# Denesting Radicals (or Unnesting Radicals)

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

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.

**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?

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.

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

*x*² = 2 − √3

−*x*² + 2 = √3

*x*⁴ − 4*x*² + 4 = 3

*x*⁴ − 4*x*² + 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:

(*x*⁴ − 2*x*² + 1) − 2*x*² = 0

(*x*² − 1)² − (√2*x*)² = 0

(*x*² − 1 + √2*x*)
(*x*² − 1 − √2*x*) = 0

(*x*² + √2*x* − 1)
(*x*² − √2*x* − 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.

“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

Leta,bbe rational numbers. Then√a± √bcan be expressed as √x± √yfor some rational numbersx,yif and only ifa> 0 anda² −bis 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.

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

*a*is rational, and*b*is positive and rational.- √
*b*is irrational. - √
*a*² −*b*is rational or is a rational multiple of the imaginary unit i. For example, √4 and √−4/9 qualify, but not √5.As Chrystal stated it,

*a*² −*b*must be positive, with a square root in the reals. However, if √*a*² −*b*equals a rational number times the imaginary unit i, the method works correctly, though the denested radicals contain complex numbers. My guess is that because Chrystal had not yet introduced complex numbers at that point in his textbook, he phrased his requirement to limit solutions to the reals.

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.

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.

- 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* - 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*a*² >*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*+*q*i and*p*−*q*i for rational*p*and*q*. Wells’s proof of the theorem above uses*x*+*y*and*x**y*, 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. - Now multiply the equation in step 1 by the equation in step 2:
√

*a*+ √*b*√*a*− √*b*= (√*x*+ √*y*) (√*x*− √*y*)Remembering that (A+B)(A−B) = (A²−B²), that becomes

√

*a*² −*b*=*x*−*y*That got rid of three of the four radical signs!

- Square the equation from step 1:
*a*+ √*b*=*x*+ 2 √*x**y*+*y* - 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√

*x**y*, also must equal each other, but we don’t need to use that.) - Solve the equations from steps 3 and 5 for
*x*and*y*:√

*a*² −*b*=*x*−*y*√

*a*² −*b*=*x*− (*a*−*x*)√

*a*² −*b*= 2*x*−*a*2

*x*=*a*+ √*a*² −*b**x*= ½ (*a*+ √*a*² −*b*)and then

*y*=*a*−*x**y*=*a*− [½ (*a*+ √*a*² −*b*)]*y*= ½ (*a*− √*a*² −*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*.

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* + √*a*² − *b*)
and
*y* = ½ (*a* − √*a*² − *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.

**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
*a*² − *b* = 4 − 3 = 1 is a
perfect square. Therefore, Method 1 will denest the
radicals.

*x* = ½ (*a* + √*a*² − *b*)

= ½ (2 + √4 − 3 = ½ (2 + 1) = 3/2

and

*x* = ½ (*a* − √*a*² − *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
*a*² − *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.
√*a*²−*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

**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 *u*² 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
*a*² − *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:

- You could tuck that 3 into the inner radical, where it becomes 3².
And 3² × 7 = 63, so the original radical becomes
√12 − √63.
*a*= 12 and*b*= 63 are positive and rational, √*b*= √63 is irrational, and*a*² −*b*= 144 − 63 = 81 is a perfect square, so Method 1 will denest this radical. - Or you could pull out the common factor of 3, changing the radical
to √3 √4 − √7.
Now

*a*= 4 and*b*= 7 are both positive and rational, √*b*= √7 is irrational, and*a*² −*b*= 16 − 7 = 9 is a perfect square, so Method 1 will denest this radical. (It’s comforting to know that both choices lead to the same result.)

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)

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 2^{1/4}:

√√392 + √360 = √ √2 (√196 + √180)

196 is 14².

√√392 + √360 = √ √2 √14 + √180

√√392 + √360 =
2^{1/4} √14 + √180

Now you have *a*² − *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 2^{1/4} times that.
So really the *final* answer 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. [Return]

Remember √2 + √2? Though that radical
looks simple, Method 1 won’t work because
*a* = 2 and *b* = 2, and
√*a*² − *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)/^{4}√2.
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.)

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When conditions are not met for Method
1 above, you can
**denest √ a + u√b to**

*a*and*u*are rational, not necessarily positive.(Two reminders:

*u*is negative if there’s a minus sign between the two terms inside the radical. And if the second term doesn’t have an explicit coefficient, then*u*= 1 or −1.)*b*is rational and positive.- √
*b*is not rational. *a*+*u*√*b*> 0.*d*= √*b*(*u*²*b*−*a*²) is rational, or is a rational multiple of the imaginary unit i. (If this condition is not met, the math is still valid but you end up with two nested radicals replacing the original one.)

Use *d* =
√*b*(*u*²*b* − *a*²), which you
computed in checking requirements.

