How do I know when a radical is fully simplified?

Run three checks. Does the radicand still have a perfect-square factor besides 1? Is there a fraction under the radical sign? Is there a radical in a denominator? If you answer no to all three, and any resulting fraction is reduced, you are done. The check people skip most often is the first one after a partial simplification, which is how answers like 2sqrt(18) get turned in.

Why is it not allowed to leave a radical in the denominator?

It is not mathematically wrong — 1sqrt(2) and sqrt(2)2 are the same number. The convention comes from the days before calculators, when dividing by a whole number was far easier than dividing by a decimal like 1.41421. It also gives everyone a single standard form, so two correct answers look identical and can be compared at a glance.

Does sqrt(x^2) always equal x?

Only when x ≥ 0. If x = -5, then sqrt(x^2) = sqrt(25) = 5, which is -x, not x. In general sqrt(x^2) = |x|. In Algebra 1 problems you are usually told to assume all variables represent nonnegative numbers, which is why writing sqrt(x^2)=x is acceptable there — but the absolute value is the honest full answer.

What is the fastest way to find the biggest perfect-square factor?

Test the perfect squares from largest to smallest against your radicand, starting with the largest square no bigger than the radicand — for 288 that means 256, then 225, 196, 169, then 144, which divides it and gives 12sqrt(2). If nothing jumps out, build a prime factorization instead. Every matched pair of primes contributes one factor outside the radical, and whatever is left unpaired stays inside. For sqrt(180): 180 = 2^2 · 3^2 · 5, so the 2s and the 3s come out as 2· 3 = 6, giving 6sqrt(5).

ALG1-9.1

Simplifying Radical Expressions

Learn to simplify square roots: pull out perfect-square factors, use the product and quotient rules, add like radicals, and rationalize denominators — with worked steps.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Simplifying Radical Expressions, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Simplifying Radical Expressions free

What this lesson covers

A radical expression like

\sqrt{72}

is a perfectly good number, but it is not in its cleanest form. Simplifying radicals is the algebra version of reducing a fraction: the value never changes, only the way it is written. Once you can rewrite

\sqrt{72}

6\sqrt{2}

, you can compare it to other radicals, add it to them, and recognize when two messy-looking expressions are actually the same number.

This lesson builds four skills that work together: extracting perfect-square factors from a radicand, applying the product and quotient rules for radicals, combining like radicals the way you combine like terms, and clearing a radical out of a denominator. You will use every one of these when you solve radical equations and apply the Pythagorean Theorem later in this unit, so it is worth getting the mechanics automatic now.

What Counts as "Simplified"

A square root expression is in simplest radical form when three things are true. First, the radicand (the number or expression under the radical sign) contains no perfect-square factor other than

1

. Second, there is no fraction under the radical. Third, there is no radical in a denominator.

The engine behind all of this is the fact that

\sqrt{a^2} = a

for

a \ge 0

. To simplify

\sqrt{72}

, hunt for the largest perfect square that divides

72

. Since

72 = 36 \cdot 2

\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36}\cdot\sqrt{2} = 6\sqrt{2}.

If you do not spot the largest perfect square, you are not stuck — you just work longer. Using

72 = 4 \cdot 18

gives

2\sqrt{18}

, and since

18 = 9 \cdot 2

still contains a perfect square, you factor again to reach

2 \cdot 3\sqrt{2} = 6\sqrt{2}

. Same answer, one extra step.

Knowing the perfect squares

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

on sight makes this fast. A backup method is a prime factor tree:

72 = 2^3 \cdot 3^2

, and every pair of identical primes escapes the radical as a single factor. The pair of

3

s gives a

3

outside, the pair of

2

s gives a

2

outside, and the leftover single

2

stays inside:

6\sqrt{2}

.

The most common error here is stopping too early — writing

2\sqrt{18}

and calling it finished. Always check the leftover radicand for another perfect-square factor.

