# Roots

### Video Walkthrough

MIT grad shows how to simplify radical expressions, specifically square root expressions, into their simplest form (“Simplified Radical Form” or “SRF Form”). To skip ahead:
1) for a PERFECT SQUARE under the root like \sqrt{16} skip to time
1:29.
2) for a SMALL number under the root that is NOT a perfect square like \sqrt{32}, skip to time
2:45.
3) for a LARGE number under the root, such as \sqrt{343}, skip to
5:14.
4) for a PRODUCT, meaning a number times a square root like 5 * \sqrt{54} skip to
7:30.
5) for a QUOTIENT, meaning a number divided by a square root like \dfrac{4}{\sqrt{2}}, and where you have to RATIONALIZE the denominator, skip to
10:10. and 6) for a more complex QUOTIENT example like \dfrac{4 + \sqrt{3}}{5 \,-\, \sqrt{3}} where you have to multiply by the CONJUGATE to simplify, skip to 12:36.

## Roots

Just like addition and subtraction “undo” each other, so do exponents and roots.

You take a root to answer questions like:
If \sqrt[\displaystyle{3}]{64} = x, what is the value of x?
You know  4 × 4 × 4 = 43 = 64,
so  \sqrt[\displaystyle{3}]{64} = 4.

The expression \sqrt{\textit{a}} is called a root or radical.  The symbol \surd is the radical symbol.
Radical symbols can have a number in front of the radical.  The 3 in \sqrt[\displaystyle{3}]{64} is asking for the cube root.  For square roots, the 2 is left off.

\sqrt[\displaystyle{3}]{8}

### Solution

\sqrt[\displaystyle{3}]{8} = 2 since 23 = 8

\sqrt[\displaystyle{3}]{-1}

### Solution

\sqrt[\displaystyle{3}]{-1} = -1 since (-1)3 = -1

## Common Values

These roots commonly appear on the GMAT, 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.)

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

#### Square Roots

\sqrt{1} = 1
\\[1ex]\sqrt{2} ≈ 1.4
\\[1ex]\sqrt{3} ≈ 1.7
\\[1ex]\sqrt{4} = 2
\\[1ex]\sqrt{5} ≈ 2.24
\\[1ex]\sqrt{9} = 3
\\[1ex]\sqrt{16} = 4
\\[1ex]\sqrt{25} = 5
\\[1ex]\sqrt{36} = 6
\\[1ex]\sqrt{49} = 7
\\[1ex]\sqrt{64} = 8
\\[1ex]\sqrt{81} = 9
\\[1ex]\sqrt{100} = 10

\sqrt{121} = 11
\\[1ex]\sqrt{144} = 12
\\[1ex]\sqrt{169} = 13
\\[1ex]\sqrt{225} = 15
\\[1ex]\sqrt{256} = 16
\\[1ex]\sqrt{400} = 20
\\[1ex]\sqrt{625} = 25
\\[1ex]\sqrt{900} = 30
\\[1ex]\sqrt{1024} = 32

#### Cube Roots

\sqrt[\displaystyle{3}]{1} = 1
\\[1ex]\sqrt[\displaystyle{3}]{-1} = -1
\\[1ex]\sqrt[\displaystyle{3}]{8} = 2
\\[1ex]\sqrt[\displaystyle{3}]{27} = 3
\\[1ex]\sqrt[\displaystyle{3}]{64} = 4
\\[1ex]\sqrt[\displaystyle{3}]{125} = 5
\\[1ex]\sqrt[\displaystyle{3}]{1000} = 10

#### Other Roots

\sqrt[\displaystyle{4}]{1} = 1
\\[1ex]\sqrt[\displaystyle{4}]{16} = 2
\\[1ex]\sqrt[\displaystyle{4}]{81} = 3
\\[1ex]\sqrt[\displaystyle{4}]{256} = 4
\\[1ex]\sqrt[\displaystyle{4}]{625} = 5
\\[1ex]\sqrt[\displaystyle{4}]{10,000} = 10

\sqrt[\displaystyle{5}]{32} = 2

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.

