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Strategy in solving quadratic equations

Based on the initial form of a quadratic equation, we can determine which solution methods are and aren't appropriate. Created by Sal Khan.

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  • sneak peak green style avatar for user Patrick
    Completing The Square seems much faster and easier then plugging numbers into the Quadratic Equation.
    When should we use the Quadratic Equation over Completing The Square? or are both valid and it's just personal preference?
    (6 votes)
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  • sneak peak blue style avatar for user Stephen Earley
    At ,(or really at ) can't we just say that (x+3) = -1? Sal says that it wouldn't work to say that (x+3) = 0 because the equation is (x+3)(x+1) = -1, and just jumping to x = -3 wouldn't work because that only works for x+3 = 0. But couldn't we just say that x = -4, because x+3=-1 = -4+3=-1?

    So we just say that x = -4 or x = -2 instead of x = -3 or x = -1?

    Does this make sense?
    (5 votes)
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    • aqualine ultimate style avatar for user CreatorOfBob
      It won't work because if x=-4, then it will be (4+3)(4+1)=-1, and 7*5 is not -1. The easiest way is to use the 0 property when factoring, because 0 times anything is 0. If not, there's going to me lots of guessing and error.

      Hope this helps!
      (1 vote)
  • blobby blue style avatar for user ✨ Sofia Utama 💯
    At about in the video for the third strategy in solving a quadratic equation, why doesn't Sal add -1 to both sides, instead of right off the bat, factoring the left hand side?
    LATER: Actually, nevermind. He goes on to say that the third example needs to be equal to 0 in order to get the solutions.

    What is the zero product property? (Sal mentions this at the last few seconds of the video.)
    (2 votes)
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  • starky tree style avatar for user D -_-
    how does

    (x+1)^2 turn into

    x^2+2x+1?

    isn't it x^2+1?

    I suck at math...
    (1 vote)
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    • mr pink green style avatar for user David Severin
      So first expand it to get (x+1)(x+1).
      FOIL method says First (x^2) + outside (1x) + inside (1x) + last (1) adding gives x^2+2x+1.
      Double distribution gives x(x+1)+1(x+1) which also gives x^2+1x+1x+1.
      So you will always end up with a middle term unless the outside and inside are additive inverses. One example is difference of perfect squares (x+1)(x-1) where you have x^2+1x-1x-1.
      You can also see that if you put -1 (which is a y intercept of the function y=(x+1)^2) would give you (-1+1)^2=0. However with your equation, you could not have a x intercept because x^2 has to be greater than or equal to 0, and if you add 1, the minimum would be where x^2=0, but 0^2+1=1. If you put in -1, you get (-1)^2 + 1 = 1+1=2. With x^+2x+1, you get (-1)^2 + 2(-1) + 1=1 -2+1=0 which again is your x intercept in a function.
      (3 votes)
  • sneak peak green style avatar for user Caleb Doughton
    At about , why can't you take x^2=-4x -4, and take the square root for

    x=+-2x+-2

    And then add or subtract the 2x from the right side to make either

    3x=+-2 (and divide both sides by 3)
    or
    -1x=+-2
    ?

    Thanks!
    (1 vote)
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  • leafers ultimate style avatar for user Marvyn Greco
    so for x^2+5x-3=-2+5x I first cancel the 5s, and then I work my way to the answer being x=±√1. Is that correct?
    (1 vote)
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  • sneak peak yellow style avatar for user level1
    How is it that (x + 1)(x - 1) = 8 multiplies to x^2 - 1 = 8 instead of (x - 1)^2 = 8?
    (1 vote)
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  • purple pi teal style avatar for user Whan Whan
    Sal, at why couldn't you just take the square root of -4x-4 and put a plus or minus sign in front? Please give me a answer.
    (1 vote)
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Video transcript

