Desnesting Radicals (or Unnesting Radicals)
Copyright © 2016–2019 by Stan Brown
Copyright © 2016–2019 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 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.
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. 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 article, 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 at all, then the following method will do it.
To unnest √a ± √b, you need:
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.
√a + √b = √x + √y
√a − √b = √x − √y
√a² − b = x − y
a + √b = x + 2 √xy + y
a = x + 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.
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. (In fact, it can’t be unnested at all.)
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 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)
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 = 21/4 √14 + √180
where 21/4 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
√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 21/4 times that. So the final solution is
√√392 + √360 = 21/4 (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 81/4 (√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.
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.
√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
x = ½ (a + √a² − c²d) and y = ½ (a − √a² − c²d)
Try this with problem 5, √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
√√1058 − √896 = 21/4 √23 − √896
x + y = 23 and x − y = 9, so x = 16 and y = 7.
√√1058 − √896 = 21/4 (4 − √7)
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