WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today we are going to be discussing the quadratic formula.
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In previous lessons, we talked about solving quadratic equations through methods such as graphing and completing the square.
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However, those methods have certain limitations that the quadratic formula does not have.
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Let's go ahead and take a look at this.
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Now, notice that the solutions of this quadratic equation are given by this formula, the quadratic formula.
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However, this specifies that the equation needs to be in standard form.
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So, make sure you have your equation in standard form before you use the quadratic formula.
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Given that it is in standard form, and has certain values of a, b, and c, this formula will give you the solutions for this equation.
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Now, you need to make sure that you know this well; just memorize it.
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And then, apply it to equations in standard form.
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Now, an example would be if I was given 2x² + x = 2.
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This is not in standard form; so first, put it in standard form by subtracting 2 from both sides.
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Now, I like to write out what a, b, and c are; that way, I can just plug them in without having to worry about making errors looking back at the equation.
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So, here a equals 1; b equals 1; and c equals -2.
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Now, using the quadratic equation, x = -b±√(b² [1 squared] - 4a (which is 2) c (which is -2), divided by 2a (which is 2).
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OK, this gives me x = -1 ±...1 squared is just 1, minus...4 times 2 is 8, times -2 gives me -16; 2 times 2 is 4.
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Here, x equals -1, plus or minus...1 minus -16 is actually positive 17, over 4.
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Therefore, x equals (-1 ±√17)/4.
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You can leave it like this, or you could break it out into x = (-1 + √17)/4, or x = (-1 - √17)/4.
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So, there are two solutions; and you found them by putting the equation in standard form, and then using the quadratic formula.
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Now, in the example I just showed you, there were actually two roots, or two solutions, to that equation.
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However, if b² - 4ac equals 0, the equation has only one rational root.
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So, let's take a look at the quadratic equation again, and figure out why that is so.
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Let's apply this to an example, x² + 2x + 1 = 0.
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This is already in standard form, so I don't need to worry about changing it at all.
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Let's take a look; and I have a = 1, b = 2, and c = 1.
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Now, notice that this b² - 4ac is just what is here under the radical sign.
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And this is called the discriminant; and we will talk more about that in a minute.
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But for right now, let's just look at it and realize that it is what is under the radical sign.
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Let's actually try to solve this out: since a equals 1, I am going to get -1 ± the square root
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of b² (which is 2²), minus 4 times a times c, over 2 times a.
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OK, what I end up with here is 4 - 4, which is 0.
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So, looking at this, when I plugged in my values, I got -1 plus or minus √(4 - 4), over 2; so that is -1 plus or minus √0, over 2.
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Well, this would come out to -1 ± 0, over 2, which just equals -1/2.
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There is only one solution--one rational root, or one rational solution.
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And the reason is because, if b² - 4ac = 0, this becomes 0.
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And the reason you have two solutions is that you take this -b/2a, and then it is plus or minus this.
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And you would get two solutions if, for example, b² - 4ac is 4.
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And if I had plus or minus 4, that is going to give me something plus 2, and then something minus 2--two different things.
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However, if I am talking about plus or minus the square root of 0, I am just going to get 0.
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I only have one number that I am going to come out with here.
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Therefore, if this discriminant, b² - 4ac, is 0, you are only going to get one root, or one solution, for your quadratic equation.
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Complex solutions may occur; and if they do, the complex solutions will actually be a pair of complex conjugates.
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Remember what a complex conjugate pair is, such as 2 + 3i and 2 - 3i; this would be an example of complex conjugates.
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Let's think about how this can occur, looking at the quadratic formula.
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I have the quadratic formula; and if this ends up being a negative number, you will end up with an imaginary number.
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And so, then you will have a real part and an imaginary part.
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And that is how you end up with solutions that are complex conjugates.
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For example, if I was given 2x² + 3x + 6 = 0, then I have a = 2, b = 3, and c = 6.
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Applying the quadratic formula to this, I am going to get -3 ± √(b²,
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which is 3², - 4 times 2 (which is a), times 6 (which is c)), all over 2 times a, which is 2.
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So, x equals -3, plus or minus 9; and 4 times 2 is 8, times 6; that is 48; so 9 minus 48, all over 4.
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This equals -3 plus or minus the square root of -39.
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You can see, right away, that this is not going to give you a real number, because it is the square root of a negative number.
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We can simplify this further by recalling that this equals this, and also recalling that the square root of -1 equals i.
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This simplifies, therefore, to (i√39)/4.
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OK, now I said that the solution is a pair of complex conjugates.
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And if we would look at this carefully, I do have a set of complex conjugates.
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because I have x = -3, plus i√39, over 4, and x = -3 minus i√39, all of this over 4.
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Complex solutions occur when this under the radical ends up being negative.
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So then, you end up with an imaginary part and a real part to this number.
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And since we have plus or minus what is under here, it ends up being a pair of complex conjugates.
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OK, I already mentioned the discriminant; and the discriminant is that expression under the square root sign.
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The discriminant is very helpful, because it can tell you the nature of the solutions of the equation.
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We already saw that; but looking a little more deeply into it, if the discriminant is positive, you get two solutions, and they are both real.
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Looking back up here at the quadratic formula, x = (-b ± √(b² - 4ac))/2a:
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this right here is the discriminant; so if this is positive, the square root of a positive number is a real number; so both solutions will be real.
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Now, we can break that down further--a subset.
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We have that, if the discriminant is positive, we get two solutions, and they are both real.
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If the discriminant is a perfect square, then the roots are rational.
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And that makes sense, because let's say I end up with a perfect square:
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if the discriminant here is a perfect square 4, then I am going to end up with 2.