In order to avoid double-value of even roots, the radical sign with even degree is a non-negative value only.
For example, \sqrt{4} = 2 \,\,(\sqrt{4} \mathrel{\char≠} -2),
\sqrt[\displaystyle{4}]{81} = 3 \,\,(\sqrt[\displaystyle{4}]{81} \mathrel{\char≠} -3).

## 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.

\textit{a}^{\displaystyle{\dfrac{1}{\textit{n}}}} = \sqrt[\displaystyle{\textit{n}}]{\textit{a}}

16^{\displaystyle{\dfrac{1}{4}}} = \sqrt[\displaystyle{4}]{16} = \sqrt[\displaystyle{4}]{2^{\displaystyle{4}}} = 2

The exponent \dfrac{1}{2} means take the square root of the base.

\textit{a}^{\displaystyle{\dfrac{1}{2}}} = \sqrt{\textit{a}}

49^{\displaystyle{\dfrac{1}{2}}} = \sqrt{49} = \sqrt{7^{\displaystyle{2}}} = 7

The exponent \dfrac{1}{3} means take the cube root of the base.

\textit{a}^{\displaystyle{\dfrac{1}{3}}} = \sqrt[\displaystyle{3}]{\textit{a}}

216^{\displaystyle{\dfrac{1}{3}}} = \sqrt[\displaystyle{3}]{216} = \sqrt[\displaystyle{3}]{(6 × 6 × 6)} = 6

The exponent \dfrac{\textit{m}}{\textit{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.

\textit{a}^{\dfrac{\textit{m}}{\text{n}}} = \sqrt[\displaystyle{\textit{n}}]{\textit{a}^{\displaystyle{\textit{m}}}}
2^{\dfrac{5}{4}} = \sqrt[\displaystyle{4}]{2^{\displaystyle{5}}} = \sqrt[\displaystyle{4}]{2^{\displaystyle{4}} × 2} = 2 \sqrt[\displaystyle{4}]{2}

The exponent \dfrac{2}{3} means square the base and take the cube root.

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

81^{\dfrac{2}{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.

\sqrt[\displaystyle{\textit{n}}]{\textit{a}} + \sqrt[\displaystyle{\textit{n}}]{\textit{b}} \mathrel{\char≠} \sqrt[\displaystyle{\textit{n}}]{\textit{a} + \textit{b}}
\sqrt{4} + \sqrt{9} \mathrel{\char≠} \sqrt{13}
\sqrt{20}\,-\,\sqrt{5} \mathrel{\char≠} \sqrt{15}

\sqrt{4} + \sqrt{9} = 2 + 3 = 5

\sqrt{20}\,-\, \sqrt{5} = \sqrt{(4 × 5)} \,-\, \sqrt{5}
= 2\sqrt{5} \,-\, \sqrt{5} = \sqrt{5}

## Roots and Negative Numbers

Not all roots are possible.  For example, \sqrt{-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.

Be careful when dealing with equations. Remember that an even root is undefined for negative numbers. A root must be positive, so
\\[2ex]\sqrt{(\textit{x}^{\displaystyle{2}})} = |\textit{x}|\,\, (NOT \sqrt{(\textit{x}^{\displaystyle{2}})} = \textit{x})

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
\sqrt{(\textit{x}^{\displaystyle{2}})} = \sqrt{4}
|x| = 2
x = 2 and x = -2 are the solutions of the equation.

## Estimating Roots

Not all roots yield an integer. For example, \sqrt{2} and \sqrt{48} do not have integer values.

Using a calculator, you get
\sqrt{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 = \sqrt{(2 × 81)} = 9\sqrt{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.

