### Definitions

For any positive integer n, the nth root of a number x is r such that rn = x.

If n is odd, then the nth root of x is single-valued. E.g. the 3rd root of -8 is -2, because (-2) × (-2) × (-2) = -8.

If n is even, then:

• The nth root of x doesn’t have a value if x is negative, because even powers result in non-negative numbers. E.g. the 2nd root of -4 doesn’t have a value, because there is no number r such that r2 = -4.
• The nth root of x has two values if x is positive, because any even power of the numbers r and –r is the same number.
E.g. the 2nd root of 4 is 2 and -2, because 22 = 4 and (-2)2 = 4.

A root is shown using a radical symbol √.  The number in front of the radical symbol is the degree of the root.  For square roots, the degree 2 is left off.

In order to avoid double-value of even roots, the radical sign with even degree is a non-negative value only.
For example, √4 = 2 (√4 ≠ -2), 4√81 = 3 (4√81 ≠ -3).

Note: On the GRE, the radical symbol will be shown in two formats. The radical expression may not have a bar across the top. Expressions with no bars often use parentheses to group terms. For example:

(√4 + 2) / √(6 – 2) = \dfrac{√4\,+\,2}{√(6\,-\,2)}

### Fractional Exponents

When a base is a non-negative number, fractional exponents are another way to indicate a root.  When the exponent is a fraction, the numerator says what power to raise the base to, and the denominator says what root to take.  You can raise the base to the power or take the root in either order.

Rules for simplifying roots are the same as for exponents – think of roots as exponential expressions with fractions as the exponents.  (Refer to the previous section for the rules of Exponents.)

When the numerator of the fractional exponent is 1, take the nth root of the base.

a1/n = \sqrt[\displaystyle{\textit{n}}]{\textit{a}}

161/4 = \sqrt[\displaystyle{4}]{16} = \sqrt[\displaystyle{4}]{2^{\displaystyle{4}}} = 2

The exponent 1/2 means take the square root of the base.

a1/2 = \sqrt{\textit{a}}

491/2 = \sqrt{49} = \sqrt{7^{\displaystyle{2}}} = 7

The exponent 1/3 means take the cube root of the base.

a1/3 = \sqrt[\displaystyle{3}]{\textit{a}}

2161/3 = \sqrt[\displaystyle{3}]{216} = \sqrt[\displaystyle{3}]{(6 × 6 × 6)} = 6

The exponent m/n means raise the base to the power m and take the nth root.
Note, that m can be any integer, while n is a positive integer only.

am/n = \sqrt[\displaystyle{\textit{n}}]{\textit{a}^{\displaystyle{\textit{m}}}}

25/4 = \sqrt[\displaystyle{4}]{2^{\displaystyle{5}}} = \sqrt[\displaystyle{4}]{2^{\displaystyle{4}} × 2} = 2 \sqrt[\displaystyle{4}]{2}

The exponent 2/3 means square the base and take the cube root.

a2/3 = \sqrt[\displaystyle{3}]{\textit{a}^{\displaystyle{2}}}

812/3 = \sqrt[\displaystyle{3}]{{(3^{\displaystyle{4}})}^{{\displaystyle{2}}}} = \sqrt[\displaystyle{3}]{3^{\displaystyle{8}}} = \sqrt[\displaystyle{3}]{(3^{\displaystyle{6}} \times 3^{\displaystyle{2}})} = 3^{\displaystyle{2}}( \sqrt[\displaystyle{3}]{3^{\displaystyle{2}}} ) = 9 \sqrt[\displaystyle{3}]{9}

Addition and subtraction of roots can lead to common mistakes.  Just like exponential expressions, you can’t just add the bases.

na + nn√(a + b)

√4 + √9 ≠ √13

√20 – √5 ≠ √15

√4 + √9 = 2 + 3 = 5

√20 – √5 = √(4 × 5) – √5 = 2√5 – √5 = √5

### Roots and Negative Numbers

Not all roots are possible.  For example, √-4 is impossible since no number raised to an even power is negative.  But odd roots of negative numbers are possible.

