BAM!  Mr. Tarrou in this math lesson
we're going to do two examples of
solving quadratic inequalities in our
first example as you can see we are
going to know our coefficients
ultimately of a B and after we get this
thing set equal to zero C and in our
second example we're going to see
something that well I didn't ever
experienced or seen any of my textbooks
that I was teaching above until I sort
of teach in IB mathematics and in
standard level and higher level IB
mathematics they'll give you questions
like this where it says something like
solve for an unknown such that this
quadratic inequality is always greater
than or less than or equal to zero
so it's a different twist on it and
we're gonna be taking a look at that
it's more it's a pretty interesting
question for a second example I have two
other lessons by the way related to this
solving higher order polynomial
inequalities and solving rational
inequalities I'll put links to those
lessons in the description of the video
I almost put timestamps two are two
examples course don't need a timestamp
for the first one we're starting it
right now we've got negative 2x plus 11
X is greater than or equal to 14 and
solving a quadratic inequality is really
just the same in the beginning as
solving for a quadratic equation you
gotta get that thing set equal to zero
so we're going to bring the 14 over we
have negative 2x squared plus 11x minus
14 is greater than or equal to zero I
don't like to solve quadratics by
factoring and this one is factorable
unless my leading coefficient is one so
we're gonna go ahead and we're going to
factor out a negative one you could also
just we're going to have our 2x squared
minus 11x plus 14 we're going to factor
this quadratic and in case you've
forgotten that's not the reason why I'm
doing this particular example because
students often struggle with factoring
if you want to factor a quadratic where
the leading coefficient is anything
other than 1 what you want to do is you
want to take your leading coefficient
and your constant and multiply those
together
and then you're gonna be looking for
factors of this product that add to the
middle term now our middle term is
negative with our last constant being
positive the factors that we use for 28
both have to have the same sign so we're
looking at two negative factors we have
negative 1 times negative 28 we have
negative 2 times negative 14 we have
negative 4 times negative 7 and here's
the period that we want to focus on
negative 4 times negative 7 is equal to
28 but they also when you add them equal
negative 11 so we're going to use these
2 factors of 28 to rewrite our middle
term as well two terms so we have
negative parentheses 2x squared we have
minus 4x minus 7x plus 14 is greater
than or equal to 0 and now that we have
our quadratic or three term quadratic
written as a with four terms we're going
to factor by grouping we're going to
look at what can we take out of our
first two terms well these both have a
common factor of 2x so we're looking at
2x times X minus 2 we're going to factor
out a negative 7 so negative 7 negative
7x divided by negative 7 is X and we
have 14 divided by negative 7 which is
equal to negative 2 and now inside this
parenthesis we have two terms 2x times X
minus 2 and a second term which is 7
times X minus 2 each of these two terms
have a common factor of X minus 2 so
factoring that common factor out of
these two terms leaves us with with the
negative out front
taking out the X I don't need another
set of four well yeah X minus 2 if we
take this factor out of our first term
that's going to leave us with 2x and if
we take this factored factoring out this
factor of X minus 2 that's going to
leave us with a negative 7 we're going
to set each of these equal to 0 given us
x equals 2 and X is equal to 7 halves
and now the 1 way of looking at this is
to think okay
at this value of x value of 2 and at
this x value of 7 halves this quadratic
is going to be equal to 0 and that
happens to be for this particular
example part of my solution so on a
number line our critical value here is x
equals 2 and X is equal to 7 halves
which is 3 & a half so we got 1 2 3
right 1 2 3 and a half and we're going
to test some intervals basically we're
looking at testing the interval from
negative infinity to 2 and then the
interval I'm using interval notation by
the way of going from 2 to 7 halves and
then I'm going to be looking at the
interval from 7 halves to infinity and
I'm going to just simply plug a number
from each of these intervals into this
expression kind of like my function if
you will negative 2x squared plus 11x
minus 14 and test these intervals and if
one number in an interval works then
that entire interval works if one number
that interval fails then the whole
interval fails and because we have an
equal sign in this inequality these are
actually going to be part of the
solutions but we want to check the
intervals between those values and see
which one's work yeah that's gonna look
like this you might be thinking of a
faster way of doing this problem by the
way and I will be going over that we're
gonna have something to the left of two
like zero
so we have negative 2 times 0 squared
plus 11 times 0 minus 14 is that greater
than or equal to 0 well negative 14 is
definitely not greater than or equal to
0 so every number between negative
infinity to 2 is not going to be a
solution to this inequality
well let's try something between 2 & 7
halves like 3 so now we're going to be
looking at negative 2 times 3 squared
plus 11 times 3 minus 14 that's going to
