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Quadratic equation part 2 | Quadratic equations | Algebra I | Khan Academy

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    Welcome to part two of
    the presentation on
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    quadratic equations.
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    Well, I think I thoroughly
    confused you the last time
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    around, so let me see if I
    can fix that a bit by doing
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    several more examples.
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    So let's just start with
    a review of what the
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    quadratic equation is.
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    The quadratic equation says, if
    I'm trying to solve the
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    equation Ax squared plus Bx
    plus C equals 0, then the
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    solution or the solutions
    because there's usually two
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    times that it intersects the
    x-axis, or two solutions for
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    this equation is x equals minus
    B plus or minus the square root
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    of B squared minus
    4 times A times C.
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    And all of that over 2A.
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    So let's do a problem and
    hopefully this should make
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    a little more sense.
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    That's a 2 on the bottom.
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    So let's say I had the equation
    minus 9x squared minus
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    9x plus 6 equals 0.
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    So in this example what's A?
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    Well, A is the coefficient
    on the x squared term.
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    The x squared term is here,
    the coefficient is minus 9.
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    So let's write that.
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    A equals minus 9.
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    What does B equal?
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    B is the coefficient on the x
    term, so that's this term here.
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    So B is also equal to minus 9.
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    And C is the constant term,
    which in this example is 6.
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    So C is equal to 6.
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    Now we just substitute these
    values into the actual
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    quadratic equation.
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    So negative B, so it's
    negative times negative 9.
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    That's B.
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    Plus or minus the square root
    of B squared, so that's 81.
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    Right?
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    Negative 9 squared.
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    Minus 4 times negative 9.
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    That's A.
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    Times C, which is 6.
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    And all of that over 2
    times negative 9, which
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    is minus 18, right?
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    2 times negative 9-- 2A.
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    Let's try to simplify
    this up here.
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    Well, negative negative
    9, that's positive 9.
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    Plus or minus the
    square root of 81.
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    Let's see.
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    This is negative 4
    times negative 9.
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    Negative 4 times negative
    9 is positive 36.
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    And then positive 36
    times 6 is-- let's see.
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    30 times 6 is 180.
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    And then 180 plus
    another 36 is 216.
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    Plus 216, is that right?
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    180 plus 36 is 216.
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    All of that over 2A.
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    2A we already said is minus 19.
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    So we simplify that more.
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    That's 9 plus or minus the
    square root 81 plus 216.
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    That's 80 plus 217.
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    That's 297.
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    And all of that over minus 18.
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    Now, this is actually-- the
    hardest part with the quadratic
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    equation is oftentimes just
    simplifying this expression.
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    We have to figure out if we
    can simplify this radical.
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    Well, let's see.
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    One way to figure out if a
    number is divisible by 9 is to
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    actually add up the digits
    and see if the digits
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    are divisible by 9.
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    In this case, it is.
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    2 plus 9 plus 7 is equal to 18.
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    So let's see how many
    times 9 goes into that.
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    I'll do it on the side here; I
    don't want to be too messy.
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    9 goes into 2 97.
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    3 times 27.
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    27-- it goes 33 times, right?
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    So this is the same thing as 9
    plus or minus the square root
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    of 9 times 33 over minus 18.
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    And 9 is a perfect square.
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    That's why I actually wanted to
    see if 9 would work because
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    that's the only way I could get
    it out of the radical, if
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    it's a perfect square.
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    As you learned in that exponent
    rules number one module.
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    So this is equal to 9 plus
    or minus 3 times the square
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    root of 33, and all of
    that over minus 18.
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    We're almost done.
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    We can actually simplify it
    because 9, 3, and minus 18
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    are all divisible by 3.
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    Let's divide everything by 3.
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    3 plus or minus the square
    root of 33 over minus 6.
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    And we are done.
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    So as you see, the hardest
    thing with the quadratic
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    equation is often just
    simplifying the expression.
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    But what we've said, I know you
    might have lost track-- we did
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    all this math-- is we said,
    this equation: minus 9x
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    squared minus 9x plus 6.
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    Now we found two x values that
    would satisfy this equation
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    and make it equal to 0.
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    One x value is x equals
    3 plus the square root
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    of 33 over minus 6.
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    And the second value is
    3 minus the square root
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    of 33 over minus 6.
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    And you might want to
    think about why we have
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    that plus or minus.
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    We have that plus or minus
    because a square root could
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    actually be a positive
    or a negative number.
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    Let's do another problem.
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    Hopefully this one will
    be a little bit simpler.
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    Let's say I wanted to
    solve minus 8x squared
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    plus 5x plus 9.
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    Now I'm going to assume that
    you've memorized the quadratic
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    equation because that's
    something you should do.
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    Or you should write it
    down on a piece of paper.
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    But the quadratic equation is
    negative B-- So b is 5, right?
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    We're trying to solve that
    equal to 0, so negative B.
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    So negative 5, plus or minus
    the square root of B squared-
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    that's 5 squared, 25.
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    Minus 4 times A,
    which is minus 8.
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    Times C, which is 9.
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    And all of that over 2 times A.
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    Well, A is minus 8, so all
    of that is over minus 16.
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    So let's simplify this
    expression up here.
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    Well, that's equal to
    minus 5 plus or minus
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    the square root of 25.
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    Let's see.
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    4 times 8 is 32 and the
    negatives cancel out, so
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    that's positive 32 times 9.
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    Positive 32 times 9, let's see.
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    30 times 9 is 270.
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    It's 288.
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    I think.
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    Right?
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    288.
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    We have all of that
    over minus 16.
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    Now simplify it more.
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    Minus 5 plus or minus the
    square root-- 25 plus
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    288 is 313 I believe.
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    And all of that over minus 16.
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    And I think, I'm not 100% sure,
    although I'm pretty sure.
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    I haven't checked it.
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    That 313 can't be factored
    into a product of a perfect
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    square and another number.
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    In fact, it actually
    might be a prime number.
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    That's something that you
    might want to check out.
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    So if that is the case and
    we've got it in completely
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    simplified form, and we say
    there are two solutions, two
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    x values that will make
    this equation true.
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    One of them is x is equal
    to minus 5 plus the square
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    root of 313 over minus 16.
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    And the other one is x is equal
    to minus 5 minus the square
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    root of 313 over minus 16.
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    Hopefully those two examples
    will give you a good
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    sense of how to use the
    quadratic equation.
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    I might add some more modules.
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    And then, once you master this,
    I'll actually teach you how to
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    solve quadratic equations when
    you actually get a negative
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    number under the radical.
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    Very interesting.
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    Anyway, I hope you can do the
    module now and maybe I'll add a
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    few more presentations because
    this isn't the easiest module.
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    But I hope you have fun.
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    Bye.
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Title:
Quadratic equation part 2 | Quadratic equations | Algebra I | Khan Academy
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Video Language:
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Duration:
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