Squares and Roots

A number is called a Perfect Square if it is a Square of another number.

For Example: If m = x², then m is the Square of x and therefore m is called a Perfect Square.

Since m = x², therefore x is the Square Root of m.

The Square Root of a number m is denoted by $\sqrt[]{m}$

Therefore since m = x², then $\sqrt[]{m}$ = x

Examples:

• $\sqrt[]{25}$ = 5

5 is the Square Root of 25

If m = x³, then m is the Cube of x and therefore x is the Cube Root of m.

The Cube Root of a number m is denoted by $\sqrt[3]{m}$

Therefore since m = x³, then $\sqrt[3]{m}$ = x

Examples:

• $\sqrt[3]{27}$ = 3

3 is the Cube Root of 27

If m = xⁿ, then x is the n-th Root of m.

The n-th Root of a number m is denoted by $\sqrt[n]{m}$

Therefore since m = xⁿ, then $\sqrt[n]{m}$ = x

Examples:

• $\sqrt[4]{16}$ = 2

2 is the 4th Root of 16

If $\sqrt[n]{m}$ = x, then m = x ^1/n

Surds

All n-th roots that are not equal to a Rational Number are called Surds.

This means if we express them as decimals, we get non-recurring or non-terminating decimals.

Examples of Surds:

• $\sqrt[]{2}$ = 1.414213562373095

• $\sqrt[]{5}$ = 2.236067977499789

• $\sqrt[]{8}$ = 2.828427124746190

Some roots are equal to Rational Numbers and therefore are not Surds.

Examples:

• $\sqrt[]{4}$ = 2

• $\sqrt[3]{27}$ = 3

• $\sqrt[4]{16}$ = 2

Simplifying n-th Roots

This is very important to know when working with n-th roots.

Multiplication of n-th Roots

$\sqrt[]{a}$ x $\sqrt[]{a}$ = $\sqrt[]{a x a}$ = $\sqrt[]{a 2}$ = a

$\sqrt[]{a}$ x $\sqrt[]{b}$ = $\sqrt[]{a x b}$ = $\sqrt[]{ab}$

$\sqrt[3]{a}$ x $\sqrt[3]{a}$ x $\sqrt[3]{a}$ = $\sqrt[3]{a x a x a}$ = $\sqrt[3]{a 3}$ = a

$\sqrt[3]{a}$ x $\sqrt[3]{b}$ = $\sqrt[3]{a x b}$ = $\sqrt[3]{ab}$

Therefore,

$\sqrt[n]{a n}$ = a

$\sqrt[n]{a}$ x $\sqrt[n]{b}$ = $\sqrt[n]{ab}$

Examples:

Given $\sqrt[]{25}$ x $\sqrt[]{4}$

• $\sqrt[]{25}$ x $\sqrt[]{4}$ = $\sqrt[]{25 x 4}$

= $\sqrt[]{100}$ = $\sqrt[]{102}$ = 10

Given $\sqrt[]{5}$ x $\sqrt[]{3}$

• $\sqrt[]{5}$ x $\sqrt[]{3}$ = $\sqrt[]{5 x 3}$ = $\sqrt[]{15}$

Given $\sqrt[3]{4}$ x $\sqrt[3]{2}$

• $\sqrt[3]{4}$ x $\sqrt[3]{2}$ = $\sqrt[3]{4 x 2}$

= $\sqrt[3]{8}$ = $\sqrt[3]{23}$ = 2

Division of n-th Roots

$\frac{\sqrt[]{a}}{\sqrt[]{b}}$ = $\sqrt[]{\frac{a}{b}}$ , where b ≠ 0

$\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$ = $\sqrt[3]{\frac{a}{b}}$ , where b ≠ 0

Therefore,

$\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$ = $\sqrt[n]{\frac{a}{b}}$ , where b ≠ 0

Examples:

Given $\frac{\sqrt[]{100}}{\sqrt[]{4}}$

• $\frac{\sqrt[]{100}}{\sqrt[]{4}}$ = $\sqrt[]{\frac{100}{4}}$ = $\sqrt[]{25}$ = $\sqrt[]{52}$ = 5

Given $\sqrt[]{\frac{5}{4}}$

• $\sqrt[]{\frac{5}{4}}$ = $\frac{\sqrt[]{5}}{\sqrt[]{4}}$ = $\frac{\sqrt[]{5}}{2}$

Given $\sqrt[3]{\frac{27}{8}}$

• $\sqrt[3]{\frac{27}{8}}$ = $\frac{\sqrt[3]{27}}{\sqrt[3]{8}}$ = $\frac{\sqrt[3]{33}}{\sqrt[3]{23}}$ = $\frac{3}{2}$

Adding and Subtracting n-th Roots

$\sqrt[n]{a}$ + $\sqrt[n]{a}$ = 2 $\sqrt[n]{a}$

2 $\sqrt[n]{a}$ + $\sqrt[n]{a}$ = 3 $\sqrt[n]{a}$

If you have $\sqrt[n]{a}$ + $\sqrt[n]{b}$ , where a ≠ b this is what you should do:

Simplify the Square Roots if it is possible. Write a in terms of b or write b in terms of a.

For Example: Write a = c x b, where c is a Natural Number.
Perform the Operation and Simplify the Final Answer

Examples:

Given $\sqrt[]{9}$ + $\sqrt[]{4}$

• $\sqrt[]{9}$ + $\sqrt[]{4}$ = $\sqrt[]{32}$ + $\sqrt[]{22}$ = 3 + 2 = 5

Given $\sqrt[]{20}$ - $\sqrt[]{5}$

• $\sqrt[]{20}$ - $\sqrt[]{5}$ = $\sqrt[]{5 x 4}$ - $\sqrt[]{5}$

= $\sqrt[]{5}$ x $\sqrt[]{4}$ - $\sqrt[]{5}$

= 2 $\sqrt[]{5}$ - $\sqrt[]{5}$ = $\sqrt[]{5}$

Given $\sqrt[3]{8}$ + $\sqrt[3]{8}$

• $\sqrt[3]{8}$ + $\sqrt[3]{8}$ = 2 $\sqrt[3]{8}$ = 2 $\sqrt[3]{23}$

= 2 x 2 = 4

Rationalising the Denominator

When you have a Fraction that contains a Surd in the Denominator, you can simplify it by changing the Denominator to a Rational Number. This is called Rationalising the Denominator.

Remember $\sqrt[]{a}$ x $\sqrt[]{a}$ = a , so If you multiply the Numerator and the Denominator by the same Surd, you are not changing the value of the number since you are multiplying by $\frac{\sqrt[]{a}}{\sqrt[]{a}}$ which is equal to 1.

Examples:

Given $\frac{\sqrt[]{3}}{\sqrt[]{2}}$

• $\frac{\sqrt[]{3}}{\sqrt[]{2}}$ = $\frac{\sqrt[]{3}}{\sqrt[]{2}}$ x $\frac{\sqrt[]{2}}{\sqrt[]{2}}$

= $\frac{\sqrt[]{3} x \sqrt[]{2}}{\sqrt[]{2} x \sqrt[]{2}}$ = $\frac{\sqrt[]{6}}{2}$

Given $\frac{2}{\sqrt[]{5}}$

• $\frac{2}{\sqrt[]{5}}$ = $\frac{2}{\sqrt[]{5}}$ x $\frac{\sqrt[]{5}}{\sqrt[]{5}}$

= $\frac{2
x \sqrt[]{5}}{\sqrt[]{5} x \sqrt[]{5}}$ = $\frac{2 \sqrt[]{5}}{5}$

Exponents

The exponent tells us how many times the number is multiplied by itself.

For Example:

a² = a x a

a² is read as a exponent 2 or a to the power of 2.

a is multiplied by itself 2 times.

a³ = a x a x a

a³ is read as a exponent 3 or a to the power of 3.

a is multiplied by itself 3 times.