\sqrt[\displaystyle{3}]{-125} = -5 since (-5)3 = -125

\sqrt[\displaystyle{5}]{-32} = -2 since (-2)5 = -32

When taking an odd root, there is only one answer.

When taking an even root, there will be two answers.

For example, the square root of 49 is both 7 and -7. Both 7 and -7 squared are 49, so both are square roots of 49.

But the GRE will only use the positive root for even roots. For √49, the only answer the GRE will use is 7.

Be careful when dealing with equations. Remember that an even root is undefined for negative numbers. A root must be positive, so
√(x2) = |x| (NOT √(x2) = x)

##### Example

If x2 = 4, what is x?

### Solution

Since 22 = 4 and (-2)2 = 4, the answer is x = 2 and x = -2.

##### Alternative solution:

x2  = 4
√(x2) = √4
|x| = 2
x = 2 and x = -2 are the solutions of the equation.

### Estimating Roots

Not all roots yield an integer. For example, √2 and √48 do not have integer values.

Using a calculator, you get √2 ≈ 1.414213562 …  (The symbol ≈ means “approximately equal to”.)

It is usually enough to just simplify expressions by factoring rather than multiplying to get a value.

162 = √(2 × 81) = 9√2

If the answer choices are numbers, you can calculate to get an estimated value. How exact the answer choices are will tell you how accurate an estimate you need to make.

One way to estimate a value for a square root is to find the closest square root.

√48 will be slightly less than 7 since √49 = 7

To get a more exact value, use decimal approximations of common roots.

First, simplify by factoring.  √48 = √(16 × 3) = 4√3

Then for √3, use an approximation.  4√3 ≈ 4(1.7) = 6.8

Knowing the approximate decimal value for these three roots will help when making estimates.

√2 ≈ 1.4

√3 ≈ 1.7

√5 ≈ 2.24

##### Example

Round to the nearest integer:

(a) √3 + √2

(b) √(25 – 10)

### Solution

(a) √3 + √2 ≈ 1.7 + 1.4 = 3.1 ≈ 3

(b) √(25 – 10) = √15  ≈ √16 = 4

### Simplifying Roots

##### Example

Simplify:

(a) √468

(b) If a and b are positive, √(56a2b3)

### Solution

(a) Find the prime factors using the rules of divisibility. One number from each pair of factors cancels out from the radical.
√468 =  √(2 × 2 × 3 × 3 × 13) = 2 × 3 × √13 = 6√13

(b) √(56a2b3) = √(22 × 2 × 7 × a2 × b2 × b) = 2ab√(14b)

##### Example

Simplify, if x and y are positive:  (\sqrt{\textit{xy}}\,)(\sqrt{\textit{x}^{\displaystyle{3}}\textit{y}}\,)

### Solution

Use the rule of exponents anbn = (ab)n.

\sqrt{\textit{xy}}\, \times \,\sqrt{\textit{x}^{\displaystyle{3}}\textit{y}} = \sqrt{(\textit{xy}\,)(\textit{x}^{\displaystyle{3}}\textit{y}}\,) = \sqrt{\textit{x}^{\displaystyle{4}}\textit{y}^{\displaystyle{2}}} = x2y

##### Example

Which expression is not equal to -3?

1. -\sqrt[\displaystyle{3}]{27}
2. \sqrt[\displaystyle{3}]{-27}
3. -\sqrt{9}
4. -91/2
5. 9-1/2

### Solution

You can calculate each value.

