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How To Solve Second Degree Equations By Factoring Or By Using The Quadratic Formula
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Feb 19, 2025
If you need tutoring, feel free to schedule an appointment via https://tayibs.com/get-math-help/ Welcome back to another math lecture. Today we will learn how to solve quadratic equations. Many students tend to get frightened when they hear words like quadratic. First we'll define what a quadratic equation is and then we'll learn how to solve it by factoring and via the quadratic formula
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0:00
hello and welcome back to another hour
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of mathematics okay today we're going to
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talk about quadratic equations
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i know people don't like quadratic
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equations because they tend to be
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a little bit confusing but we're going
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to try to uh
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get rid of that confusion because it's
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actually easy to deal with
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so first thing i want to do is i want to
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show you how to
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work with quadratic equations there's
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like multiple ways you can do it
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so we can do it by factoring we can do
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it by using the quadratic formula
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but number one thing we need to
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understand is again definition
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what is a quadratic equation that's what
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we need to understand
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now any equation in the form
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a x squared plus b
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x plus c equals
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zero is called a quadratic equation
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okay and a x squared plus b x plus c
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is a polynomial of a second degree okay
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polynomial of a second degree because
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the leading degree here is two so this
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is a second degree
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polynomial now this is a second degree
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equation so
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a x squared plus b x plus c is equal to
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zero is called
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a quadratic equation now how do you
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solve
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this quadratic equation there are
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multiple ways to do it
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number one i'm going to teach you how to
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do it by factoring
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okay now in this quadratic equation
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there are things that we need to know
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about
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those coefficients a b
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and c those are important to to know all
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right
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so now let me give you an example of a
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quadratic equation and we're going to
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try to solve this
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so let's say i have x squared
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minus um 14x
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plus 33 equals to zero
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and the question is solve for x
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so a lot of times you go i don't know
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what to do and i was just like you two i
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didn't know what to do
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until i was in high school and my
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teacher or
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actually in middle school and my teacher
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taught me how to do it
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by using the factoring so what i do here
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is this
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i look at the coefficient okay the a
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this is a here this is b and this is c
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a is equal to 1 b is equal to negative
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14
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and c is equal to 33 since a is equal to
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1
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um i can actually first what we're going
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to do by by factoring so let's let's do
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that here we have our coefficient
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i just wanted to show you what they were
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this is this qualifies as a
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second degree equation or quadratic
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equation so we're going to use the
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factoring so i'll do what
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add 1 times 33 a
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times c which is 1 multiplied by 33
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and that gives me 33. now the second
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step is
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you want to find two numbers whose
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product
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is 33 and sum is negative
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14. again i'll repeat that again you
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want to find two numbers
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who add up to negative 14 and whose
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multiplication gives you
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33. so first let's write
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this out the product p we're going to
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call the product
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p has to give you 33 and the sum
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or the addition or subtraction
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gives you negative 14. so how do you
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find these numbers
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remember i told you yesterday you have
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to know your multiplication table
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it does help a lot so you have to have
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the basics the foundation if you don't
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know your multiplication table
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you're going to have a hard time so now
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let's break down 33.
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33 can be written as what 3 times
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11 right that's the easiest way to do it
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3 times 11.
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now look at this here 3 plus 11 gives me
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14
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right if i do 3 plus 11 i get 14.
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but i don't want 14 i want negative 14.
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so what do i do
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all i have to do is this i know that if
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i do negative 3
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times negative 11 i still get what 33
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and if i do negative 3 minus 11
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i get negative 14 which is exactly what
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i want
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because i know that negative 3 plus
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negative 11
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gives me negative 14 when you add two
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negatives what do you do
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you add the numbers and you carry the
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sign in this case is a negative
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so i want negative 3 and negative 11. i
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have my two digits so if i know that
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negative 3
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times negative 11 gives me 33 and i also
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know that negative 3
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plus negative 11 gives me negative 14.
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so i found
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my two numbers i'm pretty much done with
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this problem
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so all i have to do now is factor my
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second degree equations with coefficient
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a equals one now all i have to do just
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factor it right it's going to be x minus
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3
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and then x minus 11 equals to 0
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and all i have to do now is set each one
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of those factors equal to 0 and solve
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for x so basically it's going to be x
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minus 3 equals to 0
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and x minus 11 equals to 0. so in this
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case we're going to have
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x equals 3 because you add 3 on both
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sides
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right and then here x equals 11
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and this is my solution so my solutions
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are x
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equals 3 and x equals 11. that takes me
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to our next
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problem and where a is greater than 1
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and we're going to see
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what to do let's find an example let me
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write
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3x squared plus
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20x minus 7
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and now we want to again solve this
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equation equals to zero but the first
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step is always doing what we want to
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factor
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this we want to factor it right so how
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do we factor
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we're going to use the method called uh
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factor
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by grouping we're going to group this
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bar we're going to repeat the same step
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again we need to know what
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a b and c are in this case
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a is 3 right
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a is 3 b
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is 20 and c
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is negative seven so i do a times c
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again it's the same method a times c
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so in this case a times c gives me three
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times negative seven and that gives me
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negative 21. so what did we say in the
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last problem
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so what we're going to do now is we want
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two numbers whose product
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p gives me negative 21 and sum
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s has to give me 20. so how do i find
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these numbers
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well again if you know your
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multiplication table it does help
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so i know that 3 times 7
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gives me 21. i want negative 21. so
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either negative 7
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negative 3 times 7 equals negative 21 or
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negative 7 times 3
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that gives me negative 21. but if i do
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negative
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3 plus 7 i get 4 and if i do negative 7
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plus 3 i get negative 4. so that doesn't
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work because my pro i want my my
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sum to give me 20. so let me try
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something else what about negative 20
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times one negative 20 times one gives me
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what
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it gives me uh no negative 21
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times 1 that gives me negative 21 right
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but then negative 21 plus 1 gives me
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negative 20. that's not good so all i
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have to do is just switch my
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my signs so i'm going to do 21 times
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negative 1
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that gives me negative 21 and the sum
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21 minus 1 gives me 20. so
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these two numbers are going to work
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because 21 times negative 1
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gives me negative 21 and then 21 minus 1
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gives me 20. so now that i have my two
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digits i'm going to rewrite
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this whole equation i'm going to rewrite
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it
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so this is going to turn into
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3x squared
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plus 21x minus
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1x minus 7 equals to
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0. and then the next step is we're going
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to
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factor by grouping so i'm gonna do these
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two together
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and i'm gonna do these these two
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together i need to find my gcf
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the gcf is what the greatest common
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factor
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so what is the greatest common factor
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between 3x square and 21x
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is 3x so if i plot 3x as a factor
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i'm left with x plus
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7 right now in this case
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what's my greatest common factor between
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negative 1 x and negative 7
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minus 7 is negative 1. so if i apply a
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negative 1
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i'm left with x plus 7.