\sqrt{48} will be slightly less than 7 since \sqrt{49} = 7

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

First, simplify by factoring.
\sqrt{48} = \sqrt{(16 × 3)} = 4\sqrt{3}

Then for \sqrt{3}, use an approximation.  4\sqrt{3} ≈ 4(1.7) = 6.8

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

\sqrt{2} ≈ 1.4
\sqrt{3} ≈ 1.7
\sqrt{5} ≈ 2.24

Round to the nearest integer:

\sqrt{3} + \sqrt{2}

### Solution

\sqrt{3} + \sqrt{2} ≈ 1.7 + 1.4 = 3.1 ≈ 3

Round to the nearest integer:

\sqrt{(25 \,-\, 10)}

### Solution

\sqrt{(25 \,-\, 10)} = \sqrt{15} ≈ \sqrt{16} = 4

## Simplifying Roots

\sqrt{468}

### Simplify

Find the prime factors using the rules of divisibility. One number from each pair of factors cancels out from the radical.
\\[1ex]\sqrt{468} =  \sqrt{(2 × 2 × 3 × 3 × 13)}
= 2 × 3 × \sqrt{13} = 6\sqrt{13}

If a and b are positive, \sqrt{(56\textit{a}^{\displaystyle{2}}\textit{b}^{\displaystyle{3}})}

### Simplify

\sqrt{(56\textit{a}^{\displaystyle{2}}\textit{b}^{\displaystyle{3}})}

= \sqrt{(2^{\displaystyle{2}} × 2 × 7 × \textit{a}^{\displaystyle{2}} × \textit{b}^{\displaystyle{2}} × \textit{b})}

= 2\textit{ab}\sqrt{14\textit{b}}

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

Which expression is not equal to -3?

1. -\sqrt[\displaystyle{3}]{27}
2. \sqrt[\displaystyle{3}]{-27}
3. -\sqrt{9}
4. -9^{\dfrac{1}{2}}
5. 9^{\dfrac{-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. -9^{\dfrac{1}{2}} = -\sqrt{9} = -3
5. 9^{\dfrac{-1}{2}} = \dfrac{1}{9^{\dfrac{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^{\dfrac{-1}{2}} is a positive fraction so it cannot be equal to -3.

{\Big(\dfrac{1}{8}\Big)}^{\dfrac{-2}{3}}

### Simplify

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

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

27^{\dfrac{1}{3}} × 9^{\displaystyle{3}}

### Simplify

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

27^{\dfrac{1}{3}} × 9^{\displaystyle{3}} = \sqrt[\displaystyle{3}]{3^{\displaystyle{3}}} × (32)3
= 3 × 36 = 37

\textit{y}^{\dfrac{1}{2}} × \textit{y}^{\displaystyle{3}}

### Simplify

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.

\dfrac{1}{2} + 3 = \dfrac{1}{2} + \dfrac{6}{2} = \dfrac{7}{2}

So \textit{y}^{\dfrac{1}{2}} × \textit{y}^{\displaystyle{3}} = \textit{y}^{\dfrac{7}{2}}

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

### Simplify

Use fractional exponents.

\sqrt[\displaystyle{3}]{\sqrt{\textit{y}}} = {(\textit{y}^{\dfrac{1}{2}})}^{\dfrac{1}{3}} = \textit{y}^{\dfrac{1}{6}} = \sqrt[\displaystyle{6}]{\textit{y}}

Round to the nearest integer:

\sqrt{3} \,+\,\sqrt{12}

### Solution

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

\sqrt{3}\, + \,\sqrt{12} = \sqrt{3}\, + \,\sqrt{(4 × 3)}
= \sqrt{3}\, + \,2\sqrt{3} = 3\sqrt{3} ≈ 3(1.7) ≈ 5

Round to the nearest integer:

\sqrt{(81\,–\,16)}

### Solution

Calculate inside the radical first, then estimate using the closest square root.
\sqrt{(81 \,- \,16)} = \sqrt{65} ≈ 8.
Remember that
\sqrt[\displaystyle{\textit{n}}]{\textit{a}} + \sqrt[\displaystyle{\textit{n}}]{\textit{b}} \mathrel{\char≠} \sqrt[\displaystyle{\textit{n}}]{\textit{(\textit{a} + \textit{b})}},
so \sqrt{(81\, - \,16)}\sqrt{81}\, - \,\sqrt{16}

These topics are also in Chapter 6  Algebra

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

GMAT 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.

