Rational Inequalities
Summary
TLDRThis instructional video teaches viewers how to solve rational inequalities using the test point method. The presenter simplifies the process by treating inequalities as equations to identify critical values, then selects test points on a number line to determine intervals that satisfy the inequality. The video provides step-by-step solutions to various examples, emphasizing the importance of common denominators, critical values, and sign charts in finding the solution set for each inequality. The method is demonstrated with increasing complexity, offering a clear understanding of solving rational inequalities.
Takeaways
- 📝 Rational inequalities can be solved using the test point method.
- 🔍 Treat the inequality like an equation to find critical values.
- 📊 Use test points on a number line to determine which intervals satisfy the inequality.
- ✂️ Factor the denominator and numerator to identify excluded and critical values.
- ❌ Exclude values that make the denominator zero from the solution set.
- 🔢 Set up the number line by cutting it at critical values to test intervals.
- ✅ Identify intervals where the left-hand side of the inequality is positive or negative.
- 🔄 Use extreme values and simple test points to determine the sign of intervals.
- 🔧 For more complex inequalities, rearrange terms to get zero on one side.
- 📉 Combine fractions with a common denominator when necessary.
- 🔑 Use sign charts to visualize and verify solutions for rational inequalities.
- 🔢 Critical values are points where the numerator or denominator equals zero.
- 🧠 Plug in values within intervals to test whether the inequality holds true.
- 📈 The solution set includes intervals where the inequality is satisfied, using union for multiple intervals.
- 🧩 Simplify expressions and factor to make solving and testing easier.
Q & A
What is the main method discussed in the video for solving rational inequalities?
-The main method discussed in the video for solving rational inequalities is the test point method.
What are critical values in the context of solving rational inequalities?
-Critical values are the values that make the numerator or the denominator of a rational expression zero, as they are important in determining the intervals on the number line that satisfy the inequality.
Why is it necessary to treat the inequality like an equation when finding critical values?
-Treating the inequality like an equation helps in finding the values that make the numerator or denominator zero, which are essential for identifying critical points and excluded values.
What is the first step in solving the given example inequality x - 2/x^2 - 36 > 0?
-The first step is to factor the denominator x^2 - 36 into (x + 6)(x - 6) and identify the critical values where the denominator and numerator are zero.
What are the excluded values for the example inequality x - 2/x^2 - 36 > 0?
-The excluded values for the example inequality are x = -6 and x = 6, as these values make the denominator zero.
How does the video suggest to determine the intervals that satisfy the inequality?
-The video suggests using test points on a number line between the critical values to determine which intervals make the inequality true.
What is the purpose of using test points in solving rational inequalities?
-Using test points helps to determine the sign of the rational expression in different intervals on the number line, which in turn helps to identify where the inequality holds true.
How does the video handle the case where the right-hand side of the inequality is not zero?
-The video suggests adding or subtracting the non-zero value to both sides to create a zero on one side, which simplifies the process of identifying critical values and solving the inequality.
What is the solution set for the inequality x + 9/x + 7 ≤ -4 after adjusting the inequality to have zero on one side?
-The solution set for the adjusted inequality is the interval from -37/5 to -7, including -37/5 but not including -7.
Why is it important to create a common denominator when combining fractions in an inequality?
-Creating a common denominator is important because it allows for the simplification of the inequality into a single rational expression, making it easier to identify critical values and solve the inequality.
How does the video approach the inequality 4/x - (2 - 3)/x ≥ 0?
-The video approaches this inequality by combining the fractions on the left-hand side over a common denominator and then using test points to determine where the inequality is satisfied.
What is the solution set for the inequality x > 4x/5 - x?
-The solution set for this inequality is the union of the intervals 0 to 1 and 5 to infinity.
Why is factoring useful when solving rational inequalities?
-Factoring is useful because it simplifies the process of identifying critical values and helps in creating a sign chart to determine where the inequality is true.
Outlines
📚 Introduction to Solving Rational Inequalities
In this video, the presenter introduces the concept of solving rational inequalities using a method called the Test Point method. The method involves treating the inequality as an equation to find critical values and then using test points on a number line to determine which intervals satisfy the inequality. The presenter chooses a simple inequality, x - 2/(x^2 - 36) > 0, to demonstrate the process. The critical values are identified as -6, 2, and 6, which are points where the expression is undefined or zero. The presenter then sets up a number line, marks these critical values, and uses test points to find the intervals where the left-hand side of the inequality is positive, which are the solutions to the inequality.
