# Desnesting Radicals (or Unnesting Radicals)

Copyright © 2016–2017 by Stan Brown

Copyright © 2016–2017 by Stan Brown

**Summary:**
Expressions like √2 + √2 and
√2 − √3 come up in solutions to problems,
even pretty basic algebra and trig problems. How are they simplified?
Why can some be simplified, but not others?

A bunch of odd little corners exist in algebra. They used to
be taught, back in the 1800s and early 1900s, but even by the time I
took algebra in the 1960s many had fallen by the wayside. One is the
technique of **unnesting radicals** (or
**denesting radicals**, **de-nesting radicals**,
**un-nesting radicals** — spellings vary).

It’s kind of cool, and the need arises surprisingly often. For example, sin π/8 (or sin 22½°) involves √2 − √2, and sin 15° (or sin π/12) involves √2 − √3. Can these be simplified to avoid nesting radicals?

Does it surprise you to learn that the first one can, and the second one can’t? It surprised me. Read on to learn why that’s true, or just skip to THE Method and try it yourself.

Googling turns up plenty of approaches, of varying levels of difficulty.

What we might call a brute-force approach is in Un-nesting Radicals at Ask Dr. Math, “Doctor Rob” squares the original equation, rearranges, and squares again. This leads to fourth-power equations and solution of five linear equations in five unknowns.

If you make an assumption that this simplified radical (if it
exists) will have the form
√*x* ± √*y*, the problem
reduces to finding *x* and *y*. Along the way, you may find that
*x* and *y* in fact don’t exist, or at least not as rational
numbers.

“Doctor Peterson” takes this approach in Simplifying Radicals. By making an assumption about the desired form of the solution, he makes the calculation a lot simpler than the brute-force method. (And it turns out that the assumption is valid if there’s a solution at all.) He also posted a link to a page that’s no longer there, but has been preserved at the Internet Wayback Machine.

And that page, a sci.math thread from 1999, contained an exciting article from Dave L. Renfro. You can find that article on its own at Math Forum. The exciting part is that Renfro references a textbook by Webster Wells (1904), Advanced Course in Algebra. Wells proves that if

√*a* + √*b* =
√*x* + √*y*

(for rational *a*, *b*, *x*, *y*, and
*a*² > *b*), then

√*a* − √*b* =
√*x* − √*y*

Why is this exciting? Because you can multiply the second
equation by the first, and get rid of three of the four radical
signs. And that in turn lets you proceed to a simple and direct
solution for *x* and *y*.

Not only is this article based on that 1999 article by Dave L. Renfro, but in personal correspondence (December 2016) he generously supplied additional links and information.

In
another article,
Dave L. Renfro answered the question of whether radicals of the form
√*a* ±√*b* can be denested.
He quoted George Chrystal’s Algebra: An
Elementary Text-Book (7th ed). The 7th edition listed in
WorldCat was published in 1964, but Chrystal lived 1851–1914,
so either that’s a reprint or revisions were continued by the
publisher after Chrystal’s death.

The theorem as stated (with some adjustment to notation by me)
is “Let *a*, *b* be rational numbers. Then
**√ a ± √b can be expressed as √x ± √y for some rational numbers x, y if and only if a > 0 and a² − b is a perfect square.**”
That’s not quite as restrictive as you might
think — “perfect square” can be the square of a
rational number, not necessarily an integer.

In
a follow-up article,
Renfro adds that we are meant to understand *b* must also be
> 0 and √b must be irrational. Of course if
√q is rational, there’s no need to find a denesting
technique in the first place.

Bottom line:
**If √ a ± √b can be denested at all, then the following method will do it.**

**To unnest √ a ± √b, you need:**

*a*and*b*positive and rational.*a*² −*b*is a perfect square (or √*a*−*b*is rational).- √
*b*is irrational. - If you’re denesting √
*a*− √*b*,*a*² >*b*.

Let’s start with
√*a* + √*b*. As you’ll see in
Example 3, the same method works if you replace the
plus sign with a minus.

- Assume that there exist some rational
*x*and*y*such that√

*a*+ √*b*= √*x*+ √*y* - Using the result from Wells, above,
multiply by
√

*a*− √*b*= √*x*− √*y* - This yields
√

*a*² −*b*=*x*−*y* - Square the equation from step 1:
*a*+ √*b*=*x*+ 2 √*x**y*+*y* - Set the rational parts equal:
*a*=*x*+*y* - Solve the equations from steps 2 and 5 for
*x*and*y*:*x*= ½ (*a*+ √*a*² −*b*)*y*= ½ (*a*− √*a*² −*b*)

If the original nested radical has a minus sign,
√*a* − √*b*, equations (1) and (2) come
out exactly the same, so *x* and *y* are the same.

