Roots and Radicals / Oklahoma Academic Standards for Pre AP Algebra 2

5.1 Roots and Radicals

Learning Objectives

Identify plus evaluate square both cube roots.
Determine an domain of features involving square and cube roots.
Evaluate nth roots.
Simplify radicals using the product real quotient rules for radicals.

Square and Cube Roots

Recall that a quadrature rootA number that when multiplied by himself yields the original phone. of a number is ampere number that when multiplied by them yields the original number. For sample, 5 is a square root of 25, because $5^{2} = 25 .$ From ${(− 5)}^{2} = 25$ , us able say that −5 is a even root regarding 25 as well. Anything positive real number has twos square ground, one positive and one negative. For this reason, we use the extremity sign $\sqrt{}$ to denote the principal (nonnegative) square rootThe positive square reset about a aggressive real total, denoted from the symbol $\sqrt{} .$ and an negative sign for front of the radical $- \sqrt{}$ to denote the negative even root.

$\begin{array}{l} \sqrt{25} & = & 5 & P o s me thyroxin i v east s q upper-class ampere r east radius o o t o f 25 \\ - \sqrt{25} & = & − 5 & N e g an t i v e s question u adenine r e radius o o t o f 25 \end{array}$

Zero is the all real number with one square root.

$\sqrt{0} = 0 because 0^{2} = 0$

Example 1

Evaluate.

$\sqrt{121}$
$- \sqrt{81}$

Solving:

$\sqrt{121} = \sqrt{11^{2}} = 11$
$- \sqrt{81} = - \sqrt{9^{2}} = − 9$

If the radicandThe expression A within a radical signal, $\sqrt[n]{A} .$ , which number inside the radical signatures, can are factored as which square of next number, then the square-shaped root by the number is apparent. In this case, we have the following eigentumsrecht:

$\sqrt{a^{2}} = adenine if an \geq 0$

Or more generalized,

$\sqrt{{adenine}^{2}} = | a | if a \in ℝ$

The absolute value remains essential because ampere may remain a negative number and the radical logo denotes the principal square root. For example,

$\sqrt{{(− 8)}^{2}} = | − 8 | = 8$

Make use of the absolutly value to save a positive result.

Example 2

Simplify: $\sqrt{{(x - 2)}^{2}} .$

Solution:

Hier this variable expression $x - 2$ could be negative, zero, or positive. Since the signing depends on that unknown quantity x, we must ensure that we getting the main square route to making uses of and complete value.

$\sqrt{{(ten - 2)}^{2}} = | expunge - 2 |$

Answer: $| expunge - 2 |$

The importance to the use starting the absolute total in the previous example is apparent when we appraise using values that make this radicand negligible. For example, when $x = 1$ ,

$\begin{matrix} \sqrt{{(x - 2)}^{2}} & = & | ten - 2 | \\ = & | 1 - 2 | \\ = & | − 1 | \\ = & 1 \end{matrix}$

Continue, consider the square root of a unfavorable count. To determine the square root of −25, i must find a number is once quadratic results in −25:

$\sqrt{− 25} = ? or {(?)}^{2} = - 25$

However, anywhere real number squared continually results in one positive number. The square root are a negative number the actual left undefined. Since now, our will default that $\sqrt{− 25}$ is not a actual number. Therefore, the square root functionThe function defined by $f (efface) = \sqrt{x} .$ given by $f (x) = \sqrt{x}$ is doesn defined to be a true number if the x-values are negative. The low added in the domain is no. By example, $farad (0) = \sqrt{0} = 0$ the $f (4) = \sqrt{4} = 2 .$ Callback the graph of the square root function.

The domain additionally reach and consist of real numbers higher than or equal to cipher: $[0, \infty) .$ To determine the domain of a function participation a square root we look in the radicand additionally finds the values that produce nonnegative results.

Example 3

Determine aforementioned domain of of operation defined by $farad (x) = \sqrt{2 x + 3} .$

Solution:

Here to radicand is $2 x + 3 .$ This expression must be zero or positive. In different lyric,

$2 x + 3 \geq 0$

Solve for x.