Therefore,

aⁿ is read as a exponent n or a to the power of n.

a is multiplied by itself n times.

For

aⁿ

a is the Base

n is the Exponent

aⁿ is the Power

Examples:

• 3³ , which is read as 3 to the power of 3.

3³ = 3 x 3 x 3 = 27

• 2⁴ , which is read as 2 to the power of 4.

2⁴ = 2 x 2 x 2 x 2= 16

Laws of Exponents

These laws work for exponents that are integers, rational numbers or irrational numbers

To multiply two powers with same bases, add their exponents.

b^m x bⁿ = b^{m + n}

Example:

• 2³ x 2² = 2^{3 + 2} = 2⁵

To divide two powers with same bases, subtract their exponents.

$\frac{b m}{b n}$ = b^{m - n}

Example:

• $\frac{25}{23}$ = 2^{5 - 3} = 2²

To raise a power to an exponent, multiply the exponents.

(b^m)ⁿ = b^{m x n}

(ab)ⁿ = aⁿbⁿ

( $\frac{a}{b}$ )ⁿ = $\frac{a n}{b n}$

Example:

• (2³)² = 2^{3 x 2} = 2⁶

• (5³ x 3³)² = (5³)² x (3³)² = 5⁶ x 3⁶

• ( $\frac{4}{3}$ )² = $\frac{42}{32}$

Any base raised to 0 is 1.

b⁰ = 1

Example:

• 5⁰ = 1

A positive exponent in the denominator is the same as a negative exponent in the numerator.

$\frac{1}{b n}$ = b^-n

( $\frac{a}{b}$ )^{- n} = ( $\frac{b}{a}$ )ⁿ

Example:

• $\frac{1}{23}$ = 2^-3

• ( $\frac{4}{3}$ )^{- 2} = ( $\frac{3}{4}$ )²

To find the root of a power, divide the exponents.

$\sqrt[n]{b m}$ = ( b^m )^1/n = b^m/n (n ≥ 2).

Example:

• $\sqrt[3]{34}$ = ( 3⁴ )^1/3 = 3^4/3

Algebraic Expressions with Exponents

Always remember to follow the order of operations.

B O D M A S

Brackets Of Division Multiplication Addition Subtraction

Example:

Given -3a³b² x -4a⁴b⁴

• -3a³b² x -4a⁴b⁴ = (-3 x -4)(a³ x a⁴)(b² x b⁴) = 12(a^{3 + 4})(b^{2 + 4}) = 12a⁷b⁶

Given $\frac{12x 5 y 8}{-4x 2 y 4}$ =

• $\frac{12x 5 y 8}{-4x 2 y 4}$ = ( $\frac{12}{-4}$ ) ( $\frac{x 5}{x 2}$ ) ( $\frac{y 8}{y 4}$ )

= -3x³y⁴

Given (-3a³b²)³

• (-3a³b²)³ = (-3)³(a³)³(b²)³

= (-3³)(a^{3 x 3})(b^{2 x 3})

= -27a⁹b⁶

Prime Factors

When the bases are different, we can write each base as a product of its Prime Factors.

Remember

• A Prime Number has only two different factors.

• A Composite Number has more than two factors.

The number 1 is neither a Prime Number nor a Composite Number.

Prime Numbers:

• 2; 3; 5; 7; 11; 13 . . .

Every Composite Number can be written as the product of Prime Numbers.

This helps us factorise and simplify.