🔍 Analyzing Rational Inequalities with Variations
The presenter continues by tackling variations of rational inequalities. The second example involves an inequality with a non-zero right-hand side, x + 9/(x + 7) ≤ -4. To apply the Test Point method, the presenter manipulates the inequality to have zero on the right side by adding 4, resulting in a common denominator for combining the terms. The critical values are identified as -7 and -37/5, and the presenter uses test points to find the intervals where the left-hand side is less than or equal to zero. The solution set includes the interval from -37/5 to -7, excluding -7.
📈 Solving Rational Inequalities with Common Denominators
In the third example, the presenter solves an inequality with a zero right-hand side, 4/x - 2 - 3/x ≥ 0. The presenter combines the fractions with a common denominator, x(x - 2), resulting in a simplified inequality. The critical values are 0 and 2, and the presenter uses test points to identify where the left-hand side is greater than or equal to zero. The solution includes two intervals: -6 to 0, including -6 but not 0, and 2 to infinity, excluding 2.
🎯 Advanced Techniques for Rational Inequalities
The final example presented is an inequality involving a rational expression greater than a variable, x > 4x/5 - x. The presenter discusses different approaches, including subtracting x to the right side and factoring out a negative from the denominator. The presenter then combines the fractions with a common denominator and simplifies the inequality to find critical values at 0, 1, and 5. Test points are used to determine the intervals where the left-hand side is positive, resulting in a solution set of 0 to 1 and 5 to infinity. The presenter emphasizes the importance of common denominators, critical values, factoring, and sign charts in solving rational inequalities.
Mindmap
Keywords
💡Rational Inequality
💡Test Point Method
💡Critical Values
💡Number Line
💡Undefined
💡Numerator
💡Denominator
💡Factor
💡Sign Chart
💡Union
Highlights
Introduction to solving rational inequalities using the test point method.
Treating the inequality like an equation to find critical values.
Using test points on a number line to determine intervals that satisfy the inequality.
Solving the rational inequality x - 2 / x^2 - 36 > 0 as a starting example.
Factoring the denominator x^2 - 36 to find excluded values of -6 and 6.
Identifying critical values that make the numerator or denominator zero.
Setting up a number line with critical values at -6, 2, and 6.
Using test points to determine where the left-hand side is positive.
Finding that the left-hand side is positive between -6 and 2.
Using extreme values like -1000 and 1000 as test points.
Determining the solution set for the inequality x - 2 / x^2 - 36 > 0.
Modifying the inequality x + 9 / x + 7 ≤ -4 by adding 4 to both sides.
Combining fractions with a common denominator to simplify the inequality.
Identifying new critical values -7 and -37/5 and solving the modified inequality.
Using test points to find intervals where the left-hand side is less than or equal to zero.
Solving the inequality 4 / x - (2 - 3) / x ≥ 0 by combining fractions.
Creating a sign chart to determine where the left-hand side is greater than or equal to zero.
Finding the solution set for the inequality with intervals and critical values.
Approaching the inequality x > 4x / 5 - x by moving the entire fraction to the left.
Factoring the numerator and denominator to identify critical values of 0, 1, and 5.
Using test points to find intervals where the left-hand side is positive.
Final solution set for the inequality x > 4x / 5 - x is 0 to 1 and 5 to infinity.
Emphasizing the importance of common denominators, critical values, factoring, and sign charts in solving rational inequalities.
Transcripts
hey in this video we're going to solve
rational
inequalities you know less than or
greater than with rational expressions
or fractions and the way we're going to
do this is we're going to we're going to
use what's called a test Point method so
uh we're going to we're going to treat
the inequality like an equation to find
critical values and then we're going to
use test points on a number line between
those critical values to figure out
which intervals make the inequality true
let's get into
it let's solve the rational inequality x
- 2/ x^2 - 36 is greater
than
0 and I chose a simple one to get
started the reason I'm telling you it's
simple is because the right hand side is
already zero and the left hand side has
a single fraction those are those are
things we like
now my goal is to figure out when this
left hand side is greater than zero when
when is the left hand
side positive that's what I'm looking
for
okay because and I'm trying to figure
out the X values that make that true so
what I'm going to do is treat this like
an equation and ask myself okay I need I
need to figure out the excluded values
so you know if I if I think about the
left side I can factor that denominator
x^2 - 36
as x + 6 * x -
6 and I can see from this that -6 and
positive 6 will make the denominator
zero so those are excluded values
-6 and positive
6 I don't want X to equal
those and then my numerator to make it
zero is xals 2 so that's not I can I can
have zero on the top now I'm not saying
that's the solution to the inequality
I'm just saying these three values are
my critical
values values that make the top or the
bottom zero are critical values because
they are important numbers on the number
line when we start to figure out which
parts of the number line or which
intervals on the number line make the
INE equality true so let's set up our
number line
cut at those critical values and see
what we can find so um these are X
values I need to cut them in cut it in
order so -6 is the first of those three