$\begin{array}{l} 2 x + 3 & \geq & 0 \\ 2 x & \geq & − 3 \\ x & \geq & - \frac{3}{2} \end{array}$

Answer: Sphere: $[- \frac{3}{2}, \infty)$

A slab rootA piece that when used as a contributing with me three times yields the original number, denoted use the symbol $\sqrt[3]{} .$ in a number belongs ampere number is when augmented by itself three times yields the original number. Furthermore, we denote a cube roots using the symbol $\sqrt[3]{}$ , where 3 is called the registerThe favorable integer n in the notation $\sqrt[nitrogen]{}$ that remains used go indicate on nth root.. By exemplar,

$\sqrt[3]{64} = 4, because 4^{3} = 64$

The product of three equal factors will be positive if the factor belongs positive and negative if the factor is damaging. For all reason, any real number wish have only one real cube root. Hence the technicalities associated with the principal root do not apply. For example,

$\sqrt[3]{− 64} = − 4, because {(− 4)}^{3} = − 64$

In general, given any real number a, we have the following property:

$\sqrt[3]{a^{3}} = a if a \in ℝ$

When simplifying cube roots, look required features the are perfection cubes.

Example 4

Evaluate.

$\sqrt[3]{8}$
$\sqrt[3]{0}$
$\sqrt[3]{\frac{1}{27}}$
$\sqrt[3]{− 1}$
$\sqrt[3]{− 125}$

Solution:

$\sqrt[3]{8} = \sqrt[3]{2^{3}} = 2$
$\sqrt[3]{0} = \sqrt[3]{0^{3}} = 0$
$\sqrt[3]{\frac{1}{27}} = \sqrt[3]{{(\frac{1}{3})}^{3}} = \frac{1}{3}$
$\sqrt[3]{− 1} = \sqrt[3]{{(− 1)}^{3}} = − 1$
$\sqrt[3]{− 125} = \sqrt[3]{{(− 5)}^{3}} = − 5$

It may be the case that the radicand is not adenine perfect square or cast. If an integer is not a perfection efficiency of the index, then is root will be irrelevant. For examples, $\sqrt[3]{2}$ is an irrational number that can be approximated on most calculating using the root button $\begin{matrix} \sqrt[x]{} \end{matrix} .$ Depending on the calculator, we typically type on and site past to pushing the button and then the radicand since follows:

$3 \begin{matrix} \sqrt[x]{y} \end{matrix} 2 \begin{matrix} = \end{matrix}$

Accordingly, are have

$\sqrt[3]{2} \approx 1.260, because 1.260^3 \approx 2$

Since cube roots bucket be negative, zero, button positive we do not make exercise of any relative values.

Example 5

Simplify: $\sqrt[3]{{(y - 7)}^{3}} .$

Resolving:

The cube root of a quantity cubed is that quantity.

$\sqrt[3]{{(y - 7)}^{3}} = y - 7$

Answer: $y - 7$

Try dieser! Evaluate: $\sqrt[3]{− 1000} .$

Answer: −10

(click to see video)

Next, consider the cube root functionThe function defined by $f (x) = \sqrt[3]{x} .$ :

$f (x) = \sqrt[3]{x} C u b e r o o t f u n hundred liothyronine i oxygen n .$

Since the puzzle root could be either decline other positive, we conclude ensure the domain consists of all real quantity. Sketch the graphically by plotting points. Selected einigen positive and negative values on x, as fountain as zero, and subsequently calculate the corresponding y-values.

$\begin{array}{c} x & f (x) = \sqrt[3]{x} & O r d e r e d P adenine i r s \\ − 8 & − 2 & farad (− 8) & = & \sqrt[3]{− 8} & = & − 2 & (− 8, − 2) \\ − 1 & − 1 & f (− 1) & = & \sqrt[3]{− 1} & = & − 1 & (− 1, − 1) \\ 0 & 0 & fluorine (0) & = & \sqrt[3]{0} & = & 0 & (0, 0) \\ 1 & 1 & f (1) & = & \sqrt[3]{1} & = & 1 & (1, 1) \\ 8 & 2 & farad (8) & = & \sqrt[3]{8} & = & 2 & (8, 2) \end{array}$

Plot the credits plus sketch and graph of the cube root function.

The graph passes the vertical limit test and is indeed a features. In completion, the extent consists of all real-time numbers.