If *a* is positive, then

√*a* ± *u*√*b* =
[√(*u**b* + *d*)/2 + √(*u**b* − *d*)/2] / ^{4}√*b*

If *a* is negative, then

√*a* ± *u*√*b* =
[√(*u**b* + *d*)/2 − √(*u**b* − *d*)/2] / ^{4}√*b*

**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* + *u*√*b* =
2+1√2 is positive.

*d*= √*b*(*u*²*b*−*a*²) =
√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] / ^{4}√2 =
(√1+i + √1−i) / ^{4}√2

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]/^{4}√2)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)/^{4}√2
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* + *u*√*b* =
−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*(*u*²*b*−*a*²) =
√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] / ^{4}√2

√−8 + 3√10 =
[√15 + √385 − √15 − √385] / ^{4}√2

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] / ^{4}√2

The imaginary isn’t a problem: Example 8
ended up with a complex number. The problem is that, with *d* not
rational (*d*² 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,
*d*² = *b*(*u*²*b* − *a*²) =
5(3²×5 − 5²) = 100, and *d* = 10. *a*,
*u*, and *b* are all positive rational, *d* is rational, and
*a* + *u*√*b* =
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] / ^{4}√5

√5 + 3√5 =
[√25/2 + √5/2] / ^{4}√5

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

√5 + 3√5 =
√5/2[√5 + 1] / ^{4}√5 =

√5×2/4[√5 + 1] / ^{4}√5 =
√5×2[√5 + 1] / 2^{4}√5 =

[√10 + √2] √5/2^{4}√5 =
[√10 + √2] ^{4}√5 / 2

*Note*: √5 = 5^{1/2} and
^{4}√5 = 5^{1/4}, so
√5/^{4}√5 =
5^{1/2 − 1/4} = 5^{1/4} =
^{4}√5.

**Example 11:** Denest
√−5 + 3√5.

**Solution:**
*a* = −5, *u* = 3, *b* = 5,
*d*² = *b*(*u*²*b* − *a*²) =
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* + *u*√*b* =
−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* ± *u*√*b* =
[√(*u**b* + *d*)/2 − √(*u**b* − *d*)/2] / ^{4}√*b*

√−5 ± 3√5 =
[√(3 × 5 + 10)/2 − √(3 × 5 − 10)/2] / ^{4}√5

√−5 ± 3√5 =
(√25/2 − √5/2) / ^{4}√5

Factor out √5/√2.

√−5 ± 3√5 =
(√5 − 1) √5 / (^{4}√5√2)

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

√−5 ± 3√5 =
(√5 − 1) ^{4}√5/√2

Multiply by √2/√2, using the form
^{4}√4 / √2:

√−5 ± 3√5 =
(√5 − 1) ^{4}√20 / 2

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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.

- √3 + √5
- Use your answer to the previous exercise to simplify √3 − √5.
- √2 + √5
- √2 − √5
- √5 − √21
- √2 − √2
- √20 − √144
- √10 − 4√6 (Hint: 4√6 is the square root of what number?)
- √10 + 2√6
- From Wells, page 239: √√1058 − √896

**18 Oct 2023**: A major rewrite and much new content, including:- A second method of denesting, which handles some cases that the original method can’t.
- Discovery and verification that the methods can work if the radicands are negative, producing expressions like √11+3i + √11−3i (which are actually real numbers, though they don’t look it).
- Rewrite and restructure of most of the earlier content to make the derivations clearer.
- A Quick Reference so you can avoid going through the derivations on every problem.
- More examples, more practice problems, and better explanations of the solutions.

**14 Nov 2021**: Earlier this year, the National Council of Teachers of Mathematics took over “Ask Dr. Math” a/k/a mathforum.org, and made its massive archive unavailable. (How taking useful content off the public Web enhances the teaching and learning of mathematics utterly escapes me.) I found archived pages for*some*of the articles I had cited, and replaced links to mathforum with links to the Internet Wayback Machine.- (intervening changes suppressed)
**18 Dec 2016**: New article.

**√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

**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)**√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.

**√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 − √21**Using Method 1, you get √7/2 − √3/2, or (√14 − √6) / 2

**√2 − √2**Method 2 works for this one.

*d*= √2(1² × 2 − (−2)²) = √2(2 − 4) = √−4 = 2iThen

√2 − √2 = [√(−2+2i)/2 + √(−2−2i)/2] /

^{4}√2 =(√−1+i + √−1−i) /

^{4}√2**√20 − √144**√144 is rational, so you don’t need a denesting method.

√20 − √ 144 = √20 − 12 = √8 = 2√2

**√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

**√10 + 2√6***a*² −*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.

**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 = 2

^{1/4}√23 − √448Now Method 1 can be used.

*x*= 16 and*y*= 7, so√√1058 − √896 = 2

^{1/4}(4 − √7)

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