yield us 4 times no that'd be 3 squared
is 9 times negative 2 is negative 18
plus 33 minus 14 and we're getting kind
of close here but we have negative 18
now we have negative 28 now we have
negative 32 negative 32 plus 32 let's
try that again negative 32 plus 33 is
going to be equal to 1 and yes 1 is
greater than or equal to 0 so one number
between 2 & 3 halves makes this
inequality true so every number between
2 & 3 halves is going to be a solution
to this inequality check
and finally with let's see here purple
we're going to pick a number that's
larger than 7 halves like 4 and I'm
running out of room but you've already
seen me do the factoring so we're gonna
go ahead and take this out hopefully
understood everything there I'm sure you
did and we've got negative 2 times 4
squared plus 11 times 4 minus 14 is
greater than or equal to 0 that's going
to be 16 times
negative 2 is negative 32 plus 44 minus
14 now we're looking at negative 32
negative 42 negative 46 plus 44 is
negative 2 and that is not greater than
or equal to zero so one number in this
interval failed so this whole interval
fails so what is the solution to this
quadratic inequality well because of the
equal sign our critical values of two
and seven halves are part of the
solution and every number between two
and seven halves is also a solution so
what's the answer well the answer is on
the closed interval from 2 to 7 halves
and if your textbook and teacher are not
using interval notation you might say
that the avout the solution is X is
greater than or equal to 2 and less than
or equal to 7 halves and that's how you
solve a quadratic inequality all of this
arithmetic of checking the intervals is
kind of not necessary especially if you
know how to graph your polynomial
functions you don't you know something
about end behavior and multiplicity of
zeros we talked a lot more about this
and graphing higher order polynomials
but you have an even degree function
which means that now I know this is a
quad hopefully you know this is a
parabola but you have an even degree
polynomial which means that the end
behavior of your function is either
going up to the left or as X approaches
negative infinity the function
approaches positive infinity and as X
approaches positive infinity the
function goes to infinity or it's up to
the left and up to the right but now you
have a leading coefficient which is
negative which is going to reflect the
graph over the x axis so this is a
parabola opening down opening down
forever to the left and down forever to
the right so I know my graph looks
something like this and I'm not going to
worry about finding your y-intercept
because
for the purpose of this problem that's
not all that important and when we
course now I've just said we got the
factory and done and we don't need to
see it anymore but when we had this in
factored form which was negative 2x
minus 7 and X minus 2 those zeros came
from factors that showed up an odd
number of times and this is going to
give us as a basis for something called
multiplicity hopefully you've heard
about that from your teacher and if a 0
comes from a factor that shows up an odd
number of times the graph is going to go
through the x-axis at that point if a 0
comes from a factor that shows up an
even number of times which I don't have
here but if this is a power of 2 then
the graph is not going to go through the
x value of 2 it's going to touch the x
axis at the x value of 2 but these are
both odd values or multiplicity is odd
so with a super quick sketch after the
factoring
I'm not going to worry about where the
vertex is necessarily because we only
care about trying to solve this
inequality this is the sketch not
knowing exactly what the parabola the
the vertex of the parabola is but that
orange graph is the sketch of y equals
negative 2x squared plus 11x minus 14
and certainly now that the graph is up
here I can easily identify where is this
graph this parabola given me Y values
which are greater than or equal to 0
well that's and I hope these different
colors show up on the camera that's
between I'm gonna know they show up it
hopewell it for you to really see it
well between two and three halves and
that's is how you solve a really
basically a polynomial inequality get it
set equal to zero factor set each factor
equal to zero solve it and then either
that work of testing those intervals
along the x-axis or if you know
something about end behavior and finding
those x-intercepts which you had
to do anyway and the multiplicity you
could just do a sketch and let the
sketch answer your question doing all
this work though checking the intervals
is something you have to do when you
solve rational inequalities no little
quick
well no easy way to quickly sketch a
graph and get your answer from there
what do you do when you're asked to
solve a quadratic inequality and you
don't know the coefficients coming up or
right now so for our second example and
really it's the meat's it's why I made
this video but I wanted to do that
review at the beginning find the value
of M and the quadratic inequality where
we have M squared M times x squared plus
2m X plus X plus M minus 2 is less than
or equal to 0 and this is what IB likes
to do they take what seems like a pretty
basic concept and then they write these
abstract problems that requires you to
pull from different areas of mathematics
and see how they all relate it makes
some of the questions really difficult
but ultimately it makes this curriculum
so so cool if you're you know a math
