How to simplify nested square roots, with three worked examples

What is a nested radical?

A nested radical contains one root inside another, such as $\sqrt{10-2\sqrt{21}}$. Sometimes it can be written as a sum or difference of simpler square roots. The key is to recognize a squared binomial rather than apply a special rule blindly.

Find two numbers from their sum and product

For real numbers $0\le a\le b$,

$$(\sqrt b+\sqrt a)^2=a+b+2\sqrt{ab},$$

and

$$(\sqrt b-\sqrt a)^2=a+b-2\sqrt{ab}.$$

Therefore

$$\boxed{\sqrt{a+b+2\sqrt{ab}}=\sqrt b+\sqrt a,}$$ $$\boxed{\sqrt{a+b-2\sqrt{ab}}=\sqrt b-\sqrt a.}$$

To simplify $\sqrt{10-2\sqrt{21}}$, look for two nonnegative numbers with sum $10$ and product $21$. They are $3$ and $7$, so

$$\sqrt{10-2\sqrt{21}}=\sqrt7-\sqrt3.$$

Why the order matters for subtraction

The principal square root is nonnegative. In general $\sqrt{u^2}=|u|$, so the minus case initially gives $|\sqrt b-\sqrt a|$. The assumption $b\ge a$ lets us remove the absolute value.

Writing $\sqrt3-\sqrt7$ would give a negative result and therefore cannot equal the original square root. For addition the order is immaterial, but putting the larger radicand first is a useful consistent habit.

A systematic way to find the two radicands

Suppose the expression is $\sqrt{A\pm2\sqrt B}$ and we seek nonnegative rational radicands $a,b$ with $a+b=A$ and $ab=B$. Then $a,b$ must be the roots of

$$t^2-At+B=0,$$

so

$$a,b=\frac{A\mp\sqrt{A^2-4B}}2.$$

This explains both the sum-and-product search and its limits. For example, $\sqrt{6-2\sqrt3}$ would require $a+b=6$, $ab=3$. The resulting values $3\pm\sqrt6$ are not rational. The usual technique cannot remove the nesting into square roots of rational numbers in this case; introducing those irrational radicands would merely move the nesting elsewhere.

This does not mean the expression has no other exact representation. It means the particular rational-radicand simplification we are looking for is unavailable. A perfectly valid exact answer may still contain a nested root. For more elaborate expressions, such as

$$\frac{1+\sqrt{1+8\sqrt[3]4}}{4\sqrt[3]2},$$

do not assume that the elementary sum-and-product technique will provide a simpler form.

Example 1: a direct sum-and-product match

Problem. Simplify $\sqrt{18-2\sqrt{77}}$.

We need $a+b=18$, $ab=77$. Since $7+11=18$ and $7\cdot11=77$,

$$\boxed{\sqrt{18-2\sqrt{77}}=\sqrt{11}-\sqrt7.}$$

The result is positive. Squaring it gives $18-2\sqrt{77}$, which checks the answer.

Example 2: first make the coefficient two

Problem. Simplify $\sqrt{29-3\sqrt{80}}$.

The coefficient of the inner radical is not two, so rewrite it:

$$3\sqrt{80}=6\sqrt{20}=2\sqrt{180}=2\sqrt{20\cdot9}.$$

The radicands $20$ and $9$ have sum $29$. Thus

$$\sqrt{29-3\sqrt{80}}=\sqrt{20}-\sqrt9=\boxed{2\sqrt5-3.}$$

Reversing the subtraction would make the answer negative. Always check the sign as well as the squared value.

Example 3: introduce a denominator

Problem. Simplify $\sqrt{5-\sqrt{21}}$.

Multiply the expression inside the outer root by $2/2$:

$$\sqrt{5-\sqrt{21}}=\sqrt{\frac{10-2\sqrt{21}}2}=\frac{\sqrt{10-2\sqrt{21}}}{\sqrt2}.$$

Using the earlier example,

$$\frac{\sqrt{10-2\sqrt{21}}}{\sqrt2}=\frac{\sqrt7-\sqrt3}{\sqrt2}=\boxed{\frac{\sqrt{14}-\sqrt6}{2}.}$$

The unreationalized form $(\sqrt7-\sqrt3)/\sqrt2$ is equivalent. Notice that the original expression is $5-\sqrt{21}$, not $5-3\sqrt{21}$: keeping that coefficient intact is essential.

A short checking routine

Find the sum and product, put the larger square root first in a difference, and square the proposed answer. Then check that it is nonnegative. If no nonnegative rational radicands satisfy the sum-and-product conditions, keeping the original nested radical may be the clearest exact answer.

Related: equating rational and irrational parts.