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Denesting Radicals (or Unnesting Radicals)

Copyright © 2016–2024 by Stan Brown, BrownMath.com

Summary: Expressions like √2 + √2 and √2 − √3 come up in solutions to problems, even pretty basic algebra and trig problems. How can you eliminate the nested radicals? Read on for several techniques, with examples.

Contents:

Denesting radicals — also known as unnesting radicals, de-nesting radicals, or un-nesting radicals — is one of the odd little corners that 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.

Denesting radicals is kind of cool, and they arise surprisingly often. For example, sin π/8 (sin 22½°) involves 2 − √2, and sin 15° (sin π/12) involves 2 − √3. Of course, you can evaluate them with a calculator, but how can you eliminate the nesting without resorting to a calculator for merely approximate answers?

Denesting — Some of the Hard Ways

Googling turns up plenty of approaches, some harder than others. This section shows a couple of the traditional methods, for historical interest and for contrast with the much easier ways in Method 1 and Method 2.

Brute Force, Complete with Quartic Equations

What we might call a brute-force approach is in Un-nesting Radicals. “Doctor Rob” squares the original equation, rearranges, and squares again. For example:

x = √2 − √3

x² = 2 − √3

x² + 2 = √3

x⁴ − 4x² + 4 = 3

x⁴ − 4x² + 1 = 0

As in this example, the brute-force method typically ends up with a quartic (also called biquadratic) equation. Only rarely is there any simple way to factor the left-hand side; usually you’d have to spend an afternoon solving the equation.

This particular example happens to be one of the rare ones that are fairly easy to factor:

(x⁴ − 2x² + 1) − 2x² = 0

(x² − 1)² − (√2x)² = 0

(x² − 1 + √2x) (x² − 1 − √2x) = 0

(x² + √2x − 1) (x² − √2x − 1) = 0

Then all you need to do (all!) is apply the quadratic formula to each factor, and figure out which one of the four roots is equal to the original nested radical. (Spoiler: it’s (√6 − √2)/2.) No thanks — and this is one of the easy ones!

“Doctor Rob” gives a technique for factoring the fourth-power polynomial when it doesn’t have a simple factorization like this one. That involves solving five equations in five unknowns. Then you get to use the quadratic formula on both factors and figure out which of the four solutions is correct.

Assuming √a ± √b Denests as √x ± √y if It Denests at All

“Doctor Peterson” takes this approach in Simplifying Radicals. Making that assumption reduces the problem to finding the values of x and y, which means only two equations in two unknowns and one quadratic equation, nothing with higher powers.

That’s better than brute force, certainly, but how do we know that the assumption is correct for every case — how do we know that some nested radicals don’t denest into a different form? And even if the assumption is always valid, he shows only a specific case, and it would be nice to have a formula for getting x and y.

As for the assumption, it does turn out to be correct for every case that can be denested to roots of rational numbers. In this Math Forum article, 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.

Chrystal’s theorem (with some adjustment by me to the notation) 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, unfortunately no longer available and not archived, 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.

Let’s formalize this in Method 1.

Easy Method 1 — √a ± √b when √a² − b Is Rational

Requirements for Using Method 1

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To denest √a ± √b to radicals of rational numbers, all of these must be true:

If those conditions are all met, your radical √a ± √b can be denested to √x ± √y for some x and y that are either both rational or both complex with rational real and imaginary parts.

Solution Details for Method 1

First I’ll give the derivation of Method 1, then I’ll boil it down to just the formulas. Feel free to skip ahead to that section.

  1. Assume, following Chrystal’s theorem, that there exist some x and y (either both rational, or both complex with rational real and imaginary parts) such that

    a + √b = √x + √y

  2. It would be nice to multiply that by

    a − √b = √x − √y

    But how do we know this second equation is true? Answer: “Doctor Peterson” included a link to a page by Dave Rusin that no longer exists where Peterson found it, but has been preserved by 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

    For rational a, b, x, y, with a² > b, a + √b = √x + √y if and only if a − √b = √x − √y.