The Product and Quotient Rules

Two rules let you split radicals apart or push them together. For

a \ge 0

and

b \ge 0

\sqrt{a}\cdot\sqrt{b} = \sqrt{ab} \qquad \text{and} \qquad \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, \; b \neq 0

Read them in both directions. Left to right, the product rule multiplies radicals:

\sqrt{6}\cdot\sqrt{10} = \sqrt{60} = 2\sqrt{15}

. Right to left, it is exactly the extraction move from the last section.

The quotient rule often saves work if you combine first. Instead of simplifying

\sqrt{50}

and

\sqrt{2}

separately, write

\frac{\sqrt{50}}{\sqrt{2}} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5.

Situation	Move	Result
$\sqrt{3}\cdot\sqrt{12}$	Combine, then extract	$\sqrt{36}=6$
$\sqrt{200}$	Split off $100$	$10\sqrt{2}$
$\dfrac{\sqrt{75}}{\sqrt{3}}$	Divide inside	$\sqrt{25}=5$
$\sqrt{\dfrac{9}{49}}$	Split numerator and denominator	$\dfrac{3}{7}$

Here is the misconception that causes the most damage: there is no such rule for addition.

\sqrt{a+b}

does not equal

\sqrt{a}+\sqrt{b}

. Test it with numbers:

\sqrt{9+16} = \sqrt{25} = 5

, but

\sqrt{9}+\sqrt{16} = 3+4 = 7

. Radicals distribute over multiplication and division only.

With variables, assume nonnegative values unless told otherwise, and pair the exponents:

\sqrt{x^{6}} = x^{3}

, and

\sqrt{20x^{5}} = \sqrt{4x^{4}}\cdot\sqrt{5x} = 2x^{2}\sqrt{5x}

Adding and Subtracting Like Radicals

Radicals add the same way variable terms do. Just as

3x + 5x = 8x

but

3x + 5y

cannot be combined,

3\sqrt{7} + 5\sqrt{7} = 8\sqrt{7}

while

3\sqrt{7} + 5\sqrt{2}

stays as it is. Like radicals have the identical radicand; the coefficients in front are what you add or subtract.

The catch is that two radicals can look unlike and still be like radicals in disguise. Consider

\sqrt{27} + \sqrt{12}

. Neither radicand matches, so it seems finished — but simplify first:

\sqrt{27} + \sqrt{12} = 3\sqrt{3} + 2\sqrt{3} = 5\sqrt{3}.

The rule of thumb is: simplify every radical completely, then look for matches. Students who skip the simplification step routinely report that an expression cannot be combined when it can.

The reverse error is combining radicands: writing

\sqrt{3}+\sqrt{5} = \sqrt{8}

. Check it approximately —

\sqrt{3} \approx 1.73

and

\sqrt{5} \approx 2.24

sum to about

3.97

, while

\sqrt{8} \approx 2.83

. Not equal. When you add like radicals, the radicand never changes:

2\sqrt{5}+3\sqrt{5} = 5\sqrt{5}

, not

5\sqrt{10}

.

Multiplication is different from addition here.

\sqrt{3}\cdot\sqrt{5} = \sqrt{15}

is legitimate, and

\sqrt{5}\cdot\sqrt{5} = 5

. When you multiply binomials containing radicals, distribute as usual:

(2+\sqrt{3})(1+\sqrt{3}) = 2 + 2\sqrt{3} + \sqrt{3} + 3 = 5 + 3\sqrt{3}

. Notice the

\sqrt{3}\cdot\sqrt{3}

collapsed to

3

, and the two like radical terms combined at the end.

Rationalizing the Denominator

Simplest radical form does not allow a radical in a denominator. To clear one, multiply the fraction by a clever form of

1

.

For a single square root in the denominator, multiply top and bottom by that same radical:

\frac{5}{\sqrt{3}} = \frac{5}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}.

The denominator becomes

\sqrt{3}\cdot\sqrt{3} = 3

, a whole number, and the value of the fraction is unchanged because you multiplied by

1

. Simplify the radicand before rationalizing when you can —

\frac{6}{\sqrt{8}}

becomes

\frac{6}{2\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}

, which is much less arithmetic than multiplying by

\sqrt{8}

first.

