Radicals Within Radicals: Simplifying and Solving Equations

[ExamFever]

Homework Statement

sqrt(7 + 2sqrt(6)) - sqrt(7 - 2sqrt(6))

Homework Equations

The Attempt at a Solution

In order to work out the radicals, I define the answer as "x". Now I can square both sides to get rid of some of the first radicals:

sqrt(7 + 2sqrt(6)) - sqrt(7 - 2sqrt(6)) = x
(sqrt(7 + 2sqrt(6)) - sqrt(7 - 2sqrt(6)))² = (x)²

Now, I hope this is the right way of starting on this problem, continuing we get:

(7 + 2sqrt(6)) - 2 * sqrt(7+ 2 sqrt(6)) * sqrt(7 - 2 sqrt(6)) + (7 - 2 sqrt(6))

this implies:

14 - 2 * sqrt(7+ 2 sqrt(6)) * sqrt(7 - 2 sqrt(6)) = x²

on moving the 14 to the RHS and squaring once more to get rid of more radicals:

(-2 * sqrt(7+ 2 sqrt(6)) * sqrt(7 - 2 sqrt(6)) )² = (x²-14)²

this implies

4 * (7 + 2 sqrt(6)) * (7 - 2 sqrt(6)) = (x²-14)²

because the two factors with sqrts are conjugate pairs, the LHS will look like this:

4 * (7² - (4*6) = 4 * 25 = 100

thus

100 = (x²-14)²

by taking the square root of both sides:

±10 = ± (x²-14)
±10±14 = ± x²

Now, this does not look like a proper solution. Working further I can get a couple of answers for x, but no way to check them or anything:

sqrt(24) = 2 sqrt(6) = ± x
sqrt(4) = 2 = ± x

Where do I go wrong?

 

[berkeman]

You can check the answer by using a calculator on the original problem line. In fact, you can use a calculator to check each line of your solution...

 

[ExamFever]

Yes, however, the problem states I should be able to calculate without the use of a calculator...

 

[berkeman]

Sure, I'm just saying that you can use a calculator to see if what you are doing algebraically is working...

 

[ehild]

sqrt(7 + 2sqrt(6)) - sqrt(7 - 2sqrt(6)) = x
(sqrt(7 + 2sqrt(6)) - sqrt(7 - 2sqrt(6)))² = (x)²

Now, I hope this is the right way of starting on this problem, continuing we get:

(7 + 2sqrt(6)) - 2 * sqrt(7+ 2 sqrt(6)) * sqrt(7 - 2 sqrt(6)) + (7 - 2 sqrt(6))

this implies:

14 - 2 * sqrt(7+ 2 sqrt(6)) * sqrt(7 - 2 sqrt(6)) = x²

on moving the 14 to the RHS and squaring once more to get rid of more radicals:

Why do you make it so complicated? Simplify the left side, using that

\sqrt{7+ 2 \sqrt6} \sqrt{7 - 2 \sqrt6} =\sqrt{(7+ 2 \sqrt6) (7 - 2 \sqrt6)}=?

As for the sign of x, 7+2sqrt(6)>7-2sqrt(6), and so are the square roots. The difference is positive.

ehild

 

[Hurkyl]

but no way to check them or anything:

sqrt(24) = 2 sqrt(6) = ± x
sqrt(4) = 2 = ± x

Why do you think there is no way to check? Isn't it obvious the original number is positive? And it's pretty easy to show that it's less than 4 too.

Sometimes separating the solutions via inequalities requires a lot of precision, but this one only requires very coarse estimates.

P.S. it's probably better to split into cases, rather than using ±; it's easier to avoid making mistakes that way.

 

[ExamFever]

Thanks to all of you. When I read your answers everything becomes clear to me. The only thing I do not understand sometimes is how I could be so stupid not to see it...I paid special attention to seeing the rules for radicals in here but missed the simple fact at one point that sqrt(a) * sqrt(b) = sqrt(ab)

I would practice more on my own not to ask so many seemingly stupid questions. Unfortunately I have very little time left and I need to evolve my basic understandings into a (somewhat) more advanced understanding of the basic rules.

Just want to say I really appreciate the input and patience. Feels like I found a great place to get help with math...sometime the internet is too big to find something like that, not this time luckily :)

Cheers,