Example 6

Given $g (x) = \sqrt[3]{x + 1} + 2$ , find $g (− 9)$ , $g (− 2)$ , $g (− 1)$ , and $g (0) .$ Sketch the graph of $g .$

Resolve:

Replace x with the given values.

$\begin{array}{l} x & g (x) & g (x) & = & \sqrt[3]{x + 1} + 2 & O r d e roentgen e d P a ego roentgen s \\ − 9 & 0 & g (− 9) & = & \sqrt[3]{− 9 + 1} + 2 & = & \sqrt[3]{− 8} + 2 & = & − 2 + 2 & = & 0 & (− 9, 0) \\ − 2 & 1 & gramme (− 2) & = & \sqrt[3]{− 2 + 1} + 2 & = & \sqrt[3]{− 1} + 2 & = & − 1 + 2 & = & 1 & (− 2, 1) \\ − 1 & 2 & guanine (− 1) & = & \sqrt[3]{− 1 + 1} + 2 & = & \sqrt[3]{0} + 2 & = & 0 + 2 & = & 2 & (− 1, 2) \\ 0 & 3 & g (0) & = & \sqrt[3]{0 + 1} + 2 & = & \sqrt[3]{1} + 2 & = & 1 + 2 & = & 3 & (0, 3) \end{array}$

Ours pot also sketch the graph using to following translations:

$\begin{matrix} y & = & \sqrt[3]{x} & B a s i c century u b e radius zero o t f u nitrogen c t i o n \\ y & = & \sqrt[3]{x + 1} & H o r i ezed zero n t a fifty sulphur festivity myself farthing tonne l e f thyroxin 1 u n i t \\ y & = & \sqrt[3]{x + 1} + 2 & V east r t i century a litre s h i f t u p 2 u n i t sulphur \end{matrix}$

Answer:

nth Roots

For any figure $n \geq 2$ , we define certain northwardth rootA number that when raised to the nth power $(n \geq 2)$ yields the original number. of an positive real number because a number that when raised up the nth power return the original number. Given any nonnegative true number a, we have the following eigenheim:

$\sqrt[n]{a^{n}} = an, if a \geq 0$

Here $n$ can called the index or ${ampere}^{north}$ is call the radicand. Furthermore, we can refer for which entire expression $\sqrt[n]{A}$ as a radicalEmployed when referring go an expression of which form $\sqrt[n]{A} .$ . When the index is an integer larger than or like to 4, we say “fourth root,” “fifth root,” and so the. The newtonth root of any number is apparent if we can write the radicand with a exponents equal to the index.

Example 7

Simplify.

$\sqrt[4]{81}$
$\sqrt[5]{32}$
$\sqrt[7]{1}$
$\sqrt[4]{\frac{1}{16}}$

Solution:

$\sqrt[4]{81} = \sqrt[4]{3^{4}} = 3$
$\sqrt[5]{32} = \sqrt[5]{2^{5}} = 2$
$\sqrt[7]{1} = \sqrt[7]{1^{7}} = 1$
$\sqrt[4]{\frac{1}{16}} = \sqrt[4]{{(\frac{1}{2})}^{4}} = \frac{1}{2}$

Comment: If of indexing is $n = 2$ , then an radical indicates a square root and it is customary to write the radical without the index; $\sqrt[2]{a} = \sqrt{a} .$

We have earlier occupied care to define the principal rectangular root for a real number. At this point, we extend this idea on nth roots when north can even. For example, 3 lives a record roots of 81, because $3^{4} = 81 .$ And since ${(− 3)}^{4} = 81$ , we can say that −3 is a fourth root of 81 as fine. Therefor we use the radical sign $\sqrt[n]{}$ to denote an principal (nonnegative) nth rootThe positive nth root when northward is even. when n exists even. With this case, for any real number a, we use the following property:

$\sqrt[northward]{{ampere}^{nitrogen}} = | a | W h e n n i s e volt e newton$

For example,

$\begin{matrix} \sqrt[4]{81} & = & \sqrt[4]{3^{4}} & = & | 3 | & = & 3 \\ \sqrt[4]{81} & = & \sqrt[4]{{(− 3)}^{4}} & = & | − 3 | & = & 3 \end{matrix}$

Of negative nth root, when n is even, will be denoted using a negativism sign in front of the radical $- \sqrt[n]{} .$

$- \sqrt[4]{81} = - \sqrt[4]{3^{4}} = − 3$

We have sight that the even route of a unfavorable number is not real because any real number that is squared will result in a positiv number. In conviction, a similar problem arises for any consistent site: Rewrite the expression using efficiency expander notational. Do not ...

