1.3 Radicals and rational expressions  (Page 2/11)

 Page 2 / 11

The product rule for simplifying square roots

If $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ are nonnegative, the square root of the product $\text{\hspace{0.17em}}ab\text{\hspace{0.17em}}$ is equal to the product of the square roots of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b.\text{\hspace{0.17em}}$

$\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$

Given a square root radical expression, use the product rule to simplify it.

1. Factor any perfect squares from the radicand.
3. Simplify.

Using the product rule to simplify square roots

1. $\sqrt{300}$
2. $\sqrt{162{a}^{5}{b}^{4}}$

Simplify $\text{\hspace{0.17em}}\sqrt{50{x}^{2}{y}^{3}z}.$

$5|x||y|\sqrt{2yz}.\text{\hspace{0.17em}}$ Notice the absolute value signs around x and y ? That’s because their value must be positive!

Given the product of multiple radical expressions, use the product rule to combine them into one radical expression.

1. Express the product of multiple radical expressions as a single radical expression.
2. Simplify.

Using the product rule to simplify the product of multiple square roots

$\sqrt{12}\cdot \sqrt{3}$

Simplify $\text{\hspace{0.17em}}\sqrt{50x}\cdot \sqrt{2x}\text{\hspace{0.17em}}$ assuming $\text{\hspace{0.17em}}x>0.$

$10|x|$

Using the quotient rule to simplify square roots

Just as we can rewrite the square root of a product as a product of square roots, so too can we rewrite the square root of a quotient as a quotient of square roots, using the quotient rule for simplifying square roots. It can be helpful to separate the numerator and denominator of a fraction under a radical so that we can take their square roots separately. We can rewrite $\text{\hspace{0.17em}}\sqrt{\frac{5}{2}}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}\frac{\sqrt{5}}{\sqrt{2}}.$

The quotient rule for simplifying square roots

The square root of the quotient $\text{\hspace{0.17em}}\frac{a}{b}\text{\hspace{0.17em}}$ is equal to the quotient of the square roots of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b,$ where $\text{\hspace{0.17em}}b\ne 0.$

$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

Given a radical expression, use the quotient rule to simplify it.

1. Write the radical expression as the quotient of two radical expressions.
2. Simplify the numerator and denominator.

Using the quotient rule to simplify square roots

$\sqrt{\frac{5}{36}}$

Simplify $\text{\hspace{0.17em}}\sqrt{\frac{2{x}^{2}}{9{y}^{4}}}.$

$\frac{x\sqrt{2}}{3{y}^{2}}.\text{\hspace{0.17em}}$ We do not need the absolute value signs for $\text{\hspace{0.17em}}{y}^{2}\text{\hspace{0.17em}}$ because that term will always be nonnegative.

Using the quotient rule to simplify an expression with two square roots

$\frac{\sqrt{234{x}^{11}y}}{\sqrt{26{x}^{7}y}}$

Simplify $\text{\hspace{0.17em}}\frac{\sqrt{9{a}^{5}{b}^{14}}}{\sqrt{3{a}^{4}{b}^{5}}}.$

${b}^{4}\sqrt{3ab}$

We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. For example, the sum of $\text{\hspace{0.17em}}\sqrt{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}3\sqrt{2}\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}4\sqrt{2}.\text{\hspace{0.17em}}$ However, it is often possible to simplify radical expressions, and that may change the radicand. The radical expression $\text{\hspace{0.17em}}\sqrt{18}\text{\hspace{0.17em}}$ can be written with a $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ in the radicand, as $\text{\hspace{0.17em}}3\sqrt{2},$ so $\text{\hspace{0.17em}}\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}.$

Given a radical expression requiring addition or subtraction of square roots, solve.

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.
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