Denesting Radicals (or Unnesting Radicals)

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’t be unnested, but the second one can? It surprised me. Read on to learn why that’s true, or just skip to THE Method and try them yourself.

Some Techniques

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

What we might call a brute-force approach is in Un-nesting Radicals. “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.)

Peterson included a link to a page by Dave Rusin that’s no longer there, but has been preserved at the Internet Wayback Machine. That page was a sci.math thread from 1999. In it, if you scroll about halfway down, you’ll find an exciting article from Dave L. Renfro. He referenced 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 a² > 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.

Acknowledgement

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.

Is It Doable?

In http://mathforum.org/kb/message.jspa?messageID=5451111 (accessed 2016-12-18, but no longer available and unfortunately not archived at the Internet Wayback Machine), 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. The fifth edition (1904) is on Google Books, and the relevant section is “Square Roots of Simple Surd Numbers”, starting on page 207 of Part I.

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, accessed 2016-12-18 at http://mathforum.org/kb/message.jspa?messageID=5451310 but unfortuately no longer available, Renfro adds that we are meant to understand b must also be > 0 and √b must be irrational. Of course if √b is rational, there’s no need to find a denesting technique in the first place.

Bottom line: If √a ± √b can be denested to radicals of rational numbers, then the following method will do it.

Let’s Be Rational Here

In this article, I’ll be considering only solutions that produce roots of rational numbers. Some forms, like Example 1, can be denested only if you allow roots of complex numbers, and I may go into that further in a future revision.

THE Method

To unnest √a ± √b to radicals of rational numbers, you need:

• a and b positive and rational.
• a² − b is a perfect square (or √a² − b is rational).
• √b is irrational.
• If you’re denesting √a − √b, a² > 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.

1. Assume that there exist some rational x and y such that

   √a + √b = √x + √y

2. Using the result from Wells, above, multiply by

   √a − √b = √x − √y

3. This yields

   √a² − b = x − y

4. Square the equation from step 1:

   a + √b = x + 2 √xy + y

5. Set the rational parts equal:

   a = x + y

6. Solve the equations from steps 3 and 5 for x and y:

   x = ½ (a + √a² − b)

   y = ½ (a − √a² − 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, but your answer is √x − √y.

Examples

Example 1: √2 + √2. Solution: a = 2, b = 2. a² − b = 4 − 2 = 2, and √2 is not rational, so this one can’t be simplified by this method. Therefore we know that it can’t be denested to roots of rational numbers by any method.

Interestingly, if you allow radicals of complex numbers, it can be rewritten: √2 + √2 = 2^−1/4 (√1 + i + √1 − i).

Example 2: √2 + √3. Solution: a² − 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√xy + y

On the left side, 2 is rational and √3 is irrational. On the right side, x + y is rational and 2√xy is irrational. (If √xy were rational, then √a² + 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 usually 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√xy + 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)

What about √√p ± √q?

In Advanced Course in Algebra (1904), starting on page 235. Webster Wells shows a similar method. He also points out, on the next page, that some expressions of the form √√p ± √q can be denested by the same method, if pulling out a common factor reduces them to the √a ± √b form.

Example 4 (from Wells): √√392 + √360. Solution: The key is to recognize a common factor between the two inner radicals and factor it out. If one of the resulting numbers under the radicals is a perfect square, you’re back in the √a + √b situation.

Here, 392 and 360 obviously have a common factor of 2, so pull that out:

√√392 + √360 = √ √2 (√196 + √180)

√√392 + √360 = √ √2  √14 + √180

√√392 + √360 = 2^1/4 √14 + √180

where 2^1/4 is the fourth root of 2.

Now you have 14² − 180 = 196 − 180 = 16, which is a perfect square, so the method will work. Assume

√14 + √180 = √x + √y

Multiply by

√14 − √180 = √x − √y

which gives you

√14² − 180 = √196 − 180 = √16 = x − y⇒x − y = 4

And squaring √14 + √180 = √x + √y yields

14 + √180 = x + y + 2 √xy⇒x + y = 14

Therefore x = 9 and y = 5, and

√14 + √180 = √9 + √5 = 3 + √5

But wait! The problem wasn’t √14 + √180, but 2^1/4 times that. So the final solution is

√√392 + √360 = 2^1/4 (3 + √5)

In fact, 392 and 360 have a common factor of 8. Why did Wells pull out only 2, not 8? There’s nothing wrong with factoring out the larger number, but then when you apply the method you get 8^1/4 (√9/2 + √5/2), and you have to do more work to simplify it. (Yes, the final answer is the same.) So your best strategy is to pull out the smallest common factor that leaves you with a perfect square under one of the two radicals.

Now It’s Your Turn

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.

1. √3 + √5
2. Use your answer to the previous exercise to simplify √3 − √5.
3. √2 − √5
4. √2 + √5
5. √5 − √21
6. √20 − √144
7. √10 − 4√6 (Hint: 4√6 is the square root of what number?)
8. √10 + 2√6
9. √a ± c√d for positive rational a, c, d, with irrational √d and positive rational a² − c²d. (This is an extension of what you did in problem 5.)
10. From Wells, page 239: √√1058 − √896

Solutions

1. √5/2 + √1/2. If you don’t like fractions under radicals, multiply by √2/2 (which is 1), factor out √1/4 as ½, and give your answer as (½)(√10 + √2).
2. √5/2 − √1/2, or (½)(√10 − √2).
3. 2² = 4, which is less than 5. This radical isn’t a real number.
4. The radicand is a positive number, so the radical is a real number, but still 2² − 5 = −1 is negative, so you can’t denest this one either.
5. √7/2 − √3/2, or (½) (√14 − √6)
6. √144 is rational, so you don’t need the denesting method. √20 − √ 144 = √20 − 12 = √8 = 2√2
7. 4√6 = √16 × 6 = √96. Therefore √10 − 4√6 = √10 − √96 = √6 − 2
8. a² − 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.
9. The key is that c√d = √c²d. So you are looking for

   √a ± √(c²d) = √x ± √y

   This is exactly the same as THE Method, just replacing b with c²d. You can then jump right to the solution, in step 6 of THE Method, again substituting c²d for b:

   √a ± c√d = √x ± √y

   where

   x = ½ (a + √a² − c²d) and y = ½ (a − √a² − c²d)

   Try this with problem 7, √10 − 4√6. Here a = 10, c = 4, d = 6. a² − c²d = 100 − 16 × 6 = 4 is rational, and √a² − c²d = 2. x = (½)(10 + 2) = 6, and y = (½)(10 − 2) = 4. Therefore

   √10 − 4√6 = √6 − √4

   √10 − 4√6 = √6 − 2

10. This looks worse than it is. The only common factor of 1058 and 896 is 2. 1058/2 = 529 = 23² and 896/2 = 448, so

    √√1058 − √896 = 2^1/4 √23 − √448

    Now THE Method can be used. x + y = 23 and x − y = 9, so x = 16 and y = 7.

    √√1058 − √896 = 2^1/4 (4 − √7)