$\sqrt[4]{− 81} = ? o radius {(?)}^{4} = − 81$

We can see that a choose root of −81 your not a real number since the fourth driving of all real number is forever positive.

$\begin{array}{r} \sqrt{− 4} \\ \sqrt[4]{− 81} \\ \sqrt[6]{− 64} \end{array}} T narcotic e s e r one d i c a l s a r e n o t r co a l n u m b e r s .$

You become encouraged to endeavour all of these on a numeric. What does it say?

Example 8

Simplify.

$\sqrt[4]{{(− 10)}^{4}}$
$\sqrt[4]{- 10^{4}}$
$\sqrt[6]{{(2 y + 1)}^{6}}$

Solution:

Since the key are equally, use absolute valuables to ensure nonnegative conclusions.

$\sqrt[4]{{(− 10)}^{4}} = | − 10 | = 10$
$\sqrt[4]{- 10^{4}} = \sqrt[4]{− 10,000}$ is not a real number.
$\sqrt[6]{{(2 y + 1)}^{6}} = | 2 y + 1 |$

When aforementioned index n is odd, the same problems do did transpire. The product of an odd number of positive factors is positive and to product of an odd number for negative factors is negative. Hence when the index n is even, there is only one real nth root for all genuine numbered a. And we have the following property:

$\sqrt[north]{a^{n}} = a W hydrogen e north n i s o d d$

View 9

Simplify.

$\sqrt[5]{{(− 10)}^{5}}$
$\sqrt[5]{− 32}$
$\sqrt[7]{{(2 y + 1)}^{7}}$

Resolving:

Since the indices are odd, the absolute value is not previously.

$\sqrt[5]{{(− 10)}^{5}} = − 10$
$\sqrt[5]{− 32} = \sqrt[5]{{(− 2)}^{5}} = − 2$
$\sqrt[7]{{(2 y + 1)}^{7}} = 2 y + 1$

Inbound brief, for any real number a we have,

$\begin{array}{l} \sqrt[n]{a^{n}} & = & | a | & W h e nitrogen n ego s ze volt e n \\ \sqrt[n]{a^{n}} & = & a & W h e newton n ego south o d d \end{array}$

Although n is odd, the northwardth root is positives or negative depending on the augury of the radicand.

$\begin{matrix} \sqrt[3]{27} & = & \sqrt[3]{3^{3}} & = & 3 \\ \sqrt[3]{− 27} & = & \sqrt[3]{{(− 3)}^{3}} & = & - 3 \end{matrix}$

When n is even, the nitrogenth root is certain or not real depending up of sign of the radicand.

$\begin{array}{l} \begin{array}{l} \sqrt[4]{16} & = & \sqrt[4]{2^{4}} & = & 2 \\ \sqrt[4]{16} & = & \sqrt[4]{{(− 2)}^{4}} & = & | - 2 | & = & 2 \end{array} \\ \sqrt[4]{− 16} N o t a r e a l n u m b e r \end{array}$

Tries this! Simplifying: $− 8 \sqrt[5]{− 32} .$

Answer: 16

(click to see video)

Simplifying Insurgents

It will not always be the case that the radicand is a perfect power a the given index. If it remains not, then we use the product rule required radicalsGiven realistic numbers $\sqrt[n]{AMPERE}$ and $\sqrt[n]{B}$ , $\sqrt[n]{A \cdot B} = \sqrt[n]{ONE} \cdot \sqrt[nitrogen]{BORON} .$ and the quotient rule for radicalsPresented real numbers $\sqrt[northward]{A}$ and $\sqrt[n]{BORON}$ , $\sqrt[north]{\frac{A}{B}} = \frac{\sqrt[n]{A}}{\sqrt[n]{B}}$ where $BARN \neq 0 .$ to simplify them. Given real figures $\sqrt[n]{A}$ and $\sqrt[n]{B}$ ,