Example:

• 4 = 2 x 2 = 2²

• 6 = 2 x 3

• 8 = 2 x 2 x 2 = 2³

• 9 = 3 x 3 = 3²

• 24 = 8 x 3 = 2 x 2 x 2 x 3 = 2³x3

We can simplify Equations by Expressing them in Prime Factors

Example:

Given 72^x-2

• 72^x-2 = ( 8x9 )^x-2 = ( 2³ x 3² )^x-2

= ( 2³ )^x-2 x (3² )^x-2

= 2^3x-63^2x-4

Algebraic Fractions with Exponents

Expressions with Products of Terms

• Factorise the Terms using Prime Factors

• Use Laws of Exponents

Example:

Given $\frac{5 2n x 9 2n - 3}{15 2n x 3 2n - 1}$

• $\frac{5 2n x 9 2n - 3}{15 2n x 3 2n - 1}$ = $\frac{5 2n x (3 2) 2n - 3}{(5 x 3) 2n x 3 2n - 1}$ = $\frac{5 2n x 3 4n - 6}{5 2n x 3 2n x 3 2n - 1}$ = 5^{2n - 2n} x 3^{4n - 6 - 2n - 2n + 1}

= 5⁰ x 3^{- 5} = 1 x $\frac{1}{35}$ = $\frac{1}{35}$ = $\frac{1}{243}$

Example:

Given $\frac{5 2n x 9 2n - 3}{15 2n x 3 2n - 1}$

• $\frac{5 2n x 9 2n - 3}{15 2n x 3 2n - 1}$

= $\frac{5 2n x (3 2) 2n - 3}{(5 x 3) 2n x 3 2n - 1}$

= $\frac{5 2n x 3 4n - 6}{5 2n x 3 2n x 3 2n - 1}$

= 5^{2n - 2n} x 3^{4n - 6 - 2n - 2n + 1}

= 5⁰ x 3^{- 5} = 1 x $\frac{1}{35}$

= $\frac{1}{35}$ = $\frac{1}{243}$

Example:

Given $\frac{5 2n x 9 2n - 3}{15 2n x 3 2n - 1}$

• $\frac{5 2n x 9 2n - 3}{15 2n x 3 2n - 1}$

= $\frac{5 2n x (3 2) 2n - 3}{(5 x 3) 2n x 3 2n - 1}$

= $\frac{5 2n x 3 4n - 6}{5 2n x 3 2n x 3 2n - 1}$

= 5^{2n - 2n} x 3^{4n - 6 - 2n - 2n + 1}

= 5⁰ x 3^{- 5} = 1 x $\frac{1}{35}$

= $\frac{1}{35}$ = $\frac{1}{243}$

Expressions with Terms Added or Subtracted

• First try to Factorise both the Numerator and Denominator

• Use Laws of Exponents

• Cancel any Common Factors

Example:

Given $\frac{(3 n) 2 + 3 2n - 1}{9 n}$

• $\frac{(3 n) 2 + 3 2n - 1}{9 n}$

= $\frac{3 2n + 3 2n - 1}{(3 2) n}$ = $\frac{3 2n + 3 2n x 3 - 1}{3 2n}$ = $\frac{3 2n (1 + 3 - 1)}{3 2n}$ = 1 + 3^{- 1} = 1 + $\frac{1}{3}$ = $\frac{4}{3}$

Example:

Given $\frac{(3 n) 2 + 3 2n - 1}{9 n}$

• $\frac{(3 n) 2 + 3 2n - 1}{9 n}$

= $\frac{3 2n + 3 2n - 1}{(3 2) n}$

= $\frac{3 2n + 3 2n x 3 - 1}{3 2n}$

= $\frac{3 2n (1 + 3 - 1)}{3 2n}$

= 1 + 3^{- 1} = 1 + $\frac{1}{3}$ = $\frac{4}{3}$

Example:

Given $\frac{(3 n) 2 + 3 2n - 1}{9 n}$

• $\frac{(3 n) 2 + 3 2n - 1}{9 n}$

= $\frac{3 2n + 3 2n - 1}{(3 2) n}$

= $\frac{3 2n + 3 2n x 3 - 1}{3 2n}$

= $\frac{3 2n (1 + 3 - 1)}{3 2n}$

= 1 + 3^{- 1} = 1 + $\frac{1}{3}$ = $\frac{4}{3}$

Exponential Equations

Equations where x is part of the Exponent

To Solve an Equations where x is part of the Exponent:

• Write the Powers as Products of Prime Factors

• Aim to get ONE Power with the same Base on each side of the Equation

• Equate the Exponents

• Solve for x

Example

Given 2^x = 8 solve for x

• 2^x = 8

2^x = 2³

∴ x = 3

Given 5^{2x + 1} - 125^{2x - 3}= 0 solve for x

• 5^{2x + 1} - 125^{2x - 3}= 0

5^{2x + 1} = 125^{2x - 3}

5^{2x + 1} = (5³)^{2x - 3}

5^{2x + 1} = 5^{6x - 9}

2x + 1 = 6x - 9

2x - 6x = -9 - 1

-4x = -10

∴ x = $\frac{5}{2}$

Given 2^x = 5^x solve for x

• 2^x = 5^x

$\frac{2 x}{5 x}$ = $\frac{5 x}{5 x}$

$\frac{2 x}{5 x}$ = 1

( $\frac{2}{5}$ )^x = 1

( $\frac{2}{5}$ )^x = ( $\frac{2}{5}$ )⁰

∴ x = 0

Given 3^{x + 1} - 3^{x - 1} = 216 solve for x

• 3^{x + 1} - 3^{x - 1} = 216

3^x . 3 - 3^x . 3^{- 1} = 216

3^x ( 3 - $\frac{1}{3}$ ) = 216

3^x ( $\frac{8}{3}$ ) = 216

3^x ( $\frac{8}{3}$ ) x $\frac{3}{8}$ = 216 x $\frac{3}{8}$

3^x = 81

3^x = 3⁴

∴ x = 4

Given 3^2x - 12.3^x + 27 = 0 solve for x

• 3^2x - 12.3^x + 27 = 0

(3^x)² - 12.3^x + 27 = 0

(3^x - 9)(3^x - 3) = 0

3^x - 9 = 0 or 3^x - 3 = 0

3^x - 3² = 0 or 3^x - 3¹ = 0

3^x = 3² or 3^x = 3¹

∴ x = 2 or x = 1

Equations with Rational Exponents

To Solve an Equation with a Rational Exponent:

• You must do the same operation to both sides of the equation

• Get the Variable with the Fraction Exponent on one side by itself

• Get x by itself by changing the Fraction Exponent to an Exponent of 1

• Do this by choosing an exponent for both sides, so that x^m/n becomes x¹.

• For x^m/n, if m is EVEN, then we get a Quadratic Equation with TWO possible answers.

Example

Given x^1/2 = -3 solve for x

• x^1/2 = -3

(x^1/2)² = (-3)²

x¹ = 9

∴ x = 9

Given x^3/4 = 8 solve for x

• x^3/4 = 8

(x^3/4)^4/3 = 8^4/3

x¹ = 8^4/3

x = (2³)^4/3

x = 2⁴

∴ x = 16

Given x^2/3 = 4 solve for x

• x^2/3 = 4

(x^2/3)^3/2 = ± 4^3/2

x¹ = ± (2²)^3/2

x = ± 2³

∴ x = 8 or x = - 8

Exponential Equations with Surds

To Solve an Exponential Equation containing $\sqrt[n]{x}$ :

• Isolate $\sqrt[n]{x}$

• Raise Both Sides of the Equation to the Power of n

• When Solving Exponential Equations with Surds:
YOU MUST ALWAYS CHECK IF YOUR VALUES OF x ARE A SOLUTION TO THE EQUATION.
Some might be solutions and some might not be solutions

Example

Given 3 $\sqrt[]{x + 2}$ + x = 2 solve for x

• 3 $\sqrt[]{x + 2}$ + x = 2

3 $\sqrt[]{x + 2}$ = 2 - x

( 3 $\sqrt[]{x + 2}$ )² = ( 2 - x )²

9( x + 2 ) = ( 2 - x )( 2 - x )

9x + 18 = 4 - 4x + x²

0 = x² - 4x + 4 - 9x - 18

0 = x² - 13x - 14

0 = ( x - 14 )( x + 1 )

∴ x = 14 or x = - 1

Check:

For x = 14

LHS = 3 $\sqrt[]{14 + 2}$ + 14 = 26 and RHS = 2

∴ x = 14 is not a solution.

