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# Using the Quadratic Formula

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- In this video, I'm going to expose you to what is maybe
- one of at least the top five most useful formulas in
- mathematics.
- And if you've seen many of my videos, you know that I'm not
- a big fan of memorizing things.
- But I will recommend you memorize it with the caveat
- that you also remember how to prove it, because I don't want
- you to just remember things and not know
- where they came from.
- But with that said, let me show you what I'm talking
- about: it's the quadratic formula.
- And as you might guess, it is to solve for the roots, or the
- zeroes of quadratic equations.
- So let's speak in very general terms and I'll
- show you some examples.
- So let's say I have an equation of the form ax
- squared plus bx plus c is equal to 0.
- You should recognize this.
- This is a quadratic equation where a, b and c are-- Well, a
- is the coefficient on the x squared term or the second
- degree term, b is the coefficient on the x term and
- then c, is, you could imagine, the coefficient on the x to
- the zero term, or it's the constant term.
- Now, given that you have a general quadratic equation
- like this, the quadratic formula tells us that the
- solutions to this equation are x is equal to negative b plus
- or minus the square root of b squared minus 4ac, all
- of that over 2a.
- And I know it seems crazy and convoluted and hard for you to
- memorize right now, but as you get a lot more practice you'll
- see that it actually is a pretty reasonable formula to
- stick in your brain someplace.
- And you might say, gee, this is a wacky formula, where did
- it come from?
- And in the next video I'm going to show you
- where it came from.
- But I want you to get used to using it first. But it really
- just came from completing the square on this
- equation right there.
- If you complete the square here, you're actually going to
- get this solution and that is the quadratic
- formula, right there.
- So let's apply it to some problems. Let's start off with
- something that we could have factored just to verify that
- it's giving us the same answer.
- So let's say we have x squared plus 4x minus
- 21 is equal to 0.
- So in this situation-- let me do that in a different color
- --a is equal to 1, right?
- The coefficient on the x squared term is 1.
- b is equal to 4, the coefficient on the x-term.
- And then c is equal to negative 21,
- the constant term.
- And let's just plug it in the formula, so what do we get?
- We get x, this tells us that x is going to be equal to
- negative b.
- Negative b is negative 4-- I put the negative sign in front
- of that --negative b plus or minus the
- square root of b squared.
- b squared is 16, right?
- 4 squared is 16, minus 4 times a, which is 1, times c, which
- is negative 21.
- So we can put a 21 out there and that negative sign will
- cancel out just like that with that-- Since this is the first
- time we're doing it, let me not skip too many steps.
- So negative 21, just so you can see how it fit in, and
- then all of that over 2a.
- a is 1, so all of that over 2.
- So what does this simplify, or hopefully it simplifies?
- So we get x is equal to negative 4 plus or minus the
- square root of-- Let's see we have a negative times a
- negative, that's going to give us a positive.
- And we had 16 plus, let's see this is 6, 4 times 1 is 4
- times 21 is 84.
- 16 plus 84 is 100.
- That's nice.
- That's a nice perfect square.
- All of that over 2, and so this is going to be equal to
- negative 4 plus or minus 10 over 2.
- We could just divide both of these terms by 2 right now.
- So this is equal to negative 4 divided by 2 is negative 2
- plus or minus 10 divided by 2 is 5.
- So that tells us that x could be equal to negative 2 plus 5,
- which is 3, or x could be equal to negative 2 minus 5,
- which is negative 7.
- So the quadratic formula seems to have given us
- an answer for this.
- You can verify just by substituting back in that
- these do work, or you could even just try to factor this
- right here.
- You say what two numbers when you take their product, you
- get negative 21 and when you take their sum you get
- positive 4?
- So you'd get x plus 7 times x minus 3 is
- equal to negative 21.
- Notice 7 times negative 3 is negative 21, 7 minus 3 is
- positive 4.
- You would get x plus-- sorry it's not negative --21 is
- equal to 0.
- There should be a 0 there.
- So you get x plus 7 is equal to 0, or x minus
- 3 is equal to 0.
- X could be equal to negative 7 or x could be equal to 3.
- So it definitely gives us the same answer as factoring, so
- you might say, hey why bother with this crazy mess?
- And the reason we want to bother with this crazy mess is
- it'll also work for problems that are hard to factor.
- And let's do a couple of those, let's do some
- hard-to-factor problems right now.
- So let's scroll down to get some fresh real estate.
- Let's rewrite the formula again, just in case we haven't
- had it memorized yet. x is going to be equal to negative
- b plus or minus the square root of b squared minus 4ac,
- all of that over 2a.
- I'll supply this to another problem.
- Let's say we have the equation 3x squared plus 6x is equal to
- negative 10.
- Well, the first thing we want to do is get it in the form
- where all of our terms or on the left-hand side, so let's
- add 10 to both sides of this equation.
- We get 3x squared plus the 6x plus 10 is equal to 0.
- And now we can use a quadratic formula.
- So let's apply it here.
- So a is equal to 3.
- That is a, this is b and this right here is c.
- So the quadratic formula tells us the
- solutions to this equation.
- The roots of this quadratic function, I guess
- we could call it.
- x is going to be equal to negative b.
- b is 6, so negative 6 plus or minus the
- square root of b squared.
- b is 6, so we get 6 squared minus 4 times a, which is 3
- times c, which is 10.
- Let's stretch out the radical little bit, all of that over 2
- times a, 2 times 3.
- So we get x is equal to negative 6 plus or minus the
- square root of 36 minus-- this is interesting --minus 4 times
- 3 times 10.
- So this is minus-- 4 times 3 times 10.
- So this is minus 120.
- All of that over 6.
- So this is interesting, you might already realize why it's
- interesting.
- What is this going to simplify to?