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And then, I will have -b plus or minus 2, over 2a; so that makes sense.
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Now, if the discriminant is positive, but not a perfect square, then you get two irrational roots.
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For example, if I figured out my discriminant, b² - 4ac, and I ended up with something like 2,
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(it is under the square root sign--that is my discriminant), this is not a perfect square;
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well, that is an irrational number; so I am going to get -b ± √2, over 2a.
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So, I am going to have two solutions, and they are real numbers, but they are irrational--two irrational roots.
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OK, if the discriminant is negative, which we just saw, then there are no real solutions.
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Instead, you can get a pair of complex conjugates, which is what we just saw.
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Finally, if the discriminant is 0, then you get one real solution--one real root.
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We saw that earlier today, as well, because if this comes out to be 0, you are going to get -b/2a ±0, which is just -b/2a; there is only one solution.
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So, if the discriminant is positive, then you have two real solutions.
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If there is a perfect square under here, the roots are rational.
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If it is not a perfect square, then you can get two irrational roots.
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If the discriminant is negative, there are no real solutions, but you can get a set of complex conjugates as solutions.
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If the discriminant is 0, there is one real root.
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OK, let's solve this using the quadratic formula, recalling that it is x = -b ± √(b² - 4ac), all of that over 2a.
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Here, a is 2; b is 3; and c is -4.
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This gives me x = -3 ± √b² (so that is 3²) - 4(2)(-4), over 2(2).
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So, let's see what this equals: -3 plus or minus...3 times 3 is 9, minus...4 times 2 is 8, times -4 is -32, over 2 times 2, which is 4.
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This gives me 9 minus -32; let's just say + 32, over 4.
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So, it is -3 ± √9 + 32, over 4, which equals -3 ± √...9 + 32 is 41, over 4.
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OK, so what I have here are two roots; I have x = (-3 + √41)/4, and x = (-3 - √41)/4.
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And what I ended up with are two solutions; they are both real, but they are irrational,
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because what I have in the discriminant is not a perfect square.
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Since this is not a perfect square (it is positive, so it is real, but it is not a perfect square), I end up with irrational roots.
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OK, again, solve using the quadratic formula.
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The first thing I have to see is that this is not in standard form.
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So, I am going to put it into standard form by adding 18 to both sides.
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Now, it is in standard form, but I can actually simplify this further, because I have a common factor over here (on this left side) of 2.
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So, I am just going to divide both sides by 2 to make it simpler.
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And that is going to give me x² - 6x + 9 = 0.
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OK, now I have a = 1, b = -6, and c = 9; so by dividing, I am working with smaller numbers, which is always easier.
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Recall the quadratic formula: OK.
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So, let's substitute in our values here: x equals -b, so that is -(-6), plus or minus the square root of -6 squared,
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minus 4 times a times c, all over 2 times a.
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A negative and a negative gives me a positive; plus or minus the square root of -6 squared, which is 36, minus... 4 times 1 is 4, times 9 is 36, over 2.
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So, this equals 6, plus or minus the square root of...36 minus 36 is 0, over 2.
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Now, you can probably already see that the discriminant is 0 here; so I am only going to get one real root as a solution,
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because if I say 6 plus or minus the square root of 0 (that is 0) over 2, this just comes out to 6 over 2, which is 3.
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So, the solution here is that x equals 3.
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So, we started out with something that was not in standard form; I put this quadratic equation into standard form.
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And then, to make it simpler, I divided both sides by this greatest common factor that was on the left, which is 2.
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And then, I ended up with this: x² - 6x + 9.
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I used that to solve using the quadratic formula, and I quickly found out that this discriminant was 0.
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So, I ended up with one real solution, which is x = 3.
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All right, solve using the quadratic formula again.
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x equals -b, plus or minus the square root of b² minus 4ac, over 2a.
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Now, a equals 1; b equals 1--it is already in standard form; and c equals 1.
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So, these are nice, simple numbers that I am working with.
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x equals -b, so that is -1, plus or minus the square root of b squared (that is 1 squared)
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minus 4 times 1 times 1 again (minus 4ac), all over 2 times a, which is 1.
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So, x equals -1 plus or minus the square root of 1 minus 4, all over 2.
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So, x equals -1 plus or minus the square root of 1 - 4, which is -3.
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OK, so you can see--what happened here is that I ended up with a negative number, a negative discriminant.
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Because it is asking me to take the square root of a negative number, my solutions are going to be a pair of complex conjugates.
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There are no real solutions, but there are solutions.
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Recall that I can simplify this to say -1 plus or minus √-1, times √3, over 2.
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And I am going to recall that i equals the square root of -1.
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Therefore, x equals -1...I am going to change this into i...so I had the square root of -3;
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I broke that down into the square root of -1, times the square root of 3, which gives me x = -1 ± i (since this is equal to i) √3, all over 2.
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And I can rewrite that a slightly different way, and say x = -1 + i√3, over 2, and x = -1 - i√3, all over 2.
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So, looking at this, what I have here are solutions that are complex conjugates.
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Here, I am asked to find the discriminant and describe the nature of the solutions.
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So, remember that the discriminant equals b² - 4ac.
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This is in standard form, so here I have a = 2, b = 3, and c = 7.
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So, my discriminant is going to be 3² - 4(2)(7); so the discriminant equals 9...4 times 2 is 8, times 7 is 56; so the discriminant equals -47.
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Here, the discriminant is negative, so the solutions will be a pair of complex conjugates,
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just as we saw in the last problem, because the discriminant is negative, and you are going to have
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to take the square root of that as part of the quadratic equation.
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And that is going to end up giving you complex conjugates.
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OK, that concludes this lesson of Educator.com on the quadratic formula.
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And I will see you again next time.