values that occurs and then two and then
six and I already know um I need to
figure out what happens to the left hand
side for the X values I'm going to add
here so up above X I'll write left hand
side and kind of keep track of what
happens we already know know that when X
is -6 the left hand side this fraction
is undefined so I'm going to put a u
there if x is 2 the numerator is zero
and we know that when the numerator of a
fraction is zero the whole fraction is
zero when X is six the denominator is
zero so the left hand side is
undefined so what I've done is I've cut
my numberline these critical places
these key places and now I need to look
at all right what I happens to the left
hand side in between those critical
values so what I mentioned before is
using test points so I'm going to pick
an x value in each of these intervals to
plug in to the left hand side and see
what I get so for example between -6 and
two I I would plug in
zero and if I plug in zero the top is
going to be
negative 0 - 2 is
negative the bottom left 0 + 6 is
positive the bottom right 0 - 6 is
negative so the bottom is a positive
times a negative which is negative and a
negative divided by a negative is a
positive so this the left hand side is
positive when X is between -6 and
2 now what I'm interested in when is the
left hand side positive anyway so the
left hand side is positive here so I def
Ely I'm going to include all of those X
values between -6 and 2 in my solution
set okay now I'm going to move a little
quicker now you can use a calculator to
plug these values in or you can do it
mentally like I I've done it here um on
the the ends I always like to use
extreme values so I'm going to use
negative 1,000 and positive 1,000 for my
test
points and then between two and six
maybe I'll just use three
all right if I plug in negative
1000 we get a negative over a negative
times a
negative that all simplifies to be a
negative so I'm not going to shade this
interval because I want the intervals
that are greater than
zero if I plug in three I get a positive
over a positive time a negative so
that's
negative and if I plug in 1,00 I get a
positive over a positive time a POS
positive which is a positive and so I do
want to shade all the X values bigger
than six now we have to look at the
critical values
themselves -6 if I plug that in I get
undefined that is not positive so I
don't want to include that so I'll
barricade it with this open circle if x
is 2 I get equal to zero that is not
greater than zero so I don't want to
include two in my solution either and
six I get undefined I don't want to
include that so my solution the X values
that make this true are in the
intervals -6 to 2 or 6 to infinity and
since it's this interval or this
interval I put a union between those and
that is my solution to the inequality
these are all the X values that make
this inequality
true now that was a nice one again
because we had zero on one side and
single fraction on the left side let's
let's try some variations of
this okay let's look at x + 9 over x +
7 is less than or equal to
-4 now this one almost looks as nice as
the last one and you're just saying well
ne4 is not zero that's actually a big
problem because the technique we're
using has this check the
sign positive or negative to get our
solution set and that's why we need to
know where zero is important because
zero separates the positives from the
negatives so what we have to do is add
this four over to get a zero on the the
right
side and just like when I solve rational
equations I'm going to have to combine
these two guys with a common denominator
so I'm going to call four four over
one and figure out how to get a common
denominator now on the right hand side
I've still got less than or equal to and
then
zero so my common denominator between
the two fractions well I can multiply
the second fraction on top and bottom by
x +
7 and so that gives
me the same denominator of x + 7 when I
put these together and I get x + 4x
which is 5X I'm adding my numerators now
and I'm just Distributing this four in
my head and then I get 9 + 28 which is
uh
37 and now this is set up just like the
last example we've got zero on one side
in a single fraction on the left side
and our critical values
here well my denominator critical value
is well X can't be
-7 and my numerator if I get X by itself
I'd subtract
37 and divide by 5 I get X =
-37 FS which is like
-7.4 okay so now I know where to cut my
number
line my sign chart
here I need to cut it Well - 37 fths is
further left on my number line the
-7 and I need to see what happens to the
left hand side side I'm going to go
ahead and say when X is -7 well that
makes the fraction or the left hand side
undefined and when X is -37 fths it
makes the numerator zero which makes the
entire fraction zero the the entire left
side zero and then I'm going to use test
points again so I'll plug in negative
1000 so I get a negative on top that's
pretty easy see a negative on bottom
which gives me a
positive uh between these guys maybe
I'll plug in like
-7.1 which is going to give me a let's
see here a positive on top and a
negative on bottom so I get
negative and over here if I plug in a th
I get a positive on top the positive on
bottom which is
positive and I want to find where the
left hand side is less than or equal to
zero well it's equal to zero here it's
less than zero
this interval and it's not less than or
equal to Z at -7 because it's undefined
so my solution set is the interval from
- 37 fths to -7 including - 37 fths but
not
including
-7 so that is our solution for this
inequality just a slight upgrade and
complexity by having to get zero on one
side side and combining the fractions
with a common
denominator let's do another slightly
different example let's look at 4 overx
-
2 - 3
overx is greater than or equal to