- [Instructor] In this video, we're gonna talk about a few of the pitfalls that someone might encounter while they're trying to solve a quadratic equation like this. Why is it a quadratic equation? Well, it's a quadratic because it has this second degree term right over here and it's an equation because I have something on both sides of an equal sign. So one strategy that people might try is, well, I have something squared, why don't I just try to take the square root of both sides? And if you did that, you would get the square root of x squared plus four x plus three is equal to the square root of negative one. And you immediately see a few problems. Even if this wasn't a negative one here, that's the most obvious problem. But even if this was a positive value here, how do you simplify or how do you somehow isolate the x over here? You've pretty quickly hit a dead end. So just willy nilly, taking the square root of both sides of a quadratic is not going to be too helpful. So I'll put a big X over there. Another strategy that sometimes people will try to go for is to isolate the x squared first. So you could imagine, let me just rewrite it. X squared plus four x plus three is equal to negative one. They might say, let's isolate that x squared by subtracting four x from both sides and subtracting three from both sides. And then what happens? On the left hand side, you do indeed isolate the x squared, and on the right hand side, you get negative four x minus four. And now, someone might say, if I take the square root of both sides, I could get, I'll just write that down. Square root of x squared is equal to, and you could try to take the plus of minus of one side to make sure you're hitting the negative roots. Negative four x minus four. And you could get something like this, you would get x is equal to plus or minus the square root of negative four x minus four, but this still doesn't help you. You still don't know what x is, and it's really not clear what to do with this algebraically. So this is yet another dead end. Now, there's some cases in which this strategy would have worked. In fact, it would have worked if you did not have this first degree term. If you did not have this x term, so to speak. Then this strategy would have worked assuming that there are some solutions. But if you have an x term like this and it doesn't cancel out somehow, you know, if there was another four x on the other side, then you could subtract four x from both sides, and they would disappear. But if can't make these things disappear, this strategy that I've just outlined is not going to be a productive one. Now another strategy that you'll sometimes see people use, especially when they see something like this, let me rewrite it. X squared plus four x plus three is equal to negative one. They immediately go into factoring mode. They say, hey, wait, I think I might be able to factor this. I can think of two numbers that add up to four and whose product is three. Maybe three and one. And then they immediately factor this left hand expression, say that's going to be x plus three times x plus one, and then that's going to be equal to negative one. And then they either are about to make a mistake, this is actually algebraically valid. But they either make a mistake or they realize that they're at a dead end. Because just saying that something times something is equal to negative one doesn't help you a ton. Because it's not clear yet, how you'd solve for x. Another thing, try to do is, is they'll immediately say, okay, therefore x is equal to negative three or x is equal to negative one because negative three will make this first term zero and negative one, or negative one would make the second term zero. But remember, this would only be true if you're multiplying two things and you got zero as their product, then the solutions would be anything that made either one of those terms equal to zero. But that's not what we're dealing with here. Here we're taking the product is equal to negative one. So in order to factor like this and make headway in most cases, you're going to wanna have a zero on the right hand side over here. And that's also true if you're trying to apply the quadratic formula. A lot of folks would say, okay, I see a quadratic equation right over here. Let me just apply the quadratic formula. They say, if I have something of the form ax squared plus bx plus c is equal to zero. The quadratic formula would say that the roots are gonna be x is equal to negative b plus or minus the square root of b squared minus four ac, all of that over two a. And so they'll immediately say, all right, I can recognize a here, as just being a one, there's a one coefficient implicitly there, b is four, c is three. And they'll say, okay, x is equal to negative four plus or minus the square root of b squared, which is 16, minus four times one, times three, all of that over two times one. But there's a problem. The quadratic formula applies when the left hand side is equal to zero. That's not what we have over here. So we're falling into that same pitfall. So everything I just did, none of this is a good idea. So the way to approach this, if you want zero over here, you wanna add one on the right hand side, and if you wanna maintain the equality, you have to add a one on the left hand side. And so you're going to get x squared plus four x plus four is equal to zero. And now you could use the quadratic formula or you could factor. You might recognize two plus two is equal to four. Two times two is equal to four. So you could say x plus two times x plus two is equal to zero. And so in this case, you say, all right, x could be equal to negative two or x could be equal to negative two. (chuckles) So this one only has one solution, x is equal to negative two. But the key is to recognize that you need the zero on the right hand side there, if you wanna use a quadratic formula or if you wanna use factoring and the zero product property.