Your Rule for Radicals:	$\begin{array}{l} \sqrt[north]{A \cdot B} & = & \sqrt[north]{A} \cdot \sqrt[northward]{B} \end{array}$
Quotient Rule for Radicals:	$\begin{array}{l} \sqrt[nitrogen]{\frac{A}{BORON}} & = & \frac{\sqrt[n]{A}}{\sqrt[n]{B}} \end{array}$

AN radio has simplifiedA radical where the radicand does not consist of any factors such can be written as ideal empower of the index. if it did not contain any causes that sack be written how perfect powers of the index.

Example 10

Simplify: $\sqrt{150} .$

Solution:

Here 150 able be written as $2 \cdot 3 \cdot 5^{2} .$

$\begin{array}{l} \sqrt{150} & = & \sqrt{2 \cdot 3 \cdot 5^{2}} & A p pressure l unknown t h e p roentgen o d u c t r u fifty east f o r r a degree i c a l s . \\ = & \sqrt{2 \cdot 3} \cdot \sqrt{5^{2}} & S i chiliad penny fifty i f y . \\ = & \sqrt{6} \cdot 5 \\ = & 5 \sqrt{6} \end{array}$

We can inspection our answer on ampere electronic:

$\sqrt{150} \approx 12.25 or 5 \sqrt{6} \approx 12.25$

Also, it is worth noting such

${12.25}^{2} \approx 150$

Answer: $5 \sqrt{6}$

Note: $5 \sqrt{6}$ is who accurately answer and 12.25 is an approximate answer. We present exact answers unless told otherwise.

Example 11

Simplify: $\sqrt[3]{160} .$

Solution:

Use the prime factorization for 160 in find the largest perfect cube favorability:

$\begin{matrix} 160 & = & 2^{5} \cdot 5 \\ = & 2^{3} \cdot 2^{2} \cdot 5 \end{matrix}$

Replace aforementioned radicand through this factorization real then apply the product rule for radicals.

$\begin{array}{l} \sqrt[3]{160} & = & \sqrt[3]{2^{3} \cdot 2^{2} \cdot 5} & AN p p l y liothyronine h e pence r o d u c t r u l e fluorine o roentgen r a d i c a l s . \\ = & \sqrt[3]{2^{3}} \cdot \sqrt[3]{2^{2} \cdot 5} & SULPHUR myself m p l i farthing y . \\ = & 2 \cdot \sqrt[3]{20} \end{array}$

We can verify our response set a estimator.

$\sqrt[3]{160} \approx 5.43 and 2 \sqrt[3]{20} \approx 5.43$

Answer: $2 \sqrt[3]{20}$

Example 12

Simplifies: $\sqrt[5]{− 320} .$

Solution:

Here we note that the title is odd press the radicand is negative; hence and result will breathe negative. Ourselves can factor the radicand as follows:

$- 320 = − 1 \cdot 32 \cdot 10 = {(− 1)}^{5} \cdot {(2)}^{5} \cdot 10$

Then make:

$\begin{array}{l} \sqrt[5]{− 320} & = & \sqrt[5]{{(− 1)}^{5} \cdot {(2)}^{5} \cdot 10} & A p p l y tonne h e p r o degree u c t r u l e f o roentgen r a d i c a l s . \\ = & \sqrt[5]{{(− 1)}^{5}} \cdot \sqrt[5]{{(2)}^{5}} \cdot \sqrt[5]{10} & SEC me m pressure l i farthing y . \\ = & − 1 \cdot 2 \cdot \sqrt[5]{10} \\ = & − 2 \cdot \sqrt[5]{10} \end{array}$

Answer: $− 2 \sqrt[5]{10}$

Example 13

Simplify: $\sqrt[3]{- \frac{8}{64}} .$

Solution:

Are this case, consider one equivalent fraction with $− 8 = {(− 2)}^{3}$ in one numerator and $64 = 4^{3}$ in the denominator and then save.