Always reduce the resulting fraction.

\frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}

; leaving

\frac{4\sqrt{2}}{2}

is incomplete.

When the denominator is a sum or difference such as

3 - \sqrt{5}

, multiplying by

\sqrt{5}

does not help. Use the conjugate,

3 + \sqrt{5}

\frac{2}{3-\sqrt{5}}\cdot\frac{3+\sqrt{5}}{3+\sqrt{5}} = \frac{2(3+\sqrt{5})}{9-5} = \frac{2(3+\sqrt{5})}{4} = \frac{3+\sqrt{5}}{2}.

The difference of squares

(a-b)(a+b) = a^2 - b^2

is what kills the radical. A frequent slip is multiplying only the denominator by the conjugate — that changes the value of the expression. Whatever you multiply the bottom by, the top gets it too.

Key terms

Radicand.: The number or expression written underneath the radical symbol. In $\sqrt{18}$ , the radicand is $18$ .
Principal square root.: The nonnegative square root of a number. $\sqrt{25}$ means $5$ , not $-5$ , even though both square to $25$ .
Perfect square.: A number that is the square of an integer, such as $1, 4, 9, 16, 25, 36$ . Perfect-square factors are what you extract from a radicand.
Simplest radical form.: A form in which the radicand has no perfect-square factor other than $1$ , no fraction sits under a radical, and no radical sits in a denominator.
Product rule for radicals.: For $a, b \ge 0$ , $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ . It works for multiplication only, never for addition.
Quotient rule for radicals.: For $a \ge 0$ and $b > 0$ , $\sqrt{a/b} = \sqrt{a}/\sqrt{b}$ , which lets you divide inside a single radical or split one apart.
Like radicals.: Radical terms with identical radicands, such as $4\sqrt{6}$ and $-\sqrt{6}$ . Only like radicals can be added or subtracted.
Conjugate.: The expression formed by switching the middle sign of a two-term radical expression; the conjugate of $3-\sqrt{5}$ is $3+\sqrt{5}$ . Their product contains no radical.

Worked example

Write

3\sqrt{32} - 2\sqrt{18} + \dfrac{10}{\sqrt{2}}

in simplest radical form.

Handle each term separately, then combine.

First term: find the largest perfect-square factor of

32

. Since

32 = 16\cdot 2

\sqrt{32} = \sqrt{16}\cdot\sqrt{2} = 4\sqrt{2}

. Multiply by the coefficient

3

3\sqrt{32} = 3 \cdot 4\sqrt{2} = 12\sqrt{2}

.

Second term:

18 = 9 \cdot 2

, so

\sqrt{18} = 3\sqrt{2}

and

2\sqrt{18} = 6\sqrt{2}

. Keep the subtraction sign attached to it.

Third term: there is a radical in the denominator, so rationalize. Multiply numerator and denominator by

\sqrt{2}

\frac{10}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}.

Notice the fraction reduced from

\frac{10}{2}

5

— do not leave it unreduced.

Now every term is a multiple of

\sqrt{2}

, so they are like radicals:

12\sqrt{2} - 6\sqrt{2} + 5\sqrt{2} = (12 - 6 + 5)\sqrt{2} = 11\sqrt{2}.

The radicand stays

2

; only the coefficients combine. As a check,

11\sqrt{2} \approx 15.56

, and evaluating the original terms numerically gives

3\sqrt{32} \approx 16.971

2\sqrt{18} \approx 8.485

, and

10/\sqrt{2} \approx 7.071

, so the expression is about

16.971 - 8.485 + 7.071 \approx 15.56

. The answer is

11\sqrt{2}

Practice questions

Which expression is

\sqrt{72}

written in simplest radical form?

