ 2b^2 \ = \ 11$

$\displaystyle 9 + 2b^4 \ = \ 11b^2$

$\displaystyle 2b^4 - 11b^2 + 9 \ = \ 0$

Solve this for an appropriate value(s) for b, and then substitute back to find a. (It factors.)

5. ## Re: How to turn one hard radical expression into another

Thanks a lot! Now I understand the part I mentioned. However, I still don't understand how this helps me.

Didn't you simply show that sqrt(11 + 6*sqrt(2)) is already in the form of a + b*sqrt(2)? Shouldn't I swap the 2 around, meaning show that sqrt(11 + 6*sqrt(2)) is in the form c + d*sqrt(5) and thus show that the left side is in the form of the right side? If so, wouldn't the algebra be significantly less clean? I don't think I could do it to be honest. I do suspect that the isolated square roots in either side of the equations would have something to do with it, right?

6. ## Re: How to turn one hard radical expression into another

Originally Posted by zehgs
Thanks a lot! Now I understand the part I mentioned. However, I still don't understand how this helps me.

Didn't you simply show that sqrt(11 + 6*sqrt(2)) is already in the form of a + b*sqrt(2)? Shouldn't I swap the 2 around, meaning show that sqrt(11 + 6*sqrt(2)) is in the form c + d*sqrt(5)
and thus show that the left side is in the form of the right side? If so, wouldn't the algebra be significantly less clean? I don't think I could do it to be honest. I do suspect that the
isolated square roots in either side of the equations would have something to do with it, right?
What happens if you do what I explain in my couple of posts is that you show that sqrt(11 + 6*sqrt(2)) = 3 + sqrt(2), and that sqrt(14 - 6*sqrt(5)) = 3 - sqrt(5).

Substitute these reduced quantities back into the sides of what you were given:

-sqrt(5) + (3 + sqrt(2)) $\displaystyle \ \$vs. $\displaystyle \ \$sqrt(2) + (3 - sqrt(5))

3 + sqrt(2) - sqrt(5) $\displaystyle \ = \$ 3 + sqrt(2) - sqrt(5)

7. ## Re: How to turn one hard radical expression into another

Ohhh, I got it now! Thank you so very much. Bye!

8. ## Re: How to turn one hard radical expression into another

$11+6\sqrt{2} = 9 + 6\sqrt{2} + 2 = (3 + \sqrt{2})^2 \implies -\sqrt{5} + \sqrt{11+6\sqrt{2}} = \color{red}{-\sqrt{5} + (3 + \sqrt{2})}$

$14 - 6\sqrt{5} = 9 - 6\sqrt{5} + 5 = (3 - \sqrt{5})^2 \implies \sqrt{2} + \sqrt{14-6\sqrt{5}} = \color{red}{\sqrt{2} + (3 - \sqrt{5})}$

9. ## Re: How to turn one hard radical expression into another

Thanks Cervesa! That's pretty cool to know as well.

10. ## Re: How to turn one hard radical expression into another

Hi guys! Sorry to ask for your help again but what about for this one? I've been struggling to apply the same reasoning:

sqrt(5 + 2*sqrt(5)) - sqrt(5 - 2*sqrt(5)) = sqrt(10 - 2*sqrt(5))

Thanks!

Edit: I tried squaring both sides and it worked, but what if it didn't?