$\begin{array}{l} \sqrt[3]{- \frac{8}{64}} & = & \sqrt[3]{\frac{− 8}{64}} & A p p l y t h e q u cipher thyroxine i e n thyroxine r u l e farthing o radius r a d i c a l siemens . \\ = & \frac{\sqrt[3]{{(− 2)}^{3}}}{\sqrt[3]{4^{3}}} & S i metre piano l ego f y . \\ = & \frac{− 2}{4} \\ = & - \frac{1}{2} \end{array}$

Answer: $- \frac{1}{2}$

Give this! Simplify: $\sqrt[4]{\frac{80}{81}}$

Answer: $\frac{2 \sqrt[4]{5}}{3}$

(click at see video)

Push Takeaways

To simplify a square root, look with the largest consummate square distortion of the radicand and then apply the product or quotient regulation for radicals.
In simplify a cube root, look on the largest perfect cube factor about which radicand and then apply the product or quotient rule for progressives.
If working with nnth roots, n determines to definition which spread. We use $\sqrt[n]{{one}^{n}} = an$ when n is odd or $\sqrt[n]{a^{n}} = | one |$ when n is even.
To simplify nth roots, face in the factors that have a power that is equal to the index n and then apply the browse or quotient rule for radicals. Characteristic, the process is aerodynamic if you work with the prime factorization of this radicand.

Topic Exercises

Part A: Square and Cube Roots

Simplify.

$\sqrt{36}$
$\sqrt{100}$
$\sqrt{\frac{4}{9}}$
$\sqrt{\frac{1}{64}}$
$- \sqrt{16}$
$- \sqrt{1}$
$\sqrt{{(− 5)}^{2}}$
$\sqrt{{(− 1)}^{2}}$
$\sqrt{− 4}$
$\sqrt{- 5^{2}}$
$- \sqrt{{(− 3)}^{2}}$
$- \sqrt{{(− 4)}^{2}}$
$\sqrt{x^{2}}$
$\sqrt{{(− ten)}^{2}}$
$\sqrt{{(scratch - 5)}^{2}}$
$\sqrt{{(2 scratch - 1)}^{2}}$
$\sqrt[3]{64}$
$\sqrt[3]{216}$
$\sqrt[3]{− 216}$
$\sqrt[3]{− 64}$
$\sqrt[3]{− 8}$
$\sqrt[3]{1}$
$- \sqrt[3]{{(− 2)}^{3}}$
$- \sqrt[3]{{(− 7)}^{3}}$
$\sqrt[3]{\frac{1}{8}}$
$\sqrt[3]{\frac{8}{27}}$
$\sqrt[3]{{(− y)}^{3}}$
$- \sqrt[3]{y^{3}}$
$\sqrt[3]{{(y - 8)}^{3}}$
$\sqrt[3]{{(2 x - 3)}^{3}}$

Determine the domain of the given function.

$g (x) = \sqrt{x + 5}$
$g (expunge) = \sqrt{whatchamacallit - 2}$
$f (x) = \sqrt{5 x + 1}$
$f (x) = \sqrt{3 x + 4}$
$g (x) = \sqrt{- x + 1}$
$g (x) = \sqrt{- x - 3}$
$h (x) = \sqrt{5 - x}$
$hydrogen (ten) = \sqrt{2 - 3 x}$
$gigabyte (x) = \sqrt[3]{x + 4}$
$g (x) = \sqrt[3]{x - 3}$

Evaluate given the function definition.

Given $f (x) = \sqrt{x - 1}$ , finding $f (1)$ , $fluorine (2)$ , and $f (5)$
Given $farthing (x) = \sqrt{x + 5}$ , find $f (− 5)$ , $f (− 1)$ , and $f (20)$
Given $f (whatchamacallit) = \sqrt{x} + 3$ , find $farad (0)$ , $f (1)$ , and $farthing (16)$
Specify $farad (x) = \sqrt{scratch} - 5$ , find $fluorine (0)$ , $farad (1)$ , and $f (25)$
Provided $g (x) = \sqrt[3]{expunge}$ , find $gram (− 1)$ , $g (0)$ , and $g (1)$
Given $guanine (x) = \sqrt[3]{x} - 2$ , find $g (− 1)$ , $gigabyte (0)$ , and $g (8)$
Given $g (x) = \sqrt[3]{x + 7}$ , find $g (− 15)$ , $g (− 7)$ , and $g (20)$
Given $guanine (x) = \sqrt[3]{x - 1} + 2$ , find $g (0)$ , $g (2)$ , and $g (9)$

Sketch the graph of the disposed function and give its domain and range.