zero and in this case I've got zero on
the right side that's good I just need
to combine my fractions with a common
denominator so I'll multiply top and
bottom of this fraction by X
and I'll mply top and bottom of this
fraction by x -
2 so when I put them together they're
all going to be over the common
denominator of x * x - 2 and I get 4X
here and remember this negative
distributes with the three so I get min
- 3x so 4x - 3x is just X and then
plus six out of
that and that's that's there is to it we
can see that our critical
values are 0er and two and we don't want
X to be either of those and my
numerator get xal
-6 we make a sign chart here to see what
happens to the left hand
side so let's see -6 is first and then
zero and then two
those are our X values let's see what
happens to the left hand side well if x
is -6 my numerator is zero so my
fraction is zero but if x is z or two my
denominator is zero so the left hand
side is
undefined and then we'll use some test
points let's start over here on the
right hand side by plugging in 1,00 for
X we get
positive over positive * positive so
that's
positive if we plug in one for X we get
a positive a positive * a negative
that's
negative if we plug in -3 for X we get
positive over negative * negative that's
positive and if we plug in 1000 for X we
get Negative over negative * negative
which is
negative and we're looking for where the
left hand side is greater than or equal
to
zero so it's equal to zero here it's
greater than zero there and it's greater
than zero there so let's
shade this critical value let's shade
this interval exclude this critical
value exclude this critical value and
shade this
interval so we got two intervals we
indicate that with the
Union -6 to 0 including -6 but not zero
Union 2 to Infinity not including either
one okay let's let's do one more example
just just to be thorough here I want you
to feel like you've got the idea down
really
well okay let's try this guy um X is
greater than the rational
expression 4X over 5 -
x and so we've got a couple different
ways we can do this we can subtract the
X over to the right
that's probably what I would do because
I'd rather the X be the negative then
rather make this whole fraction negative
um but let's go ahead and see what
happens if we move the entire fraction
over to the
left I need to I need to make X into a
fraction so I can combine
them and I'm just going to mention some
things here as a as a mathematician and
a
teacher we
typically we typically don't like
expressions like this I mean there's
nothing wrong with them um but it's it's
natural for uh a mathematician let me
see if I can get rid of that there we go
to think of that just off to the side
it's the same thing as Negative X +
5 and because there's a negative in the
front you know lots of times it's useful
to factor that negative
out of both terms which changes their
signs
and in a fraction that's nice because we
don't like to have a negative on the
bottom so what I'm going to
do you don't have to do this I just want
to I'm teaching math so I'm I want to
kind of tell you how MA a mathematician
might think about this I'm G to I'm
going to go ahead and write my
denominator as negative times the
quantity xus 5 I've just factored the
negative
out and in a fraction when you got a
Nega
Factor you can just pull it out of the
fraction because I don't want to get
into it too much but if you got a
negative on the top or the bottom you
can just pull it out to the front and so
if you do that minus a negative turns
into plus a positive so now we have X
over 1 + 4x over x -
5 you don't that none of that's
necessary but it's it's very common to
do that and it it just makes things more
intuitive for you usually like this 5 -
x is a bit to work with a bit weird to
work with on occasion um but if you like
it then just leave it as as
is well now now I've got two fractions
on the left I want the same denominator
so I'm going to multiply my first
fraction on top and bottom by x -
5 then I'll combine those two fractions
over that common denominator giving me
X2 I've got - 5x and plus 4X X so that's
-
x let's get rid of
this so now I actually see that my
numerator factors so let's come over
here so Factor an X Out On Top giving me
x - one and on bottom I've got x - 5 you
can see why factoring is handy because
now I can easily identify my critical
values X can't be five that would make
my denominator zero but X can be 0 or
one in my
numerator to make the numerator
zero so let's cut our number line at
0 1 and
five and see what happens to the left
hand side well if x is five the
denominator is zero so the left hand
side's undefined and if x is 0o one the
numerator is zero so the left hand side
equals
zero and then let's start over here on
the right with our test points let's
plug in a th on the left hand side we
get a positive times a positive over a
positive which simplifies to a positive
let's plug in three here we get positive
* positive over negative that's
negative let's plug in 0. five here we
get positive * negative over negative
which is a
positive and then finally let's plug in
negative 1000 we get a negative * a
negative over a negative which is
negative and we're looking for the
intervals that make the left hand side
positive so that's here and here so
we'll shade those two intervals but not
the end points because they're not
positive they're
zero and we won't shade five because
it's not positive either it's undefined
and so our solution set is 0 to 1 Union
5 to
Infinity okay so there are several
examples solving rational inequalities
common denominators are important
critical values are important so
factoring is important and then the sign
chart very important um but lots of good
examples I hope You' got it down go get
some practice done and good luck
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