Good day students welcome to mathgotserved.com
in this clip were going to be going over how
to solve quadratic inequalities algebraically
and graphically so the instructions for this
problem is us follows we are to solve the
given any quality algebraically and Om graphically
okay 
the graphical approach is much faster but
the steps involved in the algebraic solution
is very important in preparing you for some
of the procedures you be encountering in precalculus
of Olson AP calculus so let's go ahead and
take a look at the inequality that were dealing
with we have 2X square plus 3X minus one is
greater than one okay so let's say this is
the quadratic inequality we are to solve algebraically
and graphically so let's start with the first
method the algebraic solutions solus touchwood
method one is algebraic okay whenever you
think of soul mean quadratic inequalities
algebraically you want to think about the
acronym CVS okay CD S like the store you go
to to get your prescription medication think
about CVS to critical values okay critical
values and then think about S as your sign
chart so these are the two components that
are needed in order to solve quadratic inequalities
algebraically okay so let's go ahead then
I write down the inequality that with the
elites 2X square plus 3X minus one is greeted
then one now in order to find the critical
values will have to said this equals zero
so let's rewrite this inequality in standard
form first so that will help us greatly as
we proceed with the problem-solving process
so to have zero on the right side we simply
subtract one from both sides of the inequality
so we have 2X square plus 3X -2 greater than
zero okay so now we have it in this format
let's proceed with parts one of the algebraic
process which is to find the critical values
okay so what are the critical values these
are what you gets when you set your equation
equal to zero and solve what X values cause
the equation when set to zero to have a outputs
of zero that's what we looking for okay so
what will going to do is will take these quadratic
inequality that the have here 2X square plus
3X -2 greater than zero we convert it into
an equation 2X square plus 3X -2 equals zero
now we have a quadratic trinomial the different
ways that he console be quadratic trinomial
equation you can use the complete can the
square methods factoring the quadratic formula
or you can also be graphically but one attempts
use the factoring method here is hope that
works is not we will have to resort to using
the quadratic formula because that always
works but up with AC on top and be on the
bottom in this problem a is to be is three
NC is negative to so we compute AC multiply
to by negative to you have negative for NC
is on sorryb is three 
okay let's play the X game the what two numbers
multiply to give you -4 and add to give you
three if you think about the possible combinations
of numbers that you the product of four think
about one on for this works nicely but not
a to get a positive the negative product any
positive sum we need the smaller to be negative
now if you notice the value of a is not equal
to one solve it is not equal to one you have
to solve this by factoring by grouping okay
so we have to factor by grouping added been
that a was equal to one then we can skip to
the factored state but that is not the case
here a is not want so we have to factor by
grouping we have 2X square now let's insert
to two numbers that we got from our X game
negative one X plus 4X -2 equals zero now
let's partition it write down the center and
factor by grouping from the first two the
GCF is X sold factor out X it will left with
2X minus one from the next to the GCF is positive
to own factor out that out we have 2X minus
one of this point you freezing check your
work make sure that quantities in the parenthesis
are identical and that is the case here so
all factorization process is accurate okay
so let's factor out the identical quantities
2X minus one now the terms on the outside
X and positive to can be clustered together
in their own parenthesis X +2 and set equal
to zero now we're going to use the zero product
property to find our to critical values that
involves setting both quantities equal to
zero okay so let's go ahead and solve these
two resulting equations by applying the zero
product property to the previous step so here
we just add one and divide by two though give
us the first critical value let's to that
add one you have 2X equals one and then you
divide by two you have your first critical
value X equals one half these causes the equation
to attain and out the value of zero now the
second equation we can simply soul that by
subtracting two from both sides and that gives
us our second critical value X equals -2 now
recall we talked about CVS of the according
to help us organize our steps CV is the Om
critical values okay so now will proceed to
the S parts with which is the sign chart okay
so let's go ahead and do that can going to
partition my workspace let's create our sign
chart you just involves join a number line
in graphing your Om critical values on the
number line so this is the original inequality
we have 2X square plus 3X -2 is greater than
zero since there is no equal to symbol here
it's basically tells is that the critical
values are not included in the solution region
okay so let's see this is -2 right here and
then we have one half will going to use an
open circle because those two values are excluded
from our our solution region okay so we have
open circle there and am I want to partition
the region so you can clearly see what's going
on okay now how many regions do we have here
we have region one which is to any number
dots smaller than -2 we have region to between
-2 1/2 and then region three now what's we're
going to do is regroup the test values that
we can plugging into the Om inequality and
see which one generate the a true statement
okay so let's see what's we have for interval
I'll one will think about a number next easy
to work with that is less than -2 with the
use X equals Om -3 now into about to think
about a number between negative one half composite
one half that is easy to work with the keys
X equals zero it into the rights of one half
with keys X equals positive one okay now what
we're doing is one to see where this inequality
is greater than zero what do we know about
the sign of a number that is greater than
zero its positive okay so any of this input
values our results in this expression attain
any positive value Om that into of all the
interval that contains that has value will
be Valley solution region okay so it's easier
to work with the factored form of this inequality
very have it factored here so we can express
this Avenue's 2X minus one times X +2 greater
than zero okay alright let's go ahead and
proceed with our test we're going to test
into involve number one for interval I'll
one we have X equals -3 so going to plug it
into this inequality right here so is to times
-3 minus one times -3+2 is it positive that's
the question so what I really care about here
is the sign of my final result I don't care
too much about the numerical part okay so
if you work this out -3×2 is -6 minus one
is whole first parenthesis will be negative
okay is can attain a negative Valley because
you have -6 minus one which is -7 the second
parenthesis you have -3+2 that's going to
give us a negative answer if you multiply
to negative numbers you end up with a positive
number minus times minus is plus this plus
greater than zero the answer is yes this is