    — Webster Wells, Advanced Course in Algebra, page 235

    My sources all required x and y to be rationals. But in fact they could be complex, provided they are complex conjugates, numbers in the form p+qi and pqi for rational p and q. Wells’s proof of the theorem above uses x + y and xy, and when you add or multiply complex conjugates the result is real. He also separates rational and irrational parts of expressions on two sides of an equation, which works if the real and imaginary parts of x and y are rational.

    Acknowledgment: Not only was the first version of this article based on that 1999 article by Dave L. Renfro, but in personal correspondence (December 2016) he generously supplied additional links and information.

  3. Now multiply the equation in step 1 by the equation in step 2:

    a + √ba − √b = (√x + √y) (√x − √y)

    Remembering that (A+B)(A−B) = (A²−B²), that becomes

    a² − b = x − y

    That got rid of three of the four radical signs!

  4. Square the equation from step 1:

    a + √b = x + 2 √xy + y

  5. From the equation in step 4, set the rational parts equal:

    a = x + y, from which a − x = y

    (The irrational parts, √b and 2√xy, also must equal each other, but we don’t need to use that.)

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

    a² − b = x − y

    a² − b = x − (a − x)

    a² − b = 2x − a

    2x = a + √a² − b

    x = ½ (a + √a² − b)

    and then

    y = a − x

    y = a − [½ (a + √a² − b)]

    y = ½ (a − √a² − b)

If the original nested radical has a minus sign, √a − √b, the equations in steps 1 and 2 are just interchanged, so x and y in step 6 are the same, but your answer is √x − √y.

Quick Reference for Method 1

First, check that the requirements for using Method 1 on √a±√b are all met. (If not, see if you can use Method 2, below.) If the requirements for Method 1 are met, then you can denest.

Next, substitute and simplify if necessary:

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

Finally, use your x and y to finish with the appropriate one of these:

a + √b = √x + √y    or    √a − √b = √x − √y

Notice that the sign between √x and √y is the same as the sign beween the two terms of the original radical.

Examples

Example 1:2 + √3.

Solution: Always check the requirements first. Here, a = 2 and b = 3 are positive and rational, √b = √3 is irrational, and a² − b = 4 − 3 = 1 is a perfect square. Therefore, Method 1 will denest the radicals.

x = ½ (a + √a² − b)

    = ½ (2 + √4 − 3 = ½ (2 + 1) = 3/2

and

x = ½ (a − √a² − b)

    = ½ (2 − √4 − 3 = ½ (2 − 1) = 1/2

Therefore,

2 + √3 = 3/2 + √1/2

With radicals, it’s often a matter of taste which of two forms is called “simpler”. Generally, we like to see fewer operators and fewer irrational numbers, but there are doubtful cases. For example, would you prefer 1/√2 or the equivalent √2/2? Many algebra teachers say you must never have a radical in the denominator; others prefer the first alternative, or have no preference.

If you (or your teacher) prefer, you can multiply each radical on the right-hand side by √2/2 — a carefully chosen form of 1 — to get

(√3/2 + √1/2)√2/2 = (√6 + √2)/2

Example 2:2 − √3.

Solution: This is almost exactly the same deal as Example 1, so you can shortcut it. Remember that √a + √b = √x + √y if and only if √a − √b = √x − √y. You already know from Example 1 that

2 + √3 = (√6 + √2)/2

Therefore

2 − √3 = (√6 − √2)/2

Done!

Example 3:2 + √2.

Solution: Check the requirements as usual. a = 2 and b = 2 are positive and rational, √b = √2 is irrational, but a² − b = 4 − 2 = 2 is not a perfect square. Therefore, Method 1 won’t work to denest this radical. Your next step is to see whether Method 2 will conquer it. (Spoiler: it will.)

Example 4:3 + √34.

Solution: a = 3 and b = 34 are both positive and rational. √a²−b = √9−34 = 5i, a rational number times i. We’re good to go.

x = (3+5i)/2    and    y = (3−5i)/2

3+√34 = (3+5i)/2 + √(3−5i)/2

If you wish, you can rationalize the denominators, multiplying through by √2/√2:

3+√34 = √(3+5i)/2 + √(3−5i)/2 =

     √(6+10i)/4 + √(6−10i)/4 =

     (√6+10i + √6−10i) / 2

Examples — √a ± ub

Example 5:7 + 2√6.

