Question

How do you find the restricted values?

Answer 1

The values that make the denominator equal to zero for a rational expression are known as restricted values. We find these values by setting our denominator equal to zero, and solving the resulting equation.

Answer 2

So if the equation was \[\frac{ x-3 }{ x-2 }\] .. How would I do that?

Answer 3

do you know how to identify the denominator?

Answer 4

what is the denominator in the expression you have gave?

Answer 5

x-2

Answer 6

right and following what @Prettygirl_shynice said when is that equal to 0?
set equal to 0 and solve for x to find the restricted values

Answer 7

i Got x=2

Answer 8

Right we cannot plug in x=2
therefore x=2 is our restricted value
because it gives us division by 0 which is not defined

Answer 9

see plug in x=2
we get
(2-3)/(2-2)
which is -1/0
-1/0 is not defined

Answer 10

I understand! so in an equation like \[\frac{ 6-x }{x^2-7x+10 }\] I would set the bottom to 0 and then subtract 10

Answer 11

if you set the bottom equal to 0
you should see you have a quadratic equation
you could use the quadratic formula
or...you could factor
or you could complete the square in which case that would be a correct step if you wanted to complete the square that is

Answer 12

quadratic formula being the one over 2a?

Answer 13

\[ax^2+bx+c=0 \text{ is quadratic equation } \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \text{ is quadratic formula }\]

Answer 14

So, I would plug \[x^2-7x=10\] into the quadratic formula ( sorry i know im taking awhile)

Answer 15

You are fine...
If you are doing quadratic equation you want to leave in the form of ax^2+bx+c=0

Answer 16

x^2-7x+10=0
a=1
b=-7
c=10

Answer 17

ok one second..

Answer 18

I got the same equation

Answer 19

what do you mean?

Answer 20

you plug those values a=1,b=-7,c=10 into the quadratic formula ?

Answer 21

\[ax^2+bx+c=0 \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ 1x^2-7x+10=0 \\ a=1 \\ b=-7 \\ c=10 \\ x=\frac{-(-7) \pm \sqrt{(-7)^2-4(1)(10)}}{2(1)}\]....

Answer 22

and simplify

Answer 23

I see what i did wrong

Answer 24

your answers should come out nice and pretty

Answer 25

one second

Answer 26

sorry i had to pull up a graphing calcuator online because i dont have mine

Answer 27

we don't need a calculator
let's look inside the square root part
do you know what (-7)^2 is?

Answer 28

-49?

Answer 29

(-7)^2 is (-7) times (-7) which is 49 (so -49 is incorrect)
\[x=\frac{-(-7) \pm \sqrt{(-7)^2-4(1)(10)}}{2(1)} \\ x=\frac{-(-7) \pm \sqrt{49-4(1)(10)}}{2(1)}\]
how about the 4(1)(10) part?

Answer 30

that would be 4*10 which is 40*1 which is 40 so 40

Answer 31

\[x=\frac{-(-7) \pm \sqrt{ 49-40}}{2}\]

and then we do the subtract...
so 49-40 is?

Answer 32

9

Answer 33

\[x=\frac{-(-7) \pm \sqrt{9}}{2}\]
right! :)

and we know 9=3^2
so the principal square root of 9 is?

Answer 34

that im pretty positive is 3

Answer 35

\[x=\frac{-(-7) \pm 3}{2} \\ x=\frac{7 \pm 3}{2} \\ \text{ you have two restricted values.. } x=\frac{7+3}{2} \text{ or } x=\frac{7-3}{2}\]
both needs to be simplify and you are done

Answer 36

so the answer is X=5 and x=2

Answer 37

yep...
if you wanted to factor you could have found these values a little quicker
x^2-7x+10=0
(x-2)(x-5)=0
x-2=0 or x-5=0
implies
x=2 or x=5

Answer 38

completing the square one is also fun if you want me to show you that way

Answer 39

but it is also a bit longer :p

Answer 40

I never really got the hang of factoring

Answer 41

x^2-7x+10=0
when you have
x^2+bx+c=0
you find two factors of c
that first multiply to be c
and second add up to be b

so here -2(-5) is 10 and -2+(-5) is -7

so your factors for x^2-7x+10 would be (x-2)(x-5)

Answer 42

so your factors for x^2-7x+10 would be (x-2) and (x-5)

and so x^2-7x+10=(x-2)(x-5) *

I just wanted to correctly say things

Answer 43

Well that actually seems alot easier

Answer 44

it doesn't always work that well though

Answer 45

for example
x^2-3x+5
there are no two integer numbers that multiply to be 5 and also add up to be -3

so you would want to use completing the square or quadratic formula to solve x^2-3x+5=0

Answer 46

which one do you think is more efficient ?

Answer 47

Factoring if the numbers aren't too large...
Factoring can be time consuming...
It is sometimes not easy to think of two numbers that multiply to be one number and add up to be another number.
But sometimes the quadratic formula seems like overkill when the numbers are not so big and you can easily thing of the numbers mentioned above.

Answer 48

Yeah thats why i never stuck to factoring, Im not all that great with coming up with factors

Answer 49

Do you have any questions?
Did you want to try another problem?