1. -\sqrt[\displaystyle{3}]{27} = -\sqrt[\displaystyle{3}]{3^{\displaystyle{3}}} = -3
2. \sqrt[\displaystyle{3}]{-27} = \sqrt[\displaystyle{3}]{-3^{\displaystyle{3}}} = -3
3. -\sqrt{9} = -3
4. -91/2 = -\sqrt{9} = -3
5. 9-1/2 = \dfrac{1}{9^{\displaystyle{1/2}}} = \dfrac{1}{3}   The correct answer is choice (E).

You also could use a test-taking technique. Look at all of the answers before calculating each value. The negative exponent in choice (E) tells you that 9-1/2 is a positive fraction so it cannot be equal to -3.

##### Example

Simplify: {(\dfrac{1}{8})}^{\displaystyle{-2/3}}

### Solution

Use the rules of exponents \textit{a}^{\displaystyle{-\textit{n}}} = \dfrac{1}{\textit{a}^{\displaystyle{\textit{n}}}} and (am)n = amn.

{(\dfrac{1}{8})}^{\displaystyle{-2/3}} = 82/3 = (23)2/3 = 22 = 4

##### Example

Simplify: 271/3 × 93

### Solution

Factor before multiplying the terms. After factoring, these terms have the same base, 3.

271/3 × 93 = \sqrt[\displaystyle{3}]{3^{\displaystyle{3}}} × (32)3 = 3 × 36 = 37

##### Example

Simplify: y1/2 × y3

### Solution

These terms have the same base, y. Use the rule am × an = am + n and add the exponents. Exponents need to be fractions or decimals, not mixed numbers.
Add the exponents: 1/2 + 3 = 1/2 + 6/2 = 7/2
So y1/2 × y3 = y7/2

##### Example

Simplify: \sqrt[\displaystyle{3}]{\sqrt{\textit{y}}}

### Solution

Use fractional exponents.

\sqrt[\displaystyle{3}]{\sqrt{\textit{<em>y</em>}}} = (y1/2)1/3 = y1/6 = \sqrt[\displaystyle{6}]{\textit{<em>y</em>}}

##### Example

Round to the nearest integer:

(a) \sqrt{3} \,+\,\sqrt{12}

(b) \sqrt{(81\,–\,16)}

### Solution

(a) First simplify by factoring, then add terms that have the same base.

\sqrt{3}\, + \,\sqrt{12} = \sqrt{3}\, + \,\sqrt{(4 × 3)} = \sqrt{3}\, + \,\sqrt{3} = \sqrt{3} ≈ 3(1.7) ≈ 5

(b) Calculate inside the radical first, then estimate using the closest square root. \sqrt{(81 \,- \,16)} = \sqrt{65} ≈ 8

Remember that  nnb ≠ n√(a + b), so \sqrt{(81\, – \,16)}\sqrt{81}\, – \,\sqrt{16}

## Common Values

These roots commonly appear on the GRE, so you should be familiar with them. Since roots and exponents are connected, knowing the powers also tells you the value of the root.  (The list of common values of powers is in the previous section, Exponents.)

√1 = 1
√2 ≈ 1.4
√3 ≈ 1.7
√4 = 2
√5 ≈ 2.24
√9 = 3
√16 = 4
√25 = 5
√36 = 6
√49 = 7
√64 = 8
√81 = 9
√100 = 10

√121 = 11
√144 = 12
√169 = 13
√225 = 15
√256 = 16
√400 = 20
√625 = 25
√900 = 30
√1024 = 32

3√1 = 1
3√-1 = -1
3√8 = 2
3√27 = 3
3√64 = 4
3√125 = 5
3√1000 = 10

#### Other Roots

4√1 = 1
4√16 = 2
4√81 = 3
4√256 = 4
4√625 = 5
4√10,000 = 10

5√32 = 2

These topics are also in Chapter 6  Algebra

How rough can estimates be? Can we use √3 = 1.7 or should we take √3 = 1.73? Can we estimate √60 as √64?

GRE questions are usually designed such that if you simplify an expression first and then make the closest well-known approximation possible, your answer should be ok. Avoid multiplication of estimated values, which amplifies the approximation error. It’s safer to add/subtract estimates.

