# 1.3 Radicals and rational expressions  (Page 3/11)

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## Adding square roots

Add $\text{\hspace{0.17em}}5\sqrt{12}+2\sqrt{3}.$

We can rewrite $\text{\hspace{0.17em}}5\sqrt{12}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}5\sqrt{4·3}.\text{\hspace{0.17em}}$ According the product rule, this becomes $\text{\hspace{0.17em}}5\sqrt{4}\sqrt{3}.\text{\hspace{0.17em}}$ The square root of $\text{\hspace{0.17em}}\sqrt{4}\text{\hspace{0.17em}}$ is 2, so the expression becomes $\text{\hspace{0.17em}}5\left(2\right)\sqrt{3},$ which is $\text{\hspace{0.17em}}10\sqrt{3}.\text{\hspace{0.17em}}$ Now we can the terms have the same radicand so we can add.

$10\sqrt{3}+2\sqrt{3}=12\sqrt{3}$

Add $\text{\hspace{0.17em}}\sqrt{5}+6\sqrt{20}.$

$13\sqrt{5}$

## Subtracting square roots

Subtract $\text{\hspace{0.17em}}20\sqrt{72{a}^{3}{b}^{4}c}\text{\hspace{0.17em}}-14\sqrt{8{a}^{3}{b}^{4}c}.$

Rewrite each term so they have equal radicands.

$\begin{array}{ccc}\hfill 20\sqrt{72{a}^{3}{b}^{4}c}& =& 20\sqrt{9}\sqrt{4}\sqrt{2}\sqrt{a}\sqrt{{a}^{2}}\sqrt{{\left({b}^{2}\right)}^{2}}\sqrt{c}\hfill \\ & =& 20\left(3\right)\left(2\right)|a|{b}^{2}\sqrt{2ac}\hfill \\ & =& 120|a|{b}^{2}\sqrt{2ac}\hfill \end{array}$
$\begin{array}{ccc}\hfill 14\sqrt{8{a}^{3}{b}^{4}c}& =& 14\sqrt{2}\sqrt{4}\sqrt{a}\sqrt{{a}^{2}}\sqrt{{\left({b}^{2}\right)}^{2}}\sqrt{c}\hfill \\ & =& 14\left(2\right)|a|{b}^{2}\sqrt{2ac}\hfill \\ & =& 28|a|{b}^{2}\sqrt{2ac}\hfill \end{array}$

Now the terms have the same radicand so we can subtract.

$120|a|{b}^{2}\sqrt{2ac}-28|a|{b}^{2}\sqrt{2ac}=92|a|{b}^{2}\sqrt{2ac}$

Subtract $\text{\hspace{0.17em}}3\sqrt{80x}\text{\hspace{0.17em}}-4\sqrt{45x}.$

$0$

## Rationalizing denominators

When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator .

We know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of 1 that will eliminate the radical.

For a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is $\text{\hspace{0.17em}}b\sqrt{c},$ multiply by $\text{\hspace{0.17em}}\frac{\sqrt{c}}{\sqrt{c}}.$

For a denominator containing the sum or difference of a rational and an irrational term, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign of the radical portion of the denominator. If the denominator is $\text{\hspace{0.17em}}a+b\sqrt{c},$ then the conjugate is $\text{\hspace{0.17em}}a-b\sqrt{c}.$

Given an expression with a single square root radical term in the denominator, rationalize the denominator.

1. Multiply the numerator and denominator by the radical in the denominator.
2. Simplify.

## Rationalizing a denominator containing a single term

Write $\text{\hspace{0.17em}}\frac{2\sqrt{3}}{3\sqrt{10}}\text{\hspace{0.17em}}$ in simplest form.

The radical in the denominator is $\text{\hspace{0.17em}}\sqrt{10}.\text{\hspace{0.17em}}$ So multiply the fraction by $\text{\hspace{0.17em}}\frac{\sqrt{10}}{\sqrt{10}}.\text{\hspace{0.17em}}$ Then simplify.

Write $\text{\hspace{0.17em}}\frac{12\sqrt{3}}{\sqrt{2}}\text{\hspace{0.17em}}$ in simplest form.

$6\sqrt{6}$

Given an expression with a radical term and a constant in the denominator, rationalize the denominator.

1. Find the conjugate of the denominator.
2. Multiply the numerator and denominator by the conjugate.
3. Use the distributive property.
4. Simplify.

## Rationalizing a denominator containing two terms

Write $\text{\hspace{0.17em}}\frac{4}{1+\sqrt{5}}\text{\hspace{0.17em}}$ in simplest form.

Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of $\text{\hspace{0.17em}}1+\sqrt{5}\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}1-\sqrt{5}.\text{\hspace{0.17em}}$ Then multiply the fraction by $\text{\hspace{0.17em}}\frac{1-\sqrt{5}}{1-\sqrt{5}}.$

Write $\text{\hspace{0.17em}}\frac{7}{2+\sqrt{3}}\text{\hspace{0.17em}}$ in simplest form.

$14-7\sqrt{3}$

## Using rational roots

Although square roots are the most common rational roots, we can also find cube roots, 4th roots, 5th roots, and more. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. These functions can be useful when we need to determine the number that, when raised to a certain power, gives a certain number.

#### Questions & Answers

what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
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?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
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
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
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
how
Sheref
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
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