- 36 minus 120 is what?
- That's 84.
- We make this into a 10, this will become an
- 11, this is a 4.
- It is 84, so this is going to be equal to negative 6 plus or
- minus the square root of-- But not positive 84, that's if
- it's 120 minus 36.
- We have 36 minus 120.
- It's going to be negative 84 all of that 6.
- So you might say, gee, this is crazy.
- What a this silly quadratic formula you're
- introducing me to, Sal?
- It's worthless.
- It just gives me a square root of a negative number.
- It's not giving me an answer.
- And the reason why it's not giving you an answer, at least
- an answer that you might want, is because this will have no
- real solutions.
- In the future, we're going to introduce something called an
- imaginary number, which is a square root of a negative
- number, and then we can actually express this in terms
- of those numbers.
- So this actually does have solutions, but they involve
- imaginary numbers.
- So this actually has no real solutions, we're taking the
- square root of a negative number.
- So the b squared with the b squared minus 4ac, if this
- term right here is negative, then you're not going to have
- any real solutions.
- And let's verify that for ourselves.
- Let's get our graphic calculator out and let's graph
- this equation right here.
- So, let's get the graphs that y is equal to-- that's what I
- had there before --3x squared plus 6x plus 10.
- So that's the equation and we're going to see where it
- intersects the x-axis.
- Where does it equal 0?
- So let me graph it.
- Notice, this thing just comes down and then goes back up.
- Its vertex is sitting here above the x-axis and it's
- upward-opening.
- It never intersects the x-axis.
- So at no point will this expression, will this
- function, equal 0.
- At no point will y equal 0 on this graph.
- So once again, the quadratic formula seems to be working.
- Let's do one more example, you can ever see
- enough examples here.
- And I want to do ones that are, you know, maybe not so
- obvious to factor.
- So let's say we get negative 3x squared plus 12x plus 1 is
- equal to 0.
- Now let's try to do it just having the quadratic formula
- in our brain.
- So the x's that satisfy this equation are going to be
- negative b.
- This is b So negative b is negative 12 plus or minus the
- square root of b squared, of 144, that's b squared minus 4
- times a, which is negative 3 times c, which is 1, all of
- that over 2 times a, over 2 times negative 3.
- So all of that over negative 6, this is going to be equal
- to negative 12 plus or minus the square root
- of-- What is this?
- It's a negative times a negative so they cancel out.
- So I have 144 plus 12, so that is 156, right?
- 144 plus 12, all of that over negative 6.
- Now, I suspect we can simplify this 156.
- We could maybe bring some things out
- of the radical sign.
- So let's attempt to do that.
- So let's do a prime factorization of 156.
- Sometimes, this is the hardest part, simplifying the radical.
- So 156 is the same thing as 2 times 78.
- 78 is the same thing as 2 times what?
- That's 2 times 39.
- So the square root of 156 is equal to the square root of 2
- times 2 times 39 or we could say that's the square root of
- 2 times 2 times the square root of 39.
- And this, obviously, is just going to be the square root of
- 4 or this is the square root of 2 times 2 is just 2.
- 2 square roots of 39, if I did that properly, let's
- see, 4 times 39.
- Yeah, it looks like it's right.
- So this up here will simplify to negative 12 plus or minus 2
- times the square root of 39, all of that over negative 6.
- Now we can divide the numerator and the denominator
- maybe by 2.
- So this will be equal to negative 6 plus or minus the
- square root of 39 over negative 3.
- Or we could separate these two terms out.
- We could say this is equal to negative 6 over negative 3
- plus or minus the square root of 39 over negative 3.
- Now, this is just a 2 right here, right?
- These cancel out, 6 divided by 3 is 2, so we get 2.
- And now notice, if this is plus and we use this minus
- sign, the plus will become negative and the negative will
- become positive.
- But it still doesn't matter, right?
- We could say minus or plus, that's the same thing as plus
- or minus the square root of 39 nine over 3.
- I think that's about as simple as we can get this answered.
- I want to make a very clear point of what I
- did that last step.
- I did not forget about this negative sign.
- I just said it doesn't matter.
- It's going to turn the positive into the negative;
- it's going to turn the negative into the positive.
- Let me rewrite this.
- So this right here can be rewritten as 2 plus the square
- root of 39 over negative 3 or 2 minus the square root of 39
- over negative 3, right?
- That's what the plus or minus means, it could be this or
- that or both of them, really.
- Now in this situation, this negative 3 will turn into 2
- minus the square root of 39 over 3, right?
- I'm just taking this negative out.
- Here the negative and the negative will become a
- positive, and you get 2 plus the square root
- of 39 over 3, right?
- A negative times a negative is a positive.
- So once again, you have 2 plus or minus the
- square of 39 over 3.
- 2 plus or minus the square root of 39 over 3 are
- solutions to this equation right there.
- Let verify.
- I'm just curious what the graph looks like.
- So let's just look at it.
- Let me clear this.
- Where is the clear button?
- So we have negative 3 three squared plus 12x plus 1 and
- let's graph it.
- Let's see where it intersects the x-axis.
- It goes up there and then back down again.
- So 2 plus or minus the square, you see-- The square root of
- 39 is going to be a little bit more than 6, right?
- Because 36 is 6 squared.
- So it's going be a little bit more than 6, so this is going
- to be a little bit more than 2.
- A little bit more than 6 divided by 2 is a little bit
- more than 2.
- So you're going to get one value that's a little bit more
- than 4 and then another value that should be a little bit
- less than 1.
- And that looks like the case, you have 1, 2, 3, 4.
- You have a value that's pretty close to 4, and then you have
- another value that is a little bit-- It looks close to 0 but
- maybe a little bit less than that.
- So anyway, hopefully you found this application of the
- quadratic formula helpful.

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