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now look this is beautiful now i have my
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common factor
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i can just factor this right easy it
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turns into what x
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plus 7. times three x
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minus one equals to zero and then the
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next step is just to set each one of
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those equal to zero and solve for x
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so let's see raise the top and then
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we're gonna solve this problem here
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so x plus seven equals to zero
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that means x equals equal to negative
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seven and then three
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x minus one equals to zero that means
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three x
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is equal to one and if you divide by
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three
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x is equal to one third so my solutions
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are
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negative seven and one third and you are
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done
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so this is called um factoring by
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grouping
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so we learned the simple factoring when
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you when
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a is equal to one and now when a is more
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than one
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this is how you do it by grouping now
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let's say you don't like this factoring
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you just don't like this kind of stuff
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but you just want to do it
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in a different way you want to do it uh
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you said to me time i just did this
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factoring
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i don't like it i just despise it i want
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to learn how to do it another way
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because some people love formulas some
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people love
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formulas right so we're going to use the
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quadratic formula
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and i've learned this again several
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years ago the quadratic formula is
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x equals negative
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b plus or minus
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square root of b squared minus 4 ac
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over 2a again this is a ugly looking
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formula i don't like using it i like
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the factoring but some people don't like
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the formula so i'm going to show you how
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to use it
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so let's go back to the same problems
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that we had earlier the first one was
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x squared minus 14x
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plus 33 and then you want to use
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this formula to solve the problem how do
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you do it again you need to identify
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a b and c in this case a
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is 1 b is negative 14
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and c is 33. now all we have to do is
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just plug it into the equation
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into the formula so x is going to be
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negative b
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which is negative negative 14
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right negative negative 14. my 14 looks
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like uh
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my 14 looks almost like a 16 so i need
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to fix this
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let me rewrite the equation is x squared
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minus 14 x plus 33
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and x is negative b b is negative 14 so
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negative b
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will be what will be negative negative
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14 right
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plus or minus b squared which is
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negative 14 squared
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all right minus 4 times ac
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minus 4 times a is 1
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c is 33 you see
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we end up with a gigantic looking thing
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over 2a 2 times 1.
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now we need to find negative 14 squared
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14 squared
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and i have to use my calculator 14
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squared will be
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let's do that on the side since i don't
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have my calculator i'm just going to do
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it by hand
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14 squared let me multiply 14 in front
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of you
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i didn't want to do this but i have two
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four and four is sixteen
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four and one is four fifty 56 here and
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then we have
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1 and 4 1 2 6
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9 196 okay so we're going to put it in
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step by step so we have x
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equals positive 14 plus or minus
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196 minus 4 times 33
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let's do 4 times 33 on the side
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4 and 3 is 12 4 3 squared 132
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minus 132 over 2
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all right and then the next step is
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x equals 14 plus or minus
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196 minus 132 that is uh four here
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and three and nine that's 664
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over two now i don't know if you were
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listening to my
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lecture yesterday we talked about uh
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perfect
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squares okay now 64 is a perfect
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square so that's going to be easy square
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root of 64
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is 8. and i talked about what square
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root of 64
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is it means one number multiplied by
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itself
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gives you uh 64. and this is 8 so square
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root of 64 is 8. so i'm going to erase
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the top part
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and we're going to find our solution so
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this turns into x
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equals 14 plus or minus
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8 over 2 and now we have to split
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our answers so the first one is going to
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be x equals
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14 plus 8 over 2
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or x equals 14 minus 8
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over 2. our first solution is 14 plus 8
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that's 22
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22 by 2 and the last one is x equals 14
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minus 8
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that is 6 over 2 so we have x equals 11
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or x equals 3. now this is the factoring
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this is the
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finding the solutions by using the
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quadratic formula
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now this is a long process as you can
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see with the factoring is much easier
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with the factoring all i have to find
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was
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two numbers whose product is 33 and sum
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is 14
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and when we did it we had negative 3 and
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negative 11
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and then we did our factoring in this
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case it's just too long and i don't want
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to waste my time doing all these things
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this is just way too long right so i
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hope you enjoyed this
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and i'm looking forward to another um
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one of these all right again thank you
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and if you need tutoring
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please don't hesitate to reach out to me
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i'll have a link for in the description
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all right thanks for watching and stay
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tuned for more stuff
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