How To Solve Quadratic Equations By Factoring - Quick & Simple! | Algebra Online Course

00:12:29
https://www.youtube.com/watch?v=qeByhTF8WEw

概要

TLDRThis video lesson explains how to solve quadratic equations by factoring. It covers various techniques, such as using the difference of perfect squares and factoring trinomials, and provides detailed examples. For instance, it demonstrates how to solve x^2 - 49 = 0 using the difference of squares method, and how to factor equations by extracting the greatest common factor and using grouping methods when coefficients are different. The lesson also includes examples of solving quadratic equations using the quadratic formula.

収穫

  • 🔢 Identify quadratic equations that can be factored using different techniques.
  • 🧮 The difference of squares method can be quickly applied to equations like x^2 - 49 = 0.
  • 📐 Factoring helps to simplify and solve quadratic equations.
  • 📝 Each factor in a factored equation can be set to zero to find potential solutions.
  • 🔍 Understanding the GCF is crucial for simplifying before factoring.
  • 🧮 Factoring trinomials often requires finding two numbers that multiply and add to specific values.
  • 🔢 Use of the quadratic formula can verify solutions found by factoring.
  • 🔄 Always express the quadratic equation in standard form before applying formulas.
  • 🎯 Practice regularly improves recognition of factoring pattern.
  • 🧮 Pay attention to coefficients to determine the appropriate method to factor.
  • 📝 Grouping is useful to manage more complex polynomials.
  • 🔍 Validate solutions obtained with alternative methods like the quadratic formula.

タイムライン

  • 00:00:00 - 00:12:29

    In the lesson, quadratic equations are solved using factoring. Demonstrations involve using the difference of perfect squares and factoring trinomials. For example, the equation x² - 49 = 0 is factored to (x + 7)(x - 7), providing solutions x = -7 and x = 7. Another example, 3x² - 75 = 0, involves extracting the greatest common factor, leading to factoring x² - 25 into (x + 5)(x - 5) with solutions x = -5 and x = 5. Additionally, for trinomials like x² - 2x - 15, numbers that multiply to -15 and add to -2 are identified as -5 and 3, resulting in factors (x - 5)(x + 3) yielding solutions x = 5 and x = -3.

マインドマップ

ビデオQ&A

  • What is the difference of perfect squares?

    The difference of perfect squares is a technique used to factor expressions like x^2 - n^2 into (x+n)(x-n).

  • How do you factor equations where the leading coefficient is not 1?

    Equations with a leading coefficient other than 1 can be factored by multiplying the leading coefficient with the constant term, finding two numbers that multiply to this product and add to the middle term, then using grouping to factor.

  • How do you solve x^2 - 49 = 0 by factoring?

    Use the difference of perfect squares: x^2 - 49 = (x+7)(x-7), then set each factor to zero to find x = -7 and x = 7.

  • How do you find the Greatest Common Factor (GCF)?

    The GCF is the largest number that divides all terms in the expression. For example, the GCF of 3x^2 - 75 is 3.

  • How do you solve 3x^2 - 75 = 0 after factoring?

    Factor out the GCF (3), then factor the resulting quadratic as a difference of squares: (x+5)(x-5), set each factor to zero to find x = -5 and x = 5.

  • What are the solutions to 9x^2 - 64 = 0?

    Factor using the difference of squares: (3x+8)(3x-8), leading to solutions x = -8/3 and x = 8/3.

  • How do you factor a trinomial like x^2 - 2x - 15?

    Find two numbers that multiply to -15 and add to -2, which are -5 and 3, then factor into (x-5)(x+3).

  • How is the quadratic formula used to solve quadratic equations?

    Use the formula x = [-b ± √(b²-4ac)] / (2a) on the equation in standard form to find its roots.