Solution: When there’s no common factor between a (7 here) and u (2 here), all you can do is move u under the inner radical. If the result meets the requirements for Method 1, great; if not, see if you can use Method 2.

To keep the value of the overall expression unchanged, u becomes u² when it moves inside the radical:

7 + 2√6 = √7 + √(2² × 6) = √7 + √24

Can you use Method 1? Check the Requirements: a = 7 and b = 24 are positive and rational, √b = √24 is irrational, and a² − b = 49 − 24 = 25 is a perfect square. Find v and y in the usual way, and simplify:

x = ½(7 + 5) = 6    and    y = ½(7 − 5) = 1

The answer is √x + √y:

7 + 2√6 = √7 + √24 = √6 + √1 = 6 + 1

Example 6:12 − 3√7.

Solution: When you notice a common factor in both terms under the radical, there are two ways to go:

Personally I like working with smaller numbers, so

x = ½(4 + 3) = 7/2    and    y = ½(4 − 3) = 1/2

gives the answer:

12 − 3√7 = √3 √4 − √7 = 3 (√7/2 − √1/2)

It’s probably worth pulling out that common factor of √1/2:

3 (√7/2 − √1/2) = √3/2 (√7 − √1) = √3/2 (√7 − 1)

then multiplying by √2/√2 to rationalize the denominator:

3/2 (√7 − 1) = (√6/2) (√7 − 1)

Example — √p ± √q

In Advanced Course in Algebra (1904), starting on page 235, Webster Wells shows a method similar to Method 1. He also points out, on the next page, that some expressions of the form √p ± √q can be denested in the same way, if pulling out a common factor reduces them to the √a ± √b form. (If not, you may still be able to use Method 2.)

Example 7 (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.[footnote] It was inside two levels of radical, and when you pull it out it’s only inside one level, so it becomes √2, which is 21/4:

√392 + √360 = √ √2 (√196 + √180)

196 is 14².

√392 + √360 = √ √2  √14 + √180

√392 + √360 = 21/4 √14 + √180

Now you have a² − b = 14² − 180 = 196 − 180 = 16, which is a perfect square (4²), so Method 1 will work. Find x and y:

x = ½(14 + 4) = 9    and     y = ½(14 − 4) = 5

Then the final answer:

14 + √180 = √9 + √5 = 3 + √5

But wait! The problem wasn’t √14 + √180, but 21/4 times that. So really the final answer 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. [Return]

Easy Method 2 — √a + ub when √a² − u²b Isn’t Rational

Remember √2 + √2? Though that radical looks simple, Method 1 won’t work because a = 2 and b = 2, and √a² − b = √2² − 2 = √2 is irrational. But that doesn’t mean it can’t be simplified, just that it can’t be simplified to roots of rational numbers.

In fact, if you plug that nested radical into Wolfram Alpha, it gives the “alternate form” as (√1 + i + √1 − i)/42. How could you find that result on your own?

In a 2016 paper called “On the denesting of nested square roots”, Eleftherios Gkioulekas of the University of Texas gives a technique that he calls indirect denesting. It’s his theorem 2.7, on page 947 (page 6 of the PDF).

With Method 1, since materials were scattered in multiple Web pages, some no longer accessible, I thought it would be good to go through the proof in detail. But for Method 2, it’s all in one paper, so I won’t duplicate Gkioulekas’s proof and will give just the requirements and the Quick Reference. (I’ve changed his choice of variables a little bit, to show the contrasts with Method 1.)

Requirements for Using Method 2

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When conditions are not met for Method 1 above, you can denest √a + ub to radicals of rational numbers, or radicals of complex numbers with real and imaginary parts both rational, provided that:

Quick Reference for Method 2

Use d = √b(u²b − a²), which you computed in checking requirements.