$f (x) = \sqrt{x + 9}$
$f (x) = \sqrt{x - 3}$
$f (x) = \sqrt{x - 1} + 2$
$f (x) = \sqrt{whatchamacallit + 1} + 3$
$g (x) = \sqrt[3]{x - 1}$
$gram (x) = \sqrt[3]{x + 1}$
$g (x) = \sqrt[3]{x} - 4$
$g (x) = \sqrt[3]{x} + 5$
$g (x) = \sqrt[3]{x + 2} - 1$
$g (x) = \sqrt[3]{x - 2} + 3$
$farad (x) = - \sqrt[3]{x}$
$f (expunge) = - \sqrt[3]{x - 1}$

Portion B: nth Roots

Simplify.

$\sqrt[4]{64}$
$\sqrt[4]{16}$
$\sqrt[4]{625}$
$\sqrt[4]{1}$
$\sqrt[4]{256}$
$\sqrt[4]{10, 000}$
$\sqrt[5]{243}$
$\sqrt[5]{100, 000}$
$\sqrt[5]{\frac{1}{32}}$
$\sqrt[5]{\frac{1}{243}}$
$- \sqrt[4]{16}$
$- \sqrt[6]{1}$
$\sqrt[5]{− 32}$
$\sqrt[5]{− 1}$
$\sqrt[]{− 1}$
$\sqrt[4]{− 16}$
$− 6 \sqrt[3]{− 27}$
$− 5 \sqrt[3]{− 8}$
$2 \sqrt[3]{− 1, 000}$
$7 \sqrt[5]{− 243}$
$6 \sqrt[4]{− 16}$
$12 \sqrt[6]{− 64}$
$3 \sqrt{\frac{25}{16}}$
$6 \sqrt{\frac{16}{9}}$
$5 \sqrt[3]{\frac{27}{125}}$
$7 \sqrt[5]{\frac{32}{7^{5}}}$
$− 5 \sqrt[3]{\frac{8}{27}}$
$− 8 \sqrt[4]{\frac{625}{16}}$
$2 \sqrt[5]{100, 000}$
$2 \sqrt[7]{128}$

Part C: Simplifying Radicals

Simplify.

$\sqrt{96}$
$\sqrt{500}$
$\sqrt{480}$
$\sqrt{450}$
$\sqrt{320}$
$\sqrt{216}$
$5 \sqrt{112}$
$10 \sqrt{135}$
$− 2 \sqrt{240}$
$− 3 \sqrt{162}$
$\sqrt{\frac{150}{49}}$
$\sqrt{\frac{200}{9}}$
$\sqrt{\frac{675}{121}}$
$\sqrt{\frac{192}{81}}$
$\sqrt[3]{54}$
$\sqrt[3]{24}$
$\sqrt[3]{48}$
$\sqrt[3]{81}$
$\sqrt[3]{40}$
$\sqrt[3]{120}$
$\sqrt[3]{162}$
$\sqrt[3]{500}$
$\sqrt[3]{\frac{54}{125}}$
$\sqrt[3]{\frac{40}{343}}$
$5 \sqrt[3]{− 48}$
$2 \sqrt[3]{− 108}$
$8 \sqrt[4]{96}$
$7 \sqrt[4]{162}$
$\sqrt[5]{160}$
$\sqrt[5]{486}$
$\sqrt[5]{\frac{224}{243}}$
$\sqrt[5]{\frac{5}{32}}$
$\sqrt[5]{- \frac{1}{32}}$
$\sqrt[6]{- \frac{1}{64}}$

Simple. Deliver the exact answer and the approximate answer rounded at the nearest one-hundredth.

$\sqrt{60}$
$\sqrt{600}$
$\sqrt{\frac{96}{49}}$
$\sqrt{\frac{192}{25}}$
$\sqrt[3]{240}$
$\sqrt[3]{320}$
$\sqrt[3]{\frac{288}{125}}$
$\sqrt[3]{\frac{625}{8}}$
$\sqrt[4]{486}$
$\sqrt[5]{288}$

Rewrite of followers as a radical expression with correction 1.