The value of \sqrt{62} + 2\sqrt{2} is closest to

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

### Solution

\sqrt{62} + 2\sqrt{2} ≈ \sqrt{64} + 2 × 1.4
= 8 + 2.8 = 10.8 ≈ 11

The value of \sqrt{60} × (\sqrt{15} + 1) is closest to

• 38
• 40
• 44
• 46
• 48

### Solution

\sqrt{60} × (\sqrt{15} + 1)
= 2\sqrt{15} × (\sqrt{15} + 1)
= 2 × 15 + 2\sqrt{15} ≈ 30 + 2\sqrt{16} = 38

The wrong way would be:

\sqrt{60} × (\sqrt{15} + 1)
= 2\sqrt{15} × (\sqrt{15} + 1) ≈ 2\sqrt{16} × (\sqrt{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\sqrt{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\sqrt{2} is closest to” and the answer choices are 28, 29, 30, 31, 32, then we must use a closer approximation for \sqrt{2}. Its approximation to the thousandths digit is easy to remember: \sqrt{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 ≤ \sqrt{2} < 1.415.
28.1 ≤ 20\sqrt{2} < 28.3.
The correct answer is choice 28.

#### 2. You can analyze if your approximation increases or decreases the value.

2\sqrt{60} is closest to which of the following?

• 16
• 17
• 18
• 19
• 20

### Solution

If we estimate 2\sqrt{60} ≈ 2\sqrt{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.

#### 3. You can make two estimations: lower and upper.

Round \dfrac{5}{\sqrt{17}} to an integer.

### Solution

\dfrac{5}{\sqrt{25}} < \dfrac{5}{\sqrt{17}} < \dfrac{5}{\sqrt{16}}
\\[2ex]\dfrac{5}{5} < \dfrac{5}{\sqrt{17}} < \dfrac{5}{4}
\\[2ex]1 < \dfrac{5}{\sqrt{17}} < 1.25
\\[2ex]\dfrac{5}{\sqrt{17}} ≈ 1

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

\dfrac{4}{\sqrt{3}}

### Simplify

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

\dfrac{4}{\sqrt{3}} = \dfrac{4}{\sqrt{3}} × \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{4\sqrt{3}}{3}

\sqrt{\dfrac{5}{2}}

### Simplify

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

First apply the rule of exponents.
\\[1ex]{(\dfrac{\textit{\textit{a}}}{\textit{b}})}^{\displaystyle{\textit{n}}} = \dfrac{\textit{a}^{\displaystyle{\textit{n}}}}{\textit{b}^{\displaystyle{\textit{n}}}}.
\\[2ex]\sqrt{\dfrac{5}{2}} = \dfrac{\sqrt{5}}{\sqrt{2}} = \dfrac{\sqrt{5}}{\sqrt{2}} × \dfrac{\sqrt{2}}{\sqrt{2}} =\dfrac{\sqrt{10}}{2}

\dfrac{6}{3\, - \sqrt{5}}

### Simplify

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

The fraction needs to have the conjugate of the denominator.
\\[2ex]\dfrac{6}{3\, - \sqrt{5}}  = \Big(\dfrac{6}{3\, - \sqrt{5}}\Big)\Big(\dfrac{3 + \sqrt{5}}{3 + \sqrt{5}}\Big)

= \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.

How many solutions are there for the equation
\sqrt{(\textit{x}^{\displaystyle{2}} \,-\, \textit{x} \,-\, 6)} = 1 – x  ?

### Solution

WRONG SOLUTION:

\sqrt{(\textit{x}^{\displaystyle{2}} \,-\, \textit{x} \,-\, 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:

\sqrt{(\textit{x}^{\displaystyle{2}} \,-\, \textit{x} \,-\, 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.

\sqrt{(7^{\displaystyle{2}} \,-\, 7 \,-\, 6)} = 1 – 7
\\[1ex]\sqrt{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.