If a is positive, then

a ± ub = [√(ub + d)/2 + √(ub − d)/2] / 4b

If a is negative, then

a ± ub = [√(ub + d)/2 − √(ub − d)/2] / 4b

Examples

Example 8: Denest √2 + √2. (We’re revisiting Example 3, which could not be denested by Method 1.)

Solution: a = 2, u = 1, and b = 2. All three are rational and positive. √b = √2 is not rational. a + ub = 2+1√2 is positive.

d= √b(u²ba²) = √2(1²×2 − 2²) = √2(−2) = √−4 = 2i. The conditions are met for Gkioulekas’s theorem. Since a is positive,

2 + √2 = [√(1×2 + 2i)/2 + √(1×2 − 2i)/2] / 42 = (√1+i + √1−i) / 42

Strange as it looks, that sum is a real number, though the individual radicals are not. Does that seem impossible? Yes, 1+i and 1−i are complex numbers, not real, and so are their square roots. (To find roots of a complex number, see Powers and Roots of a Complex Number in my free online textbook, Trig without Tears.) But when you square them, the imaginaries drop out:

[(√1+i + √1−i]/42)² = (1+i + 2√(1+i)(1−i) + 1−i)/√2 =

    (2 + 2√(1−i²))/√2 = (2 + 2√2))/√2 = √2 + 2

which is just what you get when you square √2 + √2. Since (√1+i + √1−i)/42 is the square root of a positive number, it must be a real number.

Example 9: Denest √−8 + 3√10.

Solution: a = −8, u = 3, and b = 10. a and u are rational; b is rational and positive. √b = √10 is not rational. a + ub = −8+3√10 is positive. (You can be sure of that, even without a calculator: −8+3√10 is greater than −8+3√9, which is +1, so −8+3√10 must be positive.)

d = √b(u²ba²) = √10(3²×10 + 8²) = √1540. Unfortunately, 1540 is not a perfect square, and therefore d is not rational. The conditions are not met for Gkioulekas’s theorem, so indirect denesting is not possible for √−8 + 3√10.

Just to show you why this last requirement is a deal-breaker, why d must be rational (or a rational multiple of i, as in Example 8), look at what happens if you just blindly use the formula in the Quick Reference. Since a is negative,

−8 + 3√10 = [√(3×10 + 2√385)/2 − √(3×10 − 2√385)/2] / 42

−8 + 3√10 = [√15 + √385 − √15 − √385] / 42

The second inner radical is negative, because 15 = √225, and 385 > 225. Factor out −1, which becomes √−1 = i when you take it out of the radical.

−8 + 3√10 = [√15 + √385 − i√(√385) − 15] / 42

The imaginary isn’t a problem: Example 8 ended up with a complex number. The problem is that, with d not rational (d² not a perfect square), instead of simplifying the original expression, I made it more complicated by applying the formula where it shouldn’t be applied.

Example 10: Denest √5 + 3√5.

Solution: a = 5, u = 3, b = 5, d² = b(u²b − a²) = 5(3²×5 − 5²) = 100, and d = 10. a, u, and b are all positive rational, d is rational, and a + ub = 5 + 3√5 is positive. All requirements are met for Method 2, indirect nesting.

5 + 3√5 = [√(3×5 + 10)/2] + √(3×5 − 10)/2] / 45

5 + 3√5 = [√25/2 + √5/2] / 45

There are several paths to simplifying this expression. Here is one:

5 + 3√5 = √5/2[√5 + 1] / 45 =

     √5×2/4[√5 + 1] / 45 = √5×2[√5 + 1] / 245 =

     [√10 + √2] √5/245 = [√10 + √2] 45 / 2

Note: √5 = 51/2 and 45 = 51/4, so √5/45 = 51/2 − 1/4 = 51/4 = 45.

Example 11: Denest √−5 + 3√5.

Solution: a = −5, u = 3, b = 5, d² = b(u²b − a²) = 5(3²×5 − (−5)²) = 5(45 − 25) = 100, and d = 10. a is negative but rational; u and b are positive rational, d is rational, and a + ub = −5 + 3√5 is positive. (How do you know? −5 + 3√3 would be smaller, and it’s positive.) All requirements are met for Method 2.