**Example 1:** √2 + √2.
**Solution:** *a* = 2, *b* = 2.
*a*² − *b* = 4 − 2 = 2, and
√2 is not rational, so this one can’t be simplified
by this method. (In fact, it can’t be unnested at all.)

**Example 2:** √2 + √3.
**Solution:** *a*² − *b* =
4 − 3 = 1, and √1 is of course rational, so
this method will denest the radicals.
Assume that there exist rational *x* and *y* such
that

√2 + √3 =
√*x* + √*y*

Multiply by
√2 − √3 =
√*x* − √*y*, and you get

√4 − 3 = *x* − *y* ⇒
*x* − *y* = 1

Now go back to the original “assume that” equation, and square it:

2 − √3 =
x + 2√*x**y* + *y*

On the left side, 2 is rational and √3 is irrational.
On the right side, *x* + *y* is rational and
2√*x**y* is irrational. (If √*x**y* were
rational, then √*a*² + *b* would have been
rational and you wouldn’t have any nested radicals to begin
with.) The rational parts are equal:

2 = *x* + *y*

Solve that simultaneously with the equation
*x* − *y* = 1 that we developed a couple of
steps back, and you get

*x* = 3/2, *y* = 1/2
⇒
√2 + √3 =
√3/2 + √1/2

Algebra teachers want us to rationalize denominators, so you can multiply each radical on the right-hand side by √2/2 to get

√2 + √3 = (1/2)(√6 + √2)

**Example 3:** √2 − √3.
**Solution:** This is exactly the same deal as Example 2, so
I’ll hurry through it.

Assume: √2 − √3 =
√*x* − √*y* for rational *x*
and *y*.

Multiply by √2 + √3 =
√*x* + √*y* to obtain

√4 − 3 = 1 =
*x* − *y* ⇒
*x* − *y* = 1

This exactly matches the equation in Example 2. Now square the “Assume” equation:

2 − √3 =
*x* − 2√*x**y* + *y*

Take the rational parts:

2 = *x* + *y*

This equation, and *x* − *y* = 1, are just
the same as in Example 2, so the solution is the same except for the
signs:

√2 − √3 = (1/2)(√6 − √2)

Here are a few exercises to test your understanding. In each case, simplify (denest) the radical, or explain why it cannot be done. Try them yourself with pencil and paper, and then check your answers.

- √7 + √5
- √2 − √5
- √5 − √21
- √20 − √144
- √10 − 4√6 (Hint: 4√6 is the square root of what number?)
- √10 + 2√6
- √
*a*±*c*√*d*for positive rational*a*,*c*,*d*, with irrational √*d*and positive rational*a*² −*c*²*d*. (This is an extension of what you did in problem 5.)

- √10 + 5 √3
- 2² = 4, which is less than 5. This radical isn’t a real number.
- (1/2) (√14 − √6)
- √144 is rational, so you don’t need the denesting method. √20 − √ 144 = √20 − 12 = √8 = 2√2
- 4√6 = √16 × 6 = √96. Therefore √10 − 4√6 = √10 − √96 = √6 − 2
*a*² −*b*= 100 − 24 = 76 is not a perfect square, so this one can’t be unnested. About all you can do is factor out √2 to get √2 √5 + √6, and that hardly seems worth the effort.- The key is that
*c*√*d*= √*c*²*d*. So you are looking for√

*a*± √(*c*²*d*) = √*x*± √*y*This is exactly the same as THE Method, just replacing

*b*with*c*²*d*. You can then jump right to the solution, in step 6 of THE Method, again substituting*c*²*d*for*b*:√

*a*±*c*√*d*= √*x*± √*y*where

*x*= ½ (*a*+ √*a*² −*c*²*d*) and*y*= ½ (*a*− √*a*² −*c*²*d*)Try this with problem 5, √10 − 4√6. Here

*a*= 10,*c*= 4,*d*= 6.*a*² −*c*²*d*= 100 − 16 × 6 = 4 is rational, and √*a*² −*c*²*d*= 2.*x*= (½)(10 + 2) = 6, and*y*= (½)(10 − 2) = 4. Therefore√10 − 4√6 = √6 − √4

√10 − 4√6 = √6 − 2

**21 Dec 2016**: Added the solution for √*a*±*c*√*d*as an exercise.**18 Dec 2016**: New article.

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