##### Example

The value of √62 + 2√2 is closest to

1. 5
2. 11
3. 13
4. 16
5. 17

### Solution

√62 + 2√2 ≈ √64 + 2 × 1.4 = 8 + 2.8 = 10.8 ≈ 11

##### Example

The value of √60 × (√15 + 1) is closest to

1. 38
2. 40
3. 44
4. 46
5. 48

### Solution

√60 × (√15 + 1) = 2√15 × (√15 + 1) = 2 × 15 + 2√15 ≈ 30 + 2√16 = 38

The wrong way would be:
√60 × (√15 + 1) = 2√15 × (√15 + 1) ≈ 2√16 × (√16 + 1) = 8 × 5 = 40

But there are some tips to help you consider what approximation should be used:

#### 1. You can analyze the approximation error.

The approximation error when we have a number x rounded to some decimal is half of that decimal unit. If we round to the tenths digit the approximation error is ±0.05. What does this mean?

Let’s say some decimal is rounded to 1.4. It means that the real value lies somewhere in between 1.35 and 1.45. E.g. the numbers 1.4451, 1.36, 1.401, etc. are all rounded to the same tenth digit: 1.4.

So if we round to an integer some number x which has already been rounded to 1.4, we can be sure the original number will be rounded to 1. Because any number between 1.35 and 1.45 will be rounded to 1.

But what if we round 20x and use an approximation of x ≈ 1.4? The approximation error for 20x is 20 times greater than for x. We know that x lies somewhere in between 1.35 and 1.45 (1.35 ≤ x < 1.45), therefore 20x lies in between 27 and 29 (27 ≤ 20x < 29). So we don’t know if it rounds to 27, 28 or 29.

So if a question asks “20√2 is closest to” and the answer choices are 28, 31, 35, 36, 37, then 28 is the correct answer. But if the question asks  “20√2 is closest to” and the answer choices are 28, 29, 30, 31, 32, then we must use a closer approximation for √2. Its approximation to the thousandths digit is easy to remember: √2 ≈ 1.414. But in this case, an approximation to 1.41 is enough.

The approximation error in this case is 0.005, so 1.405 ≤√2 < 1.415.
28.1 ≤ 20√2 < 28.3.
The correct answer is choice 28.

#### Example

2√60 is closest to which of the following?

1. 16
2. 17
3. 18
4. 19
5. 20

### Solution

If we estimate 2√60 ≈ 2√64 = 16, we can be sure that the correct answer is 16, because our estimate is greater than the real value and 16 is the smallest answer choice.

#### Example

Round 5/√17 to an integer.

### Solution

5/√25 < 5/√17 < 5/√16

5/5 < 5/√17 < 5/4

1 < 5/√17 < 1.25

5/√17 ≈ 1

Terms in simplest form do not have a radical in the denominator.

##### Example

Simplify:

(a) \dfrac{4}{\sqrt{3}}

(b) \sqrt{\dfrac{5}{2}}

(c) \dfrac{6}{3\, – \sqrt{5}}

### Solution

Eliminate the radical by multiplying by a fraction equal to 1 that has the numerator and denominator equal to the radical.

(a) \dfrac{4}{\sqrt{3}} = \dfrac{4}{\sqrt{3}} × \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{4\sqrt{3}}{3}

(b) First apply the rule of exponents. {(\dfrac{\textit{\textit{a}}}{\textit{b}})}^{\displaystyle{\textit{n}}} = \dfrac{\textit{a}^{\displaystyle{\textit{n}}}}{\textit{b}^{\displaystyle{\textit{n}}}}.
\sqrt{\dfrac{5}{2}} = \dfrac{\sqrt{5}}{\sqrt{2}} = \dfrac{\sqrt{5}}{\sqrt{2}} × \dfrac{\sqrt{2}}{\sqrt{2}} =\dfrac{\sqrt{10}}{2}