\sqrt{(\textit{x}^{\displaystyle{2}} \,-\, \textit{x} \,-\, 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:

\sqrt{(\textit{x}^{\displaystyle{2}} \,-\, \textit{x} \,-\, 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.

Solve the equation:

3\sqrt{\textit{x}} \,-\, 6 = 0

### Solution

3\sqrt{\textit{x}} \,-\, 6 = 0
\\[2ex]3\sqrt{\textit{x}} = 6
\\[2ex]\sqrt{\textit{x}} = 2
\\[2ex](\sqrt{\textit{x}})^{\displaystyle{2}} = 2^{\displaystyle{2}}

x = 4

If we plug = 4, it fits:

3\sqrt{4} \,-\, 6 = 0

6 – 6 = 0

Solve the equation:

4\sqrt{\textit{x} \,-\, 5} + 6 = 18

### Solution

4\sqrt{\textit{x}\, –\, 5} + 6 = 18
\\[2ex]4\sqrt{\textit{x}\, –\, 5} = 12
\\[2ex]\sqrt{\textit{x}\, –\, 5} = 3
\\[2ex](\sqrt{\textit{x}\, –\, 5})^{\displaystyle{2}} = 32

x – 5 = 9

x = 14

If we plug x = 14, it fits:

4\sqrt{(14 \,-\, 5)} + 6 = 18

12 + 6 = 18

Solve the equation:

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

### Solution

\sqrt{9\textit{x} \,-\, 20} = \sqrt{4\textit{x} + 5}
\\[2ex](\sqrt{9\textit{x} \,-\, 20})^{\displaystyle{2}} = (\sqrt{4\textit{x} + 5})^{\displaystyle{2}}

9x – 20 = 4x + 5

9x – 4x = 5 + 20

5x = 25

x = 5

If we plug x = 5, it fits:

\sqrt{(9 × 5 \,-\, 20)}  = \sqrt{(4 × 5 + 5)}
\\[2ex]\sqrt{25} = \sqrt{25}

### Odd and Even Roots

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

\sqrt[\displaystyle{3}]{125}

### Solution

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

\\[1ex]\sqrt[\displaystyle{3}]{-125}

### Solution

\\[1ex]\sqrt[\displaystyle{3}]{-125} = \sqrt[\displaystyle{3}]{((-5) × (-5) × (-5))}
= -5

\\[1ex]\sqrt[\displaystyle{5}]{32}

### Solution

\\[1ex]\sqrt[\displaystyle{5}]{32} = \sqrt[\displaystyle{5}]{(2 × 2 × 2 × 2 × 2)}
= 2

When taking an even root, there will be two answers. But the sign \surd denotes only the non-negative answer.
For example, there are two square roots of 25: 5 and -5, since 52 = 25 and (-5)2 = 25. But \sqrt{25} has only one value: 5, \sqrt{16} = 4, \sqrt{1} = 1, etc

\sqrt{16}

### Solution

\sqrt{16} = \sqrt{(4 × 4)} = 4

\sqrt[\displaystyle{3}]{1}

### Solution

\sqrt[\displaystyle{3}]{1} = \sqrt[\displaystyle{3}]{(1 × 1 × 1)} = 1

\sqrt[\displaystyle{4}]{16}

### Solution

\sqrt[\displaystyle{4}]{16} = \sqrt[\displaystyle{4}]{(2 × 2 × 2 × 2)} = 2

Be careful when dealing with equations – remember that an even root is undefined for negative numbers.
For example, \sqrt{-4} is undefined because there is no number squared that equals -4.

If x2 = 4, what is x?

### Solution

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

Alternative solution:
x2 = 4
\sqrt{\textit{x}^{\displaystyle{2}}} = \sqrt{4}
|x| = 2
x = 2 and x = -2 are the solutions of the equation.

#### ROOTS

Best viewed in landscape mode

3 questions with video explanations

100 seconds per question

Before attempting these problems, be sure to review this section on data sufficiency questions.