For x = -1

LHS = 3 $\sqrt[]{-1 + 2}$ - 1 = 2 and RHS = 2

∴ x = -1 is a solution.

Exponents and Surds

The Number System

Natural Numbers ℕ

ℕ = { 1 , 2 , 3 , ... , ∞ }

Whole Numbers ℕ0

ℕ0 = { 0 , 1 , 2 , 3 , ... , ∞ }

Integers ℤ

ℤ = { -∞ , ... , -3 , -2 , -1 , 0 , 1 , 2 , 3 , ... , ∞ }

Rational Numbers ℚ

For Example { ℤ , -1/4 , 1/2 , 3/2 }

Irrational Numbers ℚ'

For Example the value of pi

Real Numbers ℝ

ℝ = { ℚ , ℚ' }

Non-Real Numbers

For Example any number divided by 0 or the square root of a negative number.

Exponents and Surds

Home

The Number System

QUIZ

Surds

Tutorial

Squares and Roots

For Example: If m = x 2, then m is the Square of x and therefore m is called a Perfect Square.

The Square Root of a number m is denoted by m

Therefore since m = x 2, then m = x

Examples:

• 25 = 5

The Cube Root of a number m is denoted by m 3

Therefore since m = x 3, then m 3 = x

Examples:

• 27 3 = 3

The n-th Root of a number m is denoted by m n

Therefore since m = x n, then m n = x

Examples:

• 16 4 = 2

If m n = x, then m = x 1/n

Surds

Examples of Surds:

• 2 = 1.414213562373095

• 5 = 2.236067977499789

• 8 = 2.828427124746190

Examples:

• 4 = 2

• 27 3 = 3

• 16 4 = 2

Simplifying n-th Roots

Multiplication of n-th Roots

a x a = a x a = a 2 = a

a x b = a x b = ab

a 3 x a 3 x a 3 = a x a x a 3 = a 3 3 = a

a 3 x b 3 = a x b 3 = ab 3

a n n = a

a n x b n = ab n

Examples:

Given 25 x 4

• 25 x 4 = 25 x 4 = 100 = 10 2 = 10

Given 5 x 3

• 5 x 3 = 5 x 3 = 15

Given 4 3 x 2 3

• 4 3 x 2 3 = 4 x 2 3 = 8 3 = 2 3 3 = 2

Division of n-th Roots

a b = a b , where b ≠ 0

a 3 b 3 = a b 3 , where b ≠ 0

a n b n = a b n , where b ≠ 0

Examples:

Given 100 4

• 100 4 = 100 4 = 25 = 5 2 = 5

Given 5 4

• 5 4 = 5 4 = 5 2

Given 27 8 3

• 27 8 3 = 27 3 8 3 = 3 3 3 2 3 3 = 3 2

Adding and Subtracting n-th Roots

a n + a n = 2 a n

2 a n + a n = 3 a n

If you have a n + b n , where a ≠ b this is what you should do:

For Example: Write a = c x b, where c is a Natural Number.

Examples:

Given 9 + 4

• 9 + 4 = 3 2 + 2 2 = 3 + 2 = 5

Whole Numbers ℕ₀

ℕ₀ = { 0 , 1 , 2 , 3 , ... , ∞ }

For Example: If m = x², then m is the Square of x and therefore m is called a Perfect Square.