Answer 50

I have two more on the sheet if you want to help me with those

Answer 51

Let's see them.

Answer 52

\[\frac{ 3 }{ x } * \frac{ x-2 }{ x-4}\]

Answer 53

thats one

Answer 54

Find the restricted values for both fractions to find the restricted values for the whole expression

Answer 55

im not sure, the directions says " List the restricted values of x for each expression" and its under the same category as the other one was

Answer 56

the fractions I see are 3/x and (x-2)/(x-4)

Answer 57

can you find the restricted values for 3/x?

can you find the restricted values for (x-2)/(x-4)?

Answer 58

for 3/x wouldnt i equal it to zero

Answer 59

the bottom

Answer 60

the restricted values for a/b would be when b=0

Answer 61

so for 3/x when x=?

Answer 62

x would equal 0

Answer 63

Right so one restricted value is x=0

Answer 64

now look at (x-2)/(x-4)

Answer 65

set the denominator to zero and it would be x=4

Answer 66

\[\text{ other examples } \\ \text{ restricted values for } \frac{3}{x}+\frac{x-2}{x-4} \text{ are } x=0,4 \\ \\ \text{ restricted values for } \frac{3}{x}-\frac{x-2}{x-4} \text{ are } x=0,4 \\ \text{ restricted values for } \frac{3}{x} \div \frac{x-2}{x-4} \text{ are } x=0,2,4 \\ \text{ this last one has } x=2 \text{ since } \frac{3}{x} \div \frac{x-2}{x-4} \text{ is } \frac{3}{x} \times \frac{x-4}{x-2}\]

anyways yes for your problem the restricted values are x=0,4

Answer 67

oh yay!

Answer 68

YAY! :)

Answer 69

the last one is even harder

Answer 70

k

Answer 71

go for it when you are ready

Answer 72

\[\frac{ x+1 }{ x^2-5x-24 } \div \frac{ x-8 }{ x^2-1 }\]

Answer 73

haha I just talked about a division problem
first: you don't want either of those starting fractions to have the bottom as 0

but!!!
you also don't want either of the bottoms after converting to multiplication to be 0

Answer 74

you have 3 equations to solve

Answer 75

\[\frac{a}{b} \div \frac{d}{c} \\ b \neq 0, c \neq 0 \\ \text{ but also } \frac{a}{b} \div \frac{d}{c}=\frac{a}{b} \times \frac{c}{d} , d \neq 0 \\ \text{ so resctrirected values are } b=0,c=0,d=0\]

Answer 76

Okay

Answer 77

so what three equations do you need to solve?

Answer 78

ait im so confused

Answer 79

first well look at the fractions in your division there
you don't want either of the bottoms to be zero
so you find when the bottoms are zero to exclude those values of x (aka restricted values)
then there is one more step

Answer 80

Okay, so dont set them to zero

Answer 81

you find when the bottoms are zero by setting the bottoms equal to zero
this is how you find the values to exclude from your domain (aka restricted values)

Answer 82

oh, Alright

Answer 83

just like we have been doing in previous problems

Answer 84

Yes

Answer 85

\[\frac{ x+1 }{ x^2-5x-24 } \div \frac{ x-8 }{ x^2-1 } \\ \text{ so you need to solve: }\\ x^2-5x-24=0 \\ \text{ and } x^2-1=0 \\ \text{ but also... } \\ \frac{x+1}{x^2-5x-24} \div \frac{x-8}{x^2-1}=\frac{x+1}{x^2-5x-24} \times \frac{x^2-1}{x-8} \\ \text{ so the other last equation you need to solve is } x-8=0 \\ \text{ you have three equations to solve: } \\ x^2-5x-24=0 \\ x^2-1=0 \\ x-8=0\]

Answer 86

x^2-5x=24 would be the denominator of the first one and x^2=1 would be of the second one?

Answer 87

Oh yes, ok

Answer 88

well the denominators to be looked at are
x^2-5x-24
x^2-1
x-8
and then you set equal to 0 to find the when bottoms are zero to find the non-allowed values aka restricted values

but anyways x^2-5x-24=0 is a quadratic equation
you can find two integers that multiple to be -24 and add up to be -5 pretty easily
or you can use quadratic formula

x^2-1=0 is pretty easy
you could use quadratic formula
but solving x^2=1 without quadratic formula is pretty easy
just take square root of both sides (you will get two answers here)

x-8=0 is just a one step to solve

Answer 89

okay, taking the square root of both sides i know would get rid of the squared, but the square root of one is just one

Answer 90

\[x^2=1 \\ \sqrt{x^2}=\sqrt{1} \implies x= \pm \sqrt{1} =\pm 1 \\ \text{ or you can write these solutions as } x=-1,1\]

Answer 91

Ohh, alright. im familiar with the plus/minus 1

Answer 92

since both (-1)^2=1 and (1)^2=1

Answer 93

Yes

Answer 94

two more equations to solve
I want you to try factoring on that quadratic one
can you think of two numbers that multiple to be -24 and add up to be -5?

Answer 95

well -8 times 3 is -24 and if added its -5

Answer 96

no wait

Answer 97

:)

Answer 98

so x^2-5x-24 is (x-8)(x+3)