nerd like I am well you can't solve this
inequality per se like you think you
can't have the same thought process like
we had in the previous example because
well you have two variables in the same
equation what you need to focus on here
is that it says find the value of M in
the quadratic inequality basically maybe
I could add this such that this
quadratic is always less than or equal
to zero or maybe I will say greater than
zero doesn't matter if it says less than
or was greater than the problems going
to be designed to work you just have to
find the unknown coefficient C this
isn't like the previous question where
we knew all of the coefficients and
there was a possibility you know of
their B index intercepts and well
there's a possibility of this expression
being greater than 0 or less than 0 or
whatever like I want this quadratic to
always be less than or equal to 0 I
don't there
not open for two like in the previous
example come up pass through the x-axis
come back down and pass again below it
so you need to think about oh man let's
see here
parabolas can open up and not have an
x-intercept they can open up and touch
the x-axis once since we have an
x-intercept where the graph is touching
the x axis that means there must be a
multiplicity of 2 so basically we have
the same two factors showing up because
the only quadratic the same two factors
showing up twice or you can have that
parabola opening up and crossing twice
now I'm having all these parabolas open
up but this is an inequality where the
parabola has to always be below the
x-axis that's what really what that
means so we're looking at a situation
where either the parabola looks like
this where it's always below the x-axis
we have maybe the parabola is coming up
and touching the x-axis but then going
right back down again and because we
have not less than but less than an
equal to that needs to be what happens
and then you know we can have a parabola
opening up of course and crossing the
x-axis twice
does this sketching thing over here give
you any idea of how we're going to
approach this problem what did you learn
in mathematics that helped you determine
whether a parabola crossed
twice once or 0 that doesn't ring a bell
think about how you solve quadratic
equations or find the x-intercepts for
parabolas you can complete the square
you can factor and by the way you can
take the completion of the square
process on your general a x squared plus
BX plus C equals 0 go through the
completing the square process and
actually derive for yourself if you
haven't already done in your math class
the quadratic formula and inside the
quadratic formula
X is equal to the opposite of B plus or
minus the square root of B squared minus
4
see all over 2a yeah you have got a
square root and of course when you start
talking about an even root there's a
possibility of there being imaginary
solutions yeah this discriminant is what
the problem is about if your
discriminant is which is b squared minus
4ac if that discriminant is less than
zero then there is no real solutions and
if that discriminant is equal to zero
because you may squirt a negative number
if the square if that is this the
discriminant is equal to zero then we're
going to negative be the opposite of B
plus or minus zero so really whether you
add or you subtract
it's not going to affect the value of
the numerator divided by 2a that means
you have one real solution that one real
solution with the multiplicity of two if
you have it in factored form like
parentheses X minus two close it
square it equals zero and if your
discriminant is greater than zero well
then you have two real solutions and the
only way for this parabola which I guess
must be opening down to always be less
than zero unless it just actually
touches the x axis is for that
discriminant to be less than or equal to
zero we want to focus on the less than
part because we you know we don't any
other solutions so we're going to say
that b squared minus 4ac
we need that thing to be less than or
equal to zero and if there was just a
less than zero we would not want any
real solutions and we would just be
making sure that that discriminant was
less than zero showing that there is no
x-intercept
so wherever that parabola is it's always
going to be giving you negative values
because there's no
solutions well that means that we need
to identify what a B and C are so we
need to have a coefficient of x squared
a coefficient of x + a constant anything
left that doesn't have a extra minute so
not much left to do here as far as that
goes we already have an MX squared but
we have do have two terms with X so
we're going to factor out that X and say
that we have + 2 M + 1 X and then + M
minus 2 is less than or equal to 0 that
means that our a value is M B is 2 M
plus 1 and C is M minus 2 our
discriminant is the B which is 2 M plus
1 squared minus 4a which is our M times
C which is M minus 2 and we're looking
for this to be less than or equal to 0
see we have an inequality with only one
variable now so we're looking at 4 M +
squared plus 4 M plus 1 distributing
that binomial together we've got
negative 4 M times M is negative 4m
squared we have negative 4 M times
negative 2 which is going to be plus 8 M
we're looking for that again to be less
than or equal to 0 we have the M 4 M and
the minus 4 M are cancelling out we have
4 plus 8 is 12 M plus 1 is less than or
equal to 0 12 M is less than or equal to
negative 1 and M is less than or equal
to negative 1/12 and that is the end of
my last example I miss a true Pam
go to your homework
you