The Square Root of a number m is denoted by $\sqrt[]{m}$

Therefore since m = x², then $\sqrt[]{m}$ = x

• $\sqrt[]{25}$ = 5

The Cube Root of a number m is denoted by $\sqrt[3]{m}$

Therefore since m = x³, then $\sqrt[3]{m}$ = x

• $\sqrt[3]{27}$ = 3

The n-th Root of a number m is denoted by $\sqrt[n]{m}$

Therefore since m = xⁿ, then $\sqrt[n]{m}$ = x

• $\sqrt[4]{16}$ = 2

If $\sqrt[n]{m}$ = x, then m = x ^1/n

• $\sqrt[]{2}$ = 1.414213562373095

• $\sqrt[]{5}$ = 2.236067977499789

• $\sqrt[]{8}$ = 2.828427124746190

• $\sqrt[]{4}$ = 2

• $\sqrt[3]{27}$ = 3

• $\sqrt[4]{16}$ = 2

$\sqrt[]{a}$ x $\sqrt[]{a}$ = $\sqrt[]{a x a}$ = $\sqrt[]{a 2}$ = a

$\sqrt[]{a}$ x $\sqrt[]{b}$ = $\sqrt[]{a x b}$ = $\sqrt[]{ab}$

$\sqrt[3]{a}$ x $\sqrt[3]{a}$ x $\sqrt[3]{a}$ = $\sqrt[3]{a x a x a}$ = $\sqrt[3]{a 3}$ = a

$\sqrt[3]{a}$ x $\sqrt[3]{b}$ = $\sqrt[3]{a x b}$ = $\sqrt[3]{ab}$

$\sqrt[n]{a n}$ = a

$\sqrt[n]{a}$ x $\sqrt[n]{b}$ = $\sqrt[n]{ab}$

Given $\sqrt[]{25}$ x $\sqrt[]{4}$

• $\sqrt[]{25}$ x $\sqrt[]{4}$ = $\sqrt[]{25 x 4}$

= $\sqrt[]{100}$ = $\sqrt[]{102}$ = 10

Given $\sqrt[]{5}$ x $\sqrt[]{3}$

• $\sqrt[]{5}$ x $\sqrt[]{3}$ = $\sqrt[]{5 x 3}$ = $\sqrt[]{15}$

Given $\sqrt[3]{4}$ x $\sqrt[3]{2}$

• $\sqrt[3]{4}$ x $\sqrt[3]{2}$ = $\sqrt[3]{4 x 2}$

= $\sqrt[3]{8}$ = $\sqrt[3]{23}$ = 2

$\frac{\sqrt[]{a}}{\sqrt[]{b}}$ = $\sqrt[]{\frac{a}{b}}$ , where b ≠ 0

$\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$ = $\sqrt[3]{\frac{a}{b}}$ , where b ≠ 0

$\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$ = $\sqrt[n]{\frac{a}{b}}$ , where b ≠ 0

Given $\frac{\sqrt[]{100}}{\sqrt[]{4}}$

• $\frac{\sqrt[]{100}}{\sqrt[]{4}}$ = $\sqrt[]{\frac{100}{4}}$ = $\sqrt[]{25}$ = $\sqrt[]{52}$ = 5

Given $\sqrt[]{\frac{5}{4}}$

• $\sqrt[]{\frac{5}{4}}$ = $\frac{\sqrt[]{5}}{\sqrt[]{4}}$ = $\frac{\sqrt[]{5}}{2}$

Given $\sqrt[3]{\frac{27}{8}}$

• $\sqrt[3]{\frac{27}{8}}$ = $\frac{\sqrt[3]{27}}{\sqrt[3]{8}}$ = $\frac{\sqrt[3]{33}}{\sqrt[3]{23}}$ = $\frac{3}{2}$

$\sqrt[n]{a}$ + $\sqrt[n]{a}$ = 2 $\sqrt[n]{a}$

2 $\sqrt[n]{a}$ + $\sqrt[n]{a}$ = 3 $\sqrt[n]{a}$

If you have $\sqrt[n]{a}$ + $\sqrt[n]{b}$ , where a ≠ b this is what you should do:

