1.3 Radicals and rational expressions  (Page 4/11)

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Understanding n Th roots

Suppose we know that $\text{\hspace{0.17em}}{a}^{3}=8.\text{\hspace{0.17em}}$ We want to find what number raised to the 3rd power is equal to 8. Since $\text{\hspace{0.17em}}{2}^{3}=8,$ we say that 2 is the cube root of 8.

The n th root of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is a number that, when raised to the n th power, gives $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ For example, $\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}$ is the 5th root of $\text{\hspace{0.17em}}-243\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}{\left(-3\right)}^{5}=-243.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is a real number with at least one n th root, then the principal n th root    of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is the number with the same sign as $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ that, when raised to the n th power, equals $\text{\hspace{0.17em}}a.$

The principal n th root of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is written as $\text{\hspace{0.17em}}\sqrt[n]{a},$ where $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ is a positive integer greater than or equal to 2. In the radical expression, $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ is called the index    of the radical.

Principal n Th root

If $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is a real number with at least one n th root, then the principal n th root    of $\text{\hspace{0.17em}}a,$ written as $\text{\hspace{0.17em}}\sqrt[n]{a},$ is the number with the same sign as $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ that, when raised to the n th power, equals $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ The index    of the radical is $\text{\hspace{0.17em}}n.$

Simplifying n Th roots

Simplify each of the following:

1. $\sqrt[5]{-32}$
2. $\sqrt[4]{4}\cdot \sqrt[4]{1,024}$
3. $-\sqrt[3]{\frac{8{x}^{6}}{125}}$
4. $8\sqrt[4]{3}-\sqrt[4]{48}$
1. $\sqrt[5]{-32}=-2\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}{\left(-2\right)}^{5}=-32$
2. First, express the product as a single radical expression. $\text{\hspace{0.17em}}\sqrt[4]{4,096}=8\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}{8}^{4}=4,096$

Simplify.

1. $\sqrt[3]{-216}$
2. $\frac{3\sqrt[4]{80}}{\sqrt[4]{5}}$
3. $6\sqrt[3]{9,000}+7\sqrt[3]{576}$
1. $-6$
2. $6$
3. $88\sqrt[3]{9}$

Using rational exponents

Radical expressions can also be written without using the radical symbol. We can use rational (fractional) exponents. The index must be a positive integer. If the index $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ is even, then $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ cannot be negative.

${a}^{\frac{1}{n}}=\sqrt[n]{a}$

We can also have rational exponents with numerators other than 1. In these cases, the exponent must be a fraction in lowest terms. We raise the base to a power and take an n th root. The numerator tells us the power and the denominator tells us the root.

${a}^{\frac{m}{n}}={\left(\sqrt[n]{a}\right)}^{m}=\sqrt[n]{{a}^{m}}$

All of the properties of exponents that we learned for integer exponents also hold for rational exponents.

Rational exponents

Rational exponents are another way to express principal n th roots. The general form for converting between a radical expression with a radical symbol and one with a rational exponent is

${a}^{\frac{m}{n}}={\left(\sqrt[n]{a}\right)}^{m}=\sqrt[n]{{a}^{m}}$

Given an expression with a rational exponent, write the expression as a radical.

1. Determine the power by looking at the numerator of the exponent.
2. Determine the root by looking at the denominator of the exponent.
3. Using the base as the radicand, raise the radicand to the power and use the root as the index.

Write $\text{\hspace{0.17em}}{343}^{\frac{2}{3}}\text{\hspace{0.17em}}$ as a radical. Simplify.

The 2 tells us the power and the 3 tells us the root.

${343}^{\frac{2}{3}}={\left(\sqrt[3]{343}\right)}^{2}=\sqrt[3]{{343}^{2}}$

We know that $\text{\hspace{0.17em}}\sqrt[3]{343}=7\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}{7}^{3}=343.\text{\hspace{0.17em}}$ Because the cube root is easy to find, it is easiest to find the cube root before squaring for this problem. In general, it is easier to find the root first and then raise it to a power.

${343}^{\frac{2}{3}}={\left(\sqrt[3]{343}\right)}^{2}={7}^{2}=49$

Write $\text{\hspace{0.17em}}{9}^{\frac{5}{2}}\text{\hspace{0.17em}}$ as a radical. Simplify.

${\left(\sqrt{9}\right)}^{5}={3}^{5}=243$

Write $\text{\hspace{0.17em}}\frac{4}{\sqrt[7]{{a}^{2}}}\text{\hspace{0.17em}}$ using a rational exponent.

The power is 2 and the root is 7, so the rational exponent will be $\text{\hspace{0.17em}}\frac{2}{7}.\text{\hspace{0.17em}}$ We get $\text{\hspace{0.17em}}\frac{4}{{a}^{\frac{2}{7}}}.\text{\hspace{0.17em}}$ Using properties of exponents, we get $\text{\hspace{0.17em}}\frac{4}{\sqrt[7]{{a}^{2}}}=4{a}^{\frac{-2}{7}}.$

Write $\text{\hspace{0.17em}}x\sqrt{{\left(5y\right)}^{9}}\text{\hspace{0.17em}}$ using a rational exponent.

$x{\left(5y\right)}^{\frac{9}{2}}$

An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
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12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×