a true statement okay remember we just comparing
it with zero in this positive then not it's
true is not that it's false okay now we know
that into about one is geared because the
have the positive sign in positive 18 means
that the result is greater than zero now whenever
you're solving quadratic inequalities algebraically
if there are solutions that are not a double
roots Om is how solutions then Om is either
going to be the outside the exterior into
balls or the interior is always a pattern
so since interval I'll one is already a solution
then we automatically should expect is the
ball three to be a solution and for interval
two not to be a solution so is always either
the two outer into the two outer regions or
the inner region okay plus you that in mind
so very know what the answer is technically
but let's go through the whole process test
the other values and see if our claim is true
next test is interval two will going to be
using X equals zero so for number interval
I'll to where take a look at X equals zero
will going to plug it into this inequality
so is to times zero minus one times 0+2 is
it greater than zero is it positive that is
the question so 2×00 minus one this will
have a negative answer 0+2 this will have
a positive results is negative times positive
greater than zero negative times positive
is negative is a negative number your than
zero the answer is no okay negative numbers
are smaller than zero that tells us that into
about to is no good because of its negative
sine in order to be Om greater than zero you
must have a positive sign just like we did
in interval I'll one now let's go ahead and
take a look at the last test we're going to
test Om X equals six what is the peak for
interval 310 that a plug it into the inequality
so we have two times one minus one times one
+2 is this greater than zero or is it positive
to times one is to minus one is one one times
one +2 is three is one times three greater
than zero one times threes three well let's
see were focusing on the signs… Use the
sign here so this one right here is been to
be positive and in this one right here is
positive so is a positive times positive greater
than zero positive times positive is positive
is a positive number greater than zero absolutely
so this is H restatement which tells us that
into about three is good because the have
a positive sign there okay so remember these
are our critical values the are not the solutions
okay these are critical values critical values
okay there not the answers now let's go ahead
and write on the answer desert visit with
of all one and three so what is it look like
is going to be you going to from negative
to to the left and you going from what positive
one half to the rights Laguna exclude the
endpoints okay now how do we write that in
interval notation or set or inequality notation
so for interval notation we can write ask
this is our solution region we have X is less
than -1/2 or X is greater than X is less than
negative 24X is greater than one half if you
want to write it using on interval notation
you can go from negative infinity to negative
to year neon one half all the way to infinity
okay so these are the solutions to the quadratic
inequality that the have presented in in the
ball and inequality notations okay let's proceed
to use method to is much faster Om what makes
use of the critical values so method to is
the preferred method which is the another
graphical method the reason why show the method
one F I indicated earlier is that it's preparing
you for some important proceed is that you
have to know in precalculus and AP calculus
so method to graphical liquid you start by
taking a look at the critical values the critical
values are basically what are going to use
to sketch the graph so we have X equals positive
one half -2 so critical values are X equals
-2 and positive one half let me go back to
make sure okay so we just goal we just graph
it now you graph doesn't have to be perfect
all you just want to do is take a look at
the quadratic function and see which portion
of it satisfies the condition that's been
placed on it right so what were going to do
is graph the quadratic the quadratic that
we're looking that's again is Om 2X square
plus 3X -2 is greater than zero so what were
going to do is open a graph the related function
with graph in why but a graph why is equal
to 2X square plus 3X -2 okay and then after
that's will take a look at the effects okay
Om let's go ahead and do it so we have the
critical values so is this is negative one
negative to write bear and then this is positive
one so I'll one half will be to the left of
positive one so this is one right here and
in this is one half okay so we have an open
circle here and then we also have an open
circle there goes the critical values we have
open circles because there is no inclusion
line underneath our inequality okay so if
you graph this function we know that is equal
to two which is greater than zero so them
is our parabola opens up okay so we have a
verbal looking something like this opening
up prefix this real quick okay so what's we're
looking for is where the function 2X square
plus 3X -2 is greater than zero so what portion
of this quadratic Om curve is greater than
zero now remember if you are below the x-axis
you less than zero if you are above the X
axis you have your greater than zero and and
if you're on the x-axis you have X equals
zero okay your equal to zero so what are we
looking for here greater than zero we looking
for the parts of the function that is above
be x-axis okay so let's go ahead and down
highlight those regions okay so the regions
are going to be the Queen be this region right
here that's the region it's above the x-axis
and then we have this region right here that's
above the X axis to okay so what's we're going
to do now is basically project these two regions
on to the x-axis okay these one is her on
the outside so if a project is on the X axes
to end up looking something like this you
have open circle here of course in a that
we going to the right because the have the
projection of the Om region that we just graphed
on the X axes so you have something like that
point bear and then this point going to the
right okay and of course you have an open
circle here in an open circle there soul let
me show you what I mean by projection you
take this entire in in the spread check it
down to your x-axis okay so what you end up
with is you solution region our solution region
are the portions in green on our X axis okay
so let's go ahead and write it down our solution
is going to be X is apt X is less than -2
or X is greater than positive one half and
that's our Om solution to the quadratic inequality
using the graphing method of the can see the
graph method is by far more easier and visual
Om to see but the algebraic method of those
involved that certainly teaches you some very
important skills that you be using in precalculus
and AP calculus thank you so much for taking
the time to watch this presentation really
appreciate it if our final found the contents
of this tutorial helpful helpful your study
of Om quadratic inequalities do give us a
thumbs up your positive feedback is very valuable
to us if you have any comp comments or questions
about what we covered in this presentation
just place your question in the comments section
below and will be more than glad to support
you don't forget to subscribe for channel
for updates to other tutorial such as this
more clips can be, mathgotserved.com thanks
again for watching and have a wonderful day
goodbye