Given $\sqrt[]{9}$ + $\sqrt[]{4}$

• $\sqrt[]{9}$ + $\sqrt[]{4}$ = $\sqrt[]{32}$ + $\sqrt[]{22}$ = 3 + 2 = 5

Given $\sqrt[]{20}$ - $\sqrt[]{5}$

• $\sqrt[]{20}$ - $\sqrt[]{5}$ = $\sqrt[]{5 x 4}$ - $\sqrt[]{5}$

= $\sqrt[]{5}$ x $\sqrt[]{4}$ - $\sqrt[]{5}$

= 2 $\sqrt[]{5}$ - $\sqrt[]{5}$ = $\sqrt[]{5}$

Given $\sqrt[3]{8}$ + $\sqrt[3]{8}$

• $\sqrt[3]{8}$ + $\sqrt[3]{8}$ = 2 $\sqrt[3]{8}$ = 2 $\sqrt[3]{23}$

= 2 x 2 = 4

Remember $\sqrt[]{a}$ x $\sqrt[]{a}$ = a , so If you multiply the Numerator and the Denominator by the same Surd, you are not changing the value of the number since you are multiplying by $\frac{\sqrt[]{a}}{\sqrt[]{a}}$ which is equal to 1.

Given $\frac{\sqrt[]{3}}{\sqrt[]{2}}$

• $\frac{\sqrt[]{3}}{\sqrt[]{2}}$ = $\frac{\sqrt[]{3}}{\sqrt[]{2}}$ x $\frac{\sqrt[]{2}}{\sqrt[]{2}}$

= $\frac{\sqrt[]{3} x \sqrt[]{2}}{\sqrt[]{2} x \sqrt[]{2}}$ = $\frac{\sqrt[]{6}}{2}$

Given $\frac{2}{\sqrt[]{5}}$

• $\frac{2}{\sqrt[]{5}}$ = $\frac{2}{\sqrt[]{5}}$ x $\frac{\sqrt[]{5}}{\sqrt[]{5}}$

= $\frac{2
x \sqrt[]{5}}{\sqrt[]{5} x \sqrt[]{5}}$ = $\frac{2 \sqrt[]{5}}{5}$

aⁿ

b^m x bⁿ = b^{m + n}

$\frac{b m}{b n}$ = b^{m - n}

(b^m)ⁿ = b^{m x n}

(ab)ⁿ = aⁿbⁿ

( $\frac{a}{b}$ )ⁿ = $\frac{a n}{b n}$

• (2³)² = 2^{3 x 2} = 2⁶

• (5³ x 3³)² = (5³)² x (3³)² = 5⁶ x 3⁶

• ( $\frac{4}{3}$ )² = $\frac{42}{32}$

b⁰ = 1

• 5⁰ = 1

$\frac{1}{b n}$ = b^-n

( $\frac{a}{b}$ )^{- n} = ( $\frac{b}{a}$ )ⁿ

• $\frac{1}{23}$ = 2^-3

• ( $\frac{4}{3}$ )^{- 2} = ( $\frac{3}{4}$ )²

$\sqrt[n]{b m}$ = ( b^m )^1/n = b^m/n (n ≥ 2).

• $\sqrt[3]{34}$ = ( 3⁴ )^1/3 = 3^4/3

Given -3a³b² x -4a⁴b⁴

• -3a³b² x -4a⁴b⁴ = (-3 x -4)(a³ x a⁴)(b² x b⁴) = 12(a^{3 + 4})(b^{2 + 4}) = 12a⁷b⁶

Given $\frac{12x 5 y 8}{-4x 2 y 4}$ =

• $\frac{12x 5 y 8}{-4x 2 y 4}$ = ( $\frac{12}{-4}$ ) ( $\frac{x 5}{x 2}$ ) ( $\frac{y 8}{y 4}$ )

= -3x³y⁴

Given (-3a³b²)³

• (-3a³b²)³ = (-3)³(a³)³(b²)³

= (-3³)(a^{3 x 3})(b^{2 x 3})

= -27a⁹b⁶

• 4 = 2 x 2 = 2²

• 8 = 2 x 2 x 2 = 2³

• 9 = 3 x 3 = 3²