$2\sqrt{18}$
$6\sqrt{2}$
$8\sqrt{3}$
$36\sqrt{2}$

Answer: $6\sqrt{2}$

The largest perfect-square factor of

72

36

, so

\sqrt{72} = \sqrt{36}\cdot\sqrt{2} = 6\sqrt{2}

. The choice

2\sqrt{18}

is equal in value but not simplified, because

18

still contains the perfect square

9

. The choice

8\sqrt{3}

has the wrong value entirely (

8\sqrt{3}=\sqrt{192}

), and

36\sqrt{2}

comes from forgetting to take the square root of

36

when moving it outside.

Simplify

\dfrac{\sqrt{27}}{\sqrt{12}}

completely.

Answer: $\dfrac{3}{2}$

The quotient rule lets you divide inside one radical:

\frac{\sqrt{27}}{\sqrt{12}} = \sqrt{\frac{27}{12}} = \sqrt{\frac{9}{4}} = \frac{3}{2}

. You can also simplify each radical first:

\sqrt{27}=3\sqrt{3}

and

\sqrt{12}=2\sqrt{3}

, so the quotient is

\frac{3\sqrt{3}}{2\sqrt{3}}

, and the

\sqrt{3}

factors cancel to leave

\frac{3}{2}

. Either route shows the answer contains no radical at all, which surprises students who assume every radical problem has a radical answer.

A classmate writes

\sqrt{16+9} = \sqrt{16}+\sqrt{9} = 4+3 = 7

. Explain what is wrong, give the correct value, and state the rule the classmate confused it with.

Answer: $\sqrt{16+9} = \sqrt{25} = 5$ , not $7$ . There is no rule that splits a radical over addition; the classmate misapplied the product rule $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ .

Order of operations says to add inside the radical first:

16+9=25

, and

\sqrt{25}=5

. The product and quotient rules apply to multiplication and division only, so

\sqrt{16\cdot 9} = 4\cdot 3 = 12

is valid, but

\sqrt{16+9}

cannot be broken apart. This mistake shows up again with the Pythagorean Theorem, where

c = \sqrt{a^2+b^2}

is not

a+b

— a right triangle with legs

3

and

4

has hypotenuse

5

, not

7

FAQ

How do I know when a radical is fully simplified?: Run three checks. Does the radicand still have a perfect-square factor besides $1$ ? Is there a fraction under the radical sign? Is there a radical in a denominator? If you answer no to all three, and any resulting fraction is reduced, you are done. The check people skip most often is the first one after a partial simplification, which is how answers like $2\sqrt{18}$ get turned in.
Why is it not allowed to leave a radical in the denominator?: It is not mathematically wrong — $\frac{1}{\sqrt{2}}$ and $\frac{\sqrt{2}}{2}$ are the same number. The convention comes from the days before calculators, when dividing by a whole number was far easier than dividing by a decimal like $1.41421$ . It also gives everyone a single standard form, so two correct answers look identical and can be compared at a glance.
Does $\sqrt{x^2}$ always equal $x$ ?: Only when $x \ge 0$ . If $x = -5$ , then $\sqrt{x^2} = \sqrt{25} = 5$ , which is $-x$ , not $x$ . In general $\sqrt{x^2} = |x|$ . In Algebra 1 problems you are usually told to assume all variables represent nonnegative numbers, which is why writing $\sqrt{x^2}=x$ is acceptable there — but the absolute value is the honest full answer.
What is the fastest way to find the biggest perfect-square factor?: Test the perfect squares from largest to smallest against your radicand, starting with the largest square no bigger than the radicand — for $288$ that means $256$ , then $225$ , $196$ , $169$ , then $144$ , which divides it and gives $12\sqrt{2}$ . If nothing jumps out, build a prime factorization instead. Every matched pair of primes contributes one factor outside the radical, and whatever is left unpaired stays inside. For $\sqrt{180}$ : $180 = 2^2 \cdot 3^2 \cdot 5$ , so the $2$ s and the $3$ s come out as $2\cdot 3 = 6$ , giving $6\sqrt{5}$ .

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