Question

Need help on 4 problems.. Polynomial and Rational Inequalities

Answer 1

Solve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation.

4x^2 +7x<-3

Answer 2

4x^2+7x+3<0

Answer 3

Yup I have that.

Answer 4

That's only the first step..

Answer 5

i will help if i get your number

Answer 6

No.

Answer 7

@freckles

Answer 8

@phi

Answer 9

@johnweldon1993

Answer 10

can you factor the quadratic?

Answer 11

it is a pain to factor with a 4 on the x^2
but I would first do 4*3=12
and ask if any factors of 12 add up to 7
3 and 4
work
so it does factor.

Answer 12

Sorry was doing something else.. Yes x= -3/4 and -1

Answer 13

(- infinity, -3/4) , (-3/4, -1), (-1, infinity)

Answer 14

so you know
( x+3/4) ( x+1) <0

that will be negative if either (x+3/4) is negative and (x+1) is positive
or vice versa

I would test both conditions
For example, (x+3/4) neg means (x+3/4)<0
x < -3/4
and
(x+1) pos means (x+1)>0
x>-1

Answer 15

If we do it the other way
(x+3/4)>0 and (x+1) < 0
we get
x > -3/4 and x < -1
which cannot happen i.e x cannot at the same time be below -1 and above -3/4

so only the first answer works
-1 < x < -3/4

Answer 16

I would write it
(-1, -3/4)

Answer 17

Um what? I have no clue what your doing..

Answer 18

Do you have time to go over it ?

Answer 19

My final answer is (-3/4,-1) and the system says its wrong.. I only test valued the first and last number. (-infinity, -3/4) and (-1,infinity). I did not test value (-3/4,-1)

Answer 20

would write it (-1, -3/4)

Answer 21

Why?

Answer 22

the solution is x > -1 and x < -3/4
we put the lower bound first (that is the -1) and then the upper bound
we use ( rather than [ because -1 is not part of the answer

thus
(-1, -3/4)

how we know x>-1 and x < -3/4 is another story

Answer 23

Alright.. Can you help with 3 more?

Answer 24

Solve the rational inequality and graph the solution set on a real number line. Express the solution set in interval notation.

x-9 over x-1 >0

Answer 25

I got ( -infinity,9)U(1,infinity)

Answer 26

\[ \frac{x-9}{x-1} > 0 \]
that means the left side is positive
(bigger than 0 means positive)

you can get a positive number if you divide a positive by a positive
or
negative by negative

follow so far?

Answer 27

Yeah.

Answer 28

there are two possibilities:
both top and bottom are positive
or
both top and bottom are negative.

Let's test the first case
top positive means x-9 > 0
and at the same time, the bottom must be positive, which means x-1 > 0

ok so far?

Answer 29

I guess.

Answer 30

x=9 and 1

Answer 31

don't use an = sign
x-9 > 0
add 9 to both sides
x>9

what do you get for the bottom x-1 >0

Answer 32

My examples all do an = sign.

Answer 33

we don't have = signs

Answer 34

(- infinity, 9),(9,1),(1, infinity)
^^ I thought this one was weird but its what my example did.

Answer 35

we can do solve it doing it the way I am showing you.
so far we know
x>9
and now we are doing the bottom
x-1>0
add 1 to both sides

Answer 36

Okay.

Answer 37

x-1>0
add 1 to both sides

Answer 38

x>1

Answer 39

so what we found is when
x>9 the top is positive
and when x >1 the bottom is positive

we need both top and bottom to be positive at the same time

Answer 40

Sure.

Answer 41

what x values make both top and bottom positive ?

Answer 42

What?

Answer 43

we are looking at
\[ \frac{x-9}{x-1} > 0 \]
the idea is if both the top and bottom are positive, then when we divide
a positive by a positive we get a positive (and a positive number is > 0 )

Answer 44

Does that make sense?

Answer 45

Yeah.

Answer 46

so we want to know what range of numbers x can be so that we know both the top and bottom are both positive.

Answer 47

The answer is (-infinity,1)U(9,infinity)

Answer 48

I found my mistake..

Answer 49

I have no clue as to what your doing to solve these. It doesn't follow my examples. You did not do the test value. Ill figure out the other two my self.