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  • 00:00:01
    in this lesson we're going to talk about
  • 00:00:02
    solving quadratic equations by factoring
  • 00:00:07
    so let's start with this example x
  • 00:00:08
    squared minus 49 is equal to zero
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    you can use the difference of perfect
  • 00:00:13
    squares technique for this one
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    the square root of x squared is x
  • 00:00:17
    the square root of 49 is seven
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    so it's going to be x plus seven
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    and x minus seven
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    now you need to set each factor equal to
  • 00:00:28
    zero at this point and then you could
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    find the value of x
  • 00:00:32
    so we have x plus seven is equal to zero
  • 00:00:35
    and x minus seven is equal to zero
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    the reason why we can do that is because
  • 00:00:39
    if one of these terms is equal to zero
  • 00:00:41
    then everything is zero zero times
  • 00:00:44
    anything is zero
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    so x
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    is equal to negative seven
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    and in the other equation if we add
  • 00:00:55
    seven to both sides we could see that x
  • 00:00:59
    is equal to positive 7.
  • 00:01:04
    let's try another example
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    let's say if we have 3x squared
  • 00:01:10
    minus 75
  • 00:01:12
    is equal to zero
  • 00:01:13
    what is the value of x
  • 00:01:16
    3 and 75
  • 00:01:18
    are not perfect squares
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    so we don't want to use the difference
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    of perfect squares technique yet however
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    we can take out the gcf the greatest
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    common factor
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    which is three three x squared divided
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    by three
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    is x squared
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    negative seventy-five divided by three
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    is negative twenty-five
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    now we can use the difference of perfect
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    squares technique
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    to factor x squared minus 25.
  • 00:01:43
    the square root of x squared is x the
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    square root of 25 is 5.
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    so it's going to be x plus 5 and x minus
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    5.
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    so if we set x plus five equal to zero
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    we can clearly see that x
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    will be equal to negative five and if we
  • 00:02:00
    set x minus five equal to zero
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    x is equal to plus five
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    and so that's it for that one
  • 00:02:07
    now what about this one let's say if we
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    have 9x squared
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    minus 64
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    is equal to zero
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    well first
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    we can use the difference of perfect
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    squares technique we can square root 9
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    and we can square root 64.
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    the square root of 9 is 3.
  • 00:02:29
    the square root of x squared is x
  • 00:02:31
    the square root of 64 is 8.
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    so it's going to be 3x plus 8 3x minus
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    8.
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    so if we set 3x plus 8 equal to 0
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    then we can see that 3x is equal to
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    negative 8 which means x is equal to
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    negative eight over three
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    now if we set three x minus eight equal
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    to zero and solve for x
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    x is gonna be positive eight over three
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    using the same steps
  • 00:03:02
    now what if we have a trinomial
  • 00:03:05
    x squared minus 2x minus 15.
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    and the leading coefficient is one how
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    can we factor this expression
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    all you need to do is find two numbers
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    that multiply to negative 15 but that
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    adds to negative two
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    numbers that multiply to fifteen are
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    five and three
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    so we have positive five and negative
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    three or negative five and three
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    five plus negative three adds up to
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    positive two but negative 5 plus 3 adds
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    up to negative 2.
  • 00:03:34
    so this is what we want to use
  • 00:03:36
    it turns out that
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    to factor it it's simply going to be x
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    minus
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    5 plus
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    x plus 3.
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    so if we set x minus five equal to zero
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    x will be equal to five
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    and if we set x plus three equal to zero
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    x will be equal to negative three
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    let's try another one like that
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    let's say if we have x squared plus
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    3x
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    minus 28
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    so what two numbers multiply to negative
  • 00:04:08
    28 but add to three
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    go ahead and try it
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    so if we divide 28 by 1
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    we'll get negative 28 if we divide
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    negative 28 by 2
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    negative 14 3 doesn't go into it if we
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    divide it by 4 we'll get negative 7. 4
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    and negative 7 differs by three if we
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    add them it's negative three
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    so we need to change the sign
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    so it's going to be x minus four times x
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    plus seven
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    which means that x
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    is equal to positive four
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    and negative seven
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    here's another problem
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    so how can we factor this trinomial when
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    the leading coefficient is not one
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    so what we need to do in this problem we
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    need to multiply eight and negative
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    fifteen
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    eight times negative 15
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    is negative 120.
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    now what two numbers multiply to
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    negative 120 but add to two
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    if you're not sure make a list
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    let's start with one we have one in 120
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    two and sixty
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    three and forty
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    four and thirty
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    five and twenty four
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    six and twenty
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    eight and fifteen
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    now 10 and 12 seem promising