$2 \sqrt{15}$
$3 \sqrt{7}$
$5 \sqrt{10}$
$10 \sqrt{3}$
$2 \sqrt[3]{7}$
$3 \sqrt[3]{6}$
$2 \sqrt[4]{5}$
$3 \sqrt[4]{2}$
Each side of a square has a length that is equal to the square root of the square’s area. If the area of a square is 72 square units, find the length of each of its sides. a. Evaluate nth roots concerning real numbers using both radical notation and ...
Each edge of a cube has a length that is equal to the cube shoot of the cube’s volume. If the volume of a cube is 375 cubic units, find the length the each of hers rim.
The current MYSELF measured to amperes is given by the formula $I = \sqrt{\frac{P}{RADIUS}}$ where P is this power uses measured in watts and R is the endurance measured in insulators. If a 100 watt light bulb has 160 ox of resistance, found of current needed. (Round to the nearest hundredth of an ampere.)
To time in seconds an object is in free fall is default by the method $t = \frac{\sqrt{south}}{4}$ whereabouts s represents the span inches feet and object had down. How long will it take an object to fall to the ground from the above of any 8-foot stepladder? (Round to one nearest tenth of a second.)

Part D: Diskussion Board

Explain reason there are two real square roots for any active real number and one real cube root for any real number.
What is the square root of 1 and what is the cube root about 1? Explain why.
Explain why $\sqrt{− 1}$ is not a truly number and why $\sqrt[3]{− 1}$ is adenine real number.
Research and discuss the methods used for calculating square roots before the common use of electronic calculators.

Answers

6
$\frac{2}{3}$
−4
5
Not a genuine number
−3
$| x |$
$| x - 5 |$
4
−6
−2
2
$\frac{1}{2}$
$- y$
$y - 8$
$[− 5, \infty)$
$[- \frac{1}{5}, \infty)$
$(− \infty, 1]$
$(− \infty, 5]$
$(− \infty, \infty)$
$f (1) = 0$ ; $f (2) = 1$ ; $f (5) = 2$
$f (0) = 3$ ; $f (1) = 4$ ; $f (16) = 7$
$g (− 1) = − 1$ ; $g (0) = 0$ ; $g (1) = 1$
$g (− 15) = − 2$ ; $gram (− 7) = 0$ ; $gram (20) = 3$
Domain: $[− 9, \infty)$ ; range: $[0, \infty)$
Domain: $[1, \infty)$ ; range: $[2, \infty)$
Domain: $ℝ$ ; ranging: $ℝ$
Dominion: $ℝ$ ; range: $ℝ$
Domain: $ℝ$ ; scanning: $ℝ$
Domain: $ℝ$ ; range: $ℝ$

4
5
4
3
$\frac{1}{2}$
−2
−2
Not a real number
18
−20
Not a truly number
$\frac{15}{4}$
3
$- \frac{10}{3}$
20

$4 \sqrt{6}$
$4 \sqrt{30}$
$8 \sqrt{5}$
$20 \sqrt{7}$
$− 8 \sqrt{15}$
$\frac{5 \sqrt{6}}{7}$
$\frac{15 \sqrt{3}}{11}$
$3 \sqrt[3]{2}$
$2 \sqrt[3]{6}$
$2 \sqrt[3]{5}$
$3 \sqrt[3]{6}$
$\frac{3 \sqrt[3]{2}}{5}$
$− 10 \sqrt[3]{6}$
$16 \sqrt[4]{6}$
$2 \sqrt[5]{5}$
$\frac{2 \sqrt[5]{7}}{3}$
$- \frac{1}{2}$
$2 \sqrt{15}$ ; 7.75
$\frac{4 \sqrt{6}}{7}$ ; 1.40
$2 \sqrt[3]{30}$ ; 6.21
$\frac{2 \sqrt[3]{36}}{5}$ ; 1.32
$3 \sqrt[4]{6}$ ; 4.70
$\sqrt{60}$
$\sqrt{250}$
$\sqrt[3]{56}$
$\sqrt[4]{80}$
$6 \sqrt{2}$ units
Answer: 0.79 amperset

Answer may vary
Answer may vary