(c) The fraction needs to have the conjugate of the denominator.
\dfrac{6}{3\, – \sqrt{5}}  = (\dfrac{6}{3\, – \sqrt{5}})(\dfrac{3 + \sqrt{5}}{3 + \sqrt{5}}) = \dfrac{18 \,–\, 6\sqrt{5}}{9 + 3\sqrt{5} –\, 3\sqrt{5}\, – \,5} = \dfrac{2(9 + 3\sqrt{5})}{9 \,- \,5} = \dfrac{2(9 + 3\sqrt{5})}{4} = \dfrac{9 + 3\sqrt{5}}{2}

When we deal with equations that contain variables under radical signs with an even degree (usually square radicals), we must keep in mind that such radicals can be calculated only for non-negative values. We also must keep in mind that such radicals yield a non-negative value.

There are two ways to deal with this issue:

• Plug the solution values into the original equation (or at least into the radicals in the original equation).
• Define the range of possible values for the variable right from the beginning.
##### Example

How many solutions are there for the equation √(x2x – 6) = 1 – x  ?

### Solution

WRONG SOLUTION:
√(x2x – 6) = 1 – x
x2x – 6 = (1 – x)2
x2x – 6 = x2 – 2x + 1
x = 7
The equation has one solution.

It might seem that everything is OK in the WRONG solution provided above. But let’s take a look at the proper solution.
PROPER SOLUTION:

√(x2x – 6) = 1 – x
x2x – 6 = (1 – x)2
x2x – 6 = x2– 2x + 1
x = 7

Let’s plug x = 7 in the original equation to see if it fits.
√(72 – 7 – 6) = 1 – 7
√36 = -6

We see that x = 7 doesn’t fit. Therefore this question has NO solution.

Alternative way:

Let’s define the range of possible values for x.

√(x2x – 6) = 1 – x, so 1 – x ≥ 0
1 ≥ x

On the other hand, x2x – 6 ≥ 0
(x – 3)(x + 2) ≥ 0

So x ≥ 3 or x ≤ -2. If we combine this statement with the earlier one, 1 ≥ x, we get: x ≤ -2.
x ≤ -2 is the range of possible values for x.

Let’s find all possible solutions:

√(x2x – 6) = 1 – x
x2 – x – 6 = (1 – x)2
x2x – 6 = x2 – 2x + 1
x = 7

We see that x = 7 doesn’t fit the range of possible values for x: x ≤ -2.
Therefore the equation has NO solutions.

#### Example

Solve the equation:

(a) 3√x – 6 = 0

(b) 4\sqrt{\textit{x} \,-\, 5} + 6 = 18

(c) \sqrt{9\textit{x} \,-\, 20} = \sqrt{4\textit{x} + 5}

### Solution

(a)

3√x – 6 = 0
3√x = 6
x = 2
(√x)2 = 22
x = 4
If we plug = 4, it fits:
3√4 – 6 = 0
6 – 6 = 0

(b)

4\sqrt{\textit{x}\, –\, 5} + 6 = 18
4\sqrt{\textit{x}\, –\, 5} = 12
\sqrt{\textit{x}\, –\, 5} = 3
(\sqrt{\textit{x}\, –\, 5})2 = 32
x – 5 = 9
x = 14
If we plug x = 14, it fits:
4√(14 – 5)  + 6 = 18
12 + 6 = 18

(c)

\sqrt{9\textit{x} \,-\, 20} = \sqrt{4\textit{x} + 5}
(\sqrt{9\textit{x} \,-\, 20})2 = (\sqrt{4\textit{x} + 5})2
9x – 20 = 4x + 5
9x – 4x = 5 + 20
5x = 25
x = 5
If we plug x = 5, it fits:
√(9 × 5 – 20)  = √(4 × 5 + 5)
√25 = √25