De-Nesting Radicals: Solving & Simplifying Solutions

[bamajon1974]

I want to de-nest the following radical:
(1) $$\sqrt{3+2\sqrt{2}}$$
Into the general simplified form:
(2) $$a+b\sqrt{2}$$
Equating (1) with (2),
(3) $$\sqrt{3+2\sqrt{2}} = a+b\sqrt{2}$$
and squaring both sides:
(4) $$3+2\sqrt{2} = a^2 + 2b^2 + 2ab\sqrt{2}$$
generates a system of two equations with two unknowns, a and b, after equating the rational and irrational parts:
(5) $$3 = a^2 + 2b^2$$
(6) $$2\sqrt{2} = 2ab\sqrt{2}$$
Simplifying (6) and solving for b:
(7) $$b=\frac{1}{a}$$
Substituting (7) into (5) yields:
(8) $$3 = a^2 + 2\frac{1}{a^2}$$
Clearing the denominator and moving non-zero terms to one side generates a quartic equation:
(9) $$0 = a^4 - 3a^2 +2$$
that can be made into a quadratic equation with the substitution:
(10) $$x=a^2, x^2 = (a^2)^2 = a^4, x = \pm \sqrt{a}$$
(11) $$0 = x^2 - 3x + 2$$
The square root of the discriminant is an integer, 1, which presumably makes simplification of the nested radical possible. Finding the roots, x, of (11) gives
(12)$$x=1, x =\sqrt{2}$$
Substituting the roots in (12) into (10) gives a:
(13) $$a = \pm \sqrt{2} , a = \pm 1$$
Then b is found from (7):
(14) $$b = \pm \frac{1}{\sqrt{2}} , b = \pm 1$$
Using the positive values of a and b, the de-nested radical is:
(15) $$\sqrt{3+2\sqrt{2}} = 1+\sqrt{2}$$

My questions are:
(1) Is this approach for simplifying nested radical correct?

(2) The positive values of a and b produce the correct simplified form in (15) while the negative values of a and b do not. Is there a way to figure out which roots from (9) are correct and which to reject other than calculating the numerical value of (15) with both positive and negative a's and b's to see which is equal to the nested form?

(3) I have also seen the general simplified de-nested form as:
(16) $$\sqrt{a} + \sqrt{b}$$
Going through an analogous process as above, it generates the correct simplified form in (15) as well. Is one form (2) or (15) better than the other? Or is it just a personal preference which one to use?

Thanks!

 

[symbolipoint]

The square root radical symbol means, the expression under the radical raised to the one-half power.

$$\sqrt{3+2\sqrt{2}}$$
$$(3+2\sqrt{2})^(1/2)$$
$$(3+2(2)^(1/2))^(1/2)$$

( I KNOW what I'm trying to do but still cannot make it appear correctly)

 

[SteamKing]

The square root radical symbol means, the expression under the radical raised to the one-half power.

$$\sqrt{3+2\sqrt{2}}$$
$$(3+2\sqrt{2})^{(1/2)}$$
$$(3+2(2)^{(1/2}))^{(1/2)}$$

( I KNOW what I'm trying to do but still cannot make it appear correctly)

You mean like this? Just enclose your expression for the exponent in a pair of curly braces {}.

 

[Mentallic]

(12)$$x=1, x =\sqrt{2}$$

Your first mistake.

Substituting the roots in (12) into (10) gives a:
(13) $$a = \pm \sqrt{2} , a = \pm 1$$

Your second mistake (assuming your first mistake was correct).

I'll give you a chance to look at these again. You've ended up with the correct answer however because you made two errors that both canceled each other out, so that was a lucky one on your part ;)

My questions are:
(1) Is this approach for simplifying nested radical correct?

Yes.

(2) The positive values of a and b produce the correct simplified form in (15) while the negative values of a and b do not. Is there a way to figure out which roots from (9) are correct and which to reject other than calculating the numerical value of (15) with both positive and negative a's and b's to see which is equal to the nested form?

You'll often have situations arise whereby you need to find, say, the length of an object, but the equations you're dealing with are quadratics and will thus give you two solutions, a positive and a negative one. It suffices to toss away the negative value because you're solving a problem that only makes sense with a positive answer.

It's similar in this case. You're trying to simplify a square root. You know that square roots don't give negative results so you're allowed to simply toss out the result $-1-\sqrt{2}$ because that is clearly negative and can't be the answer. You don't need to look at the numerical values of each result with a calculator if you simply use a bit of intuition.

(3) I have also seen the general simplified de-nested form as:
(16) $$\sqrt{a} + \sqrt{b}$$
Going through an analogous process as above, it generates the correct simplified form in (15) as well. Is one form (2) or (15) better than the other? Or is it just a personal preference which one to use?

Thanks!

I prefer yours merely because the $\sqrt{2}$ result cancels and you end up with a nice result in (7). Neither are incorrect, but also both are assumptions. You assumed that the answer is of the form $a+b\sqrt{2}$ and the "general" simplified de-nested form $\sqrt{a}+\sqrt{b}$ assumes that the result is the sum of two square roots. What if it were $\sqrt{a}+\sqrt{b}+\sqrt{c}$ instead? Or something even more complicated?
So yes, I prefer your assumption over the other because yours is easier to work with.

 

[bamajon1974]

I see my error:
12)
$$x=1, x=\sqrt{2}$$
Should be
$$x=1, x=2$$
My bad. Although in my defense that was an error I didn't catch after typing so much unfamiliar latex code into these lines.

I had neglected to mention that in the if you try to de-nest
$$\sqrt{3+2\sqrt{2}}$$
by assuming it equals the other simplified form (16)
$$\sqrt{a} + \sqrt{b}$$
Equating the nested radical with the assumed simplified form and squaring both sides:
$$3 + 2\sqrt{2} = a + b + 2\sqrt{ab}$$
Equating rational and irrational parts generates 2 equations with 2 unknowns:
$$3 = a + b$$
$$2\sqrt{2} = 2\sqrt{ab}$$
Solving for b in the second equation, substituting into the first and moving all non-zero terms to one side of the equation generates a quadratic:
$$0=a^2+3a+2$$
of which the roots, a, are
$$a= 1, a=2$$
So a and b are
$$a=1, b=2$$
or
$$a=2, b=1$$
So the simplified form is
$$1+\sqrt{2}$$
identical to above, but I didn't have any negative roots to discard. So in this particular case, the second form was a little easier.

I see why rejecting the negative roots is justified as well.