Since a is negative,

a ± ub = [√(ub + d)/2 − √(ub − d)/2] / 4b

−5 ± 3√5 = [√(3 × 5 + 10)/2 − √(3 × 5 − 10)/2] / 45

−5 ± 3√5 = (√25/2 − √5/2) / 45

Factor out √5/√2.

−5 ± 3√5 = (√5 − 1) √5 / (452)

According to It’s the Law — the Laws of Exponents, √5 is 25 = 51/2, and 45 = 51/4. Therefore √5/45 = 51/2/51/4 = 51/2 − 1/4 = 51/4 or 45.

−5 ± 3√5 = (√5 − 1) 45/√2

Multiply by √2/√2, using the form 44 / √2:

−5 ± 3√5 = (√5 − 1) 420 / 2

Now It’s Your Turn

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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.

Caution! Don’t just start in substituting in a formula. Check the preconditions for Method 1 and for Method 2 or you’ll get nonsense 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. 2 − √2
  7. 20 − √144
  8. 10 − 4√6 (Hint: 4√6 is the square root of what number?)
  9. 10 + 2√6
  10. From Wells, page 239: √√1058 − √896

What’s New?

Solutions to Practice Problems

  1. 3 + √5

    This meets the requirements for Method 1, and evaluates to 5/2 + √1/2. If you don’t like fractions under radicals (or if your teacher doesn’t), multiply by √2/2 (which is 1), factor out √1/4 as ½, and give your answer as (√10 + √2) / 2

  2. Use your answer to the previous exercise to simplify √3 − √5.

    Remember that changing √a + √b to √a − √b changes the denested form √x + √y to √x − √y. Your answer is 5/2 − √1/2, or (√10 − √2)

  3. 2 + √5

    The radicand is a positive number, so the radical is a real number, but still 2² − 5 = −1 is the negative of a perfect square, so you can use Method 1.

    x = ½(2 + √−1) = (2+i)/2    and    y = ½(2 − √−1) = (2−i)/2.

    That gives

    2 + √5 = √(2+i)/2 + √(2−i)/2

        = (√4+2i + √4−2i)/2

    You can verify that this answer is correct. Squaring the right-hand side, you get

    [(√4+2i + √4−2i)/2]² = (4+2i + 2√16−4i² + 4−2i)/4

        = (8 + 2√20)/4 = (8 + 4√5)/4 = 2 + √5

    which is the square of the original nested radical.

  4. 2 − √5

    The radicand is negative, so there’s no way to get a purely real solution to this one. However, you could factor −1 out of the radical:

    2 − √5 = √(−1) (−2 + √5) = i √−2 + √5

    and then proceed by Method 1, to get (i/2)(√−4+2i + √−4−2i). (That’s i/2, not 1/2.) Square the answer to verify that you get 2 − √5.

  5. 5 − √21

    Using Method 1, you get √7/2 − √3/2, or (√14 − √6) / 2

  6. 2 − √2

    Method 2 works for this one.

    d = √2(1² × 2 − (−2)²) = √2(2 − 4) = √−4 = 2i

    Then

    2 − √2 = [√(−2+2i)/2 + √(−2−2i)/2] / 42 =

         (√−1+i + √−1−i) / 42

  7. 20 − √144

    144 is rational, so you don’t need a denesting method.

    20 − √ 144 = √20 − 12 = √8 = 2√2

  8. 10 − 4√6 (Hint: 4√6 is the square root of what number?)

    4√6 = √16 × 6 = √96. Therefore √10 − 4√6 = √10 − √96, and by Method 1 that equals 6 − 2

  9. 10 + 2√6

    a² − b = 100 − 24 = 76 is not a perfect square, so this one can’t be denested by Method 1. And in Method 2, d = √−456 is not a rational multiple of i, so you’re blocked there too.

    About all you can do is factor out √2 to get √2 √5 + √6, and that hardly seems worth the effort.

  10. From Wells, page 239: √√1058 − √896

    This one 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 = 21/4 √23 − √448

    Now Method 1 can be used. x = 16 and y = 7, so

    √1058 − √896 = 21/4 (4 − √7)

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