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    10 and negative 12 differ by negative 2
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    but positive 12 and negative 10
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    adds up to positive 2.
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    so what we're going to do in this
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    problem is we're going to replace 2x
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    with 12x and negative 10x
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    and then factor by grouping
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    in the first two terms let's take out
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    the gcf
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    which is going to be 4x
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    8x squared divided by 4x
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    is 2x
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    and 12x divided by 4x
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    is 3.
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    and the last two terms
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    take out the greatest common factor in
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    this case negative 5.
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    negative 10x divided by negative 5
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    is 2x
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    negative 15 divided by negative 5 that's
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    plus 3.
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    now if you get two common terms
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    that means you're on the right track you
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    can write it once
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    in
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    a parenthesis in the next line
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    now the stuff on the outside 4x and
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    negative 5 that's going to go in the
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    second parentheses
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    so that's what we have
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    now let's set two x plus three equal to
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    zero
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    and 4x minus 5 equal to 0.
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    so in the first equation let's subtract
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    3 from both sides
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    so 2x is equal to negative 3.
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    and then let's divide by 2.
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    so the first answer x is equal to
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    negative three over two
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    now let's find the other answer
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    so let's add five to both sides
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    so we can see that four x is equal to
  • 00:07:27
    five
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    and then let's divide both sides by four
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    so x
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    is equal to five over four
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    and that's it for this problem
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    now let's get some of the answers
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    to the quadratic equations that we had
  • 00:07:47
    in the last lesson
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    so for this particular problem
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    when we factor it we got a solution of 5
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    and negative 3
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    in less than 10.2
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    but now let's use the quadratic equation
  • 00:08:03
    to get those same answers
  • 00:08:06
    so x is equal to negative b
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    plus or minus the square root of b
  • 00:08:12
    squared minus 4ac
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    divided by 2a that's the quadratic
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    formula
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    and you need the quadratic equation in
  • 00:08:21
    standard form
  • 00:08:24
    so we can see that a is equal to one
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    b
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    is the number in front of x b is
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    negative two
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    and c
  • 00:08:34
    is negative fifteen
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    so let's replace b with negative two
  • 00:08:42
    b squared or negative two squared
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    negative two times negative two is four
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    a is one and c is negative fifteen
  • 00:08:51
    divided by two a or two times one which
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    is two
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    negative times negative two is positive
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    two
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    and then we have four negative four
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    times negative fifteen that's positive
  • 00:09:03
    sixty
  • 00:09:04
    and sixty plus four is sixty-four
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    now the square root of sixty-four is
  • 00:09:10
    eight
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    so we have two plus or minus eight
  • 00:09:13
    divided by two
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    two plus eight is ten
  • 00:09:18
    ten divided by two is five that gives us
  • 00:09:20
    the first answer the next one is two
  • 00:09:23
    minus eight divided by two
  • 00:09:26
    two minus eight is negative six negative
  • 00:09:29
    six divided by two is negative three
  • 00:09:31
    which gives us the second answer
  • 00:09:34
    so you can solve a quadratic equation by
  • 00:09:36
    factoring
  • 00:09:38
    or by using the quadratic formula
  • 00:09:41
    now let's try another example
  • 00:09:43
    eight x squared plus two x
  • 00:09:47
    minus fifteen
  • 00:09:49
    use the quadratic equation to find the
  • 00:09:50
    values of x
  • 00:09:53
    so we can see that a is equal to eight
  • 00:09:56
    b is the number in front of x that's two
  • 00:09:59
    c is negative 15.
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    so using the quadratic formula x equals
  • 00:10:04
    negative b
  • 00:10:06
    plus or minus the square root
  • 00:10:08
    of b squared minus 4ac
  • 00:10:11
    divided by 2a
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    so b is 2
  • 00:10:16
    which means b squared that's going to be
  • 00:10:18
    positive 4
  • 00:10:20
    minus 4 times a a is 8 c is negative 15
  • 00:10:25
    divided by 2 a or 2 times 8 which is 16.
  • 00:10:30
    so this is negative 2 plus or minus
  • 00:10:32
    square root
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    4.
  • 00:10:34
    now negative 4
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    times
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    negative 15 is positive 60 60 times 8
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    that's 480
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    so we have 4 plus four eighty
  • 00:10:47
    so this is negative two plus or minus
  • 00:10:50
    the square root of four hundred and
  • 00:10:51
    eighty four
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    the square root of four eighty four is
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    twenty two
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    so now we have negative 2
  • 00:11:00
    plus or minus
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    22 over 16.
  • 00:11:04
    so now what we're going to do at this
  • 00:11:05
    point is separate that into two
  • 00:11:07
    fractions
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    but let's just uh
  • 00:11:10
    let's make some space first
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    so this is negative 2
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    plus 22 over 16
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    or
  • 00:11:21
    negative 2 minus 22 over 16.
  • 00:11:24
    negative 2 plus 22 that's positive
  • 00:11:27
    twenty
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    and twenty over sixteen
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    both numbers are divisible by four
  • 00:11:34
    twenty divided by four is five
  • 00:11:36
    sixteen divided by four is four
  • 00:11:39
    so the first answer is
  • 00:11:41
    five divided by four
  • 00:11:43
    negative two minus twenty two
  • 00:11:45
    is negative twenty four
  • 00:11:47
    twenty four and sixteen are both
  • 00:11:50
    divisible by eight
  • 00:11:53
    negative twenty four divided by eight is
  • 00:11:55
    negative 3
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    16 divided by 8 is 2.
  • 00:11:58
    and so that's the other answer
  • 00:12:00
    negative 3 over 2.
  • 00:12:02
    so now you know how to use the quadratic
  • 00:12:04
    formula to solve quadratic equations
  • 00:12:28
    you
タグ
  • quadratic equations
  • factoring
  • difference of squares
  • trinomial
  • quadratic formula
  • GCF
  • roots
  • equations
  • mathematics
  • algebra