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How To Approach Quadratic Equations With Brackets

  1. Jul 21, 2009 #1
    First of all, I searched the forums and couldn't really find one on how to approach a quadratic equation that has brackets.
    Here is the equation: y=(x+1)2 - 2
    My teacher told me to factorize by sight in pairs and i used this equation for it: ax2+bx+c.
    I think c=-2 but I don't know how to get "a" and "b". When I worked on the brackets it came to (x+1)2=2x+x2+1 and I don't know which is "a" and "b"
    2. Relevant equations. Oops! I forgot that the purpose is to find the turning point for the quadratic function. I really HATE the ones with brackets. If anyone can help me that would be very much appreciated.
  2. jcsd
  3. Jul 21, 2009 #2


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    When in the form

    y= a(x+h)2+k

    the turning point occurs when the part in brackets = 0. (so can you get the 'y' value at the minimum point?)

    Now, if a is positive, the curve has a minimum point
    if a is negative then curve has a maximum point.
  4. Jul 21, 2009 #3
    Huh? I'm not sure what you mean but I know that if "a" is positive, the curve has a minimum point and if "a" is negative then curve has a maximum point. I just need to know "a" and "b" from the equation y=(x+1)2-2 or if (x+1)2= 2x+x2+1. Forgive me if you don't understand what I'm saying, I'm learning it the Homo way. I know there is a much shorter way of doing it but I'm not allowed to do that.
  5. Jul 21, 2009 #4


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    Well, if you substitute (x+1)2= 2x+x2+1 into y=(x+1)2-2 what do you get? Can you simplify that so it obtains the form ax2 + bx + c ?
  6. Jul 21, 2009 #5
    Huh? my answers were -1,-2 and the answers in the book are -1,2
    How did I get a -2 instead of 2?
    I subbed the x into the equation y=(x+1)2-2 by using the formula x=-b/2a to find x.
    I got the x right, but not the y.
  7. Jul 21, 2009 #6


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    There are a few ways to do this. There is an easy way for this kind of question with "the brackets", but you might get confused since it isn't the usual method that you would use to solve quadratics.

    Ok so I'll try and explain it to the best of my ability the simple way first:

    You have [tex]y=(x+1)^2-2[/tex]

    Now, you know when a number is squared [tex]a^2[/tex] the value will always be zero or positive, because whether "a" is positive or negative, square it and it becomes positive.

    This means that the [tex](x+1)^2[/tex] will always be 0 or more. When is it the smallest though? Well that happens when the part inside the brackets is equal to zero. So [tex]x+1=0[/tex] and thus [tex]x=-1[/tex]

    Now since every other value of x will make the part inside the bracket not zero, when you square this non-zero number it will be positive. This means that the y value will grow larger as x moves away from -1. So that means the parabola has a minimum turning point when the y value is the smallest. And this happens when [tex]x=-1[/tex]. So what is the smallest y value? Well when [tex]x=-1[/tex] the bracket part is 0 and all that is left is [tex]y=0-2=-2[/tex]

    So now you have the minimum turning point (-1,-2)

    If you're still reading up to this point, then you have determination, and I'll keep going :biggrin:
    As for the way the teacher was asking you to do it, which I think is not the way you were trying to solve the quadratic.
    Your teacher said "factorize by sight in pairs"


    Notice how this is equivalent to [tex]y=a^2-b^2[/tex]
    And if you factorize by difference of 2 squares: [tex]y=(a+b)(a-b)[/tex]

    We just need to find what a and b are.

    Well if [tex](x+1)^2-2\equiv a^2-b^2[/tex]

    Then [tex](x+1)^2=a^2[/tex]

    So [tex]x+1=a[/tex]

    And [tex]2=b^2[/tex]

    So [tex]b=\sqrt{2}[/tex]

    (don't worry about the [tex]\pm[/tex] here though, as you'll notice later)

    So substituting the a and b values into the general difference of 2 squares formula thingy:


    So you've now factorized the quadratic (don't worry, it will become way easier than it looks here right now)

    So find the roots of the quadratic (or the zeros, simply where it cuts the x-axis).
    This is when y=0, so [tex]0=(x+1+\sqrt{2})(x+1-\sqrt{2})[/tex]

    Now solve for x like you would've done with all the other quadratics. Don't be afraid of the irrationals, they work just like any other number.

    And notice that if you also take the negative of [tex](x+1)[/tex] and [tex]\sqrt{2}[/tex] because of the [tex]\pm[/tex] that we ignored earlier, it will give you the 2 same answers.

    Now that you have the roots, find the midpoint to give you the x-value where the turning point is and then substitute that x value into the quadratic to give you the y value.

    And finally, just to clear you up on the way you were trying to do it.

    The formula [tex]y=ax^2+bx+c[/tex] needs to be followed in the exact way it is shown. What I mean is, all the values with [tex]x^2[/tex] need to be together, all the values with x need to be together, and all the constants need to be together.

    e.g. [tex]y=2x^2+3x+x^2-5-\sqrt{5}x+1+\sqrt{2}[/tex]

    You need to collect all co-efficients (numbers in front of) the [itex]x^2[/itex]s, [itex]x[/itex]s and all the numbers. (EVEN if they cannot be expressed as 1 number).

    So, putting all equal variables (x is the variable) together:


    And obviously things like [tex]2x^2+x^2=3x^2[/tex] but it would be best to factorize x^2 out first just to get the idea:


    and adding terms as much as possible:


    Now look at the formula again: [tex]y=ax^2+bx+c[/tex]

    The values a, b and c must be equivalent so:

    [tex]a=3[/tex] [tex]b=3-\sqrt{5}[/tex] [tex]c=-4+\sqrt{2}[/tex]

    This is how you will now substitute into the quadratic formula to find the roots of the quadratic. When you try it for your question, you first need to expand and then collect all the terms and put them into the same form as I've shown.

    Ok I think I've written enough... *walks away quietly*
  8. Jul 21, 2009 #7
    Thank you very much for that. So I need to expand and collect all the terms and put all the 'x' variables together then sub in the values a,b and c then work it out with the quadratic formula. Also, can you go though the shortest way so that I can write down the steps?
  9. Jul 21, 2009 #8


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    No problem :smile:
    Yes that is exactly how it is done. The quadratic always has to be expressed in the general form before using the quadratic formula. Do you mean the shortest way for this particular question (which would be the first method I posted), or how to take quick steps using the quadratic formula?
    But I suggest you answer a few questions by each of these methods the long way first. i.e. Take the steps slowly just like I posted, then you'll figure out for yourself what steps you find necessary and what are just a waste of time that can be skipped.
  10. Jul 24, 2009 #9
    Yeah I think its the first one. The only problem is that I won't get enough CAJ for it (communication and justification). Doesn't matter, I know two methods now. Something my other classmates don't know.:biggrin:
  11. Jul 24, 2009 #10
    Maybe I'm missing something, but was it necessary to put the quadratic in standard form in order to find the "turning point" (= vertex?)? The equation is already in vertex form, which is
    [tex]f(x) = a(x - h)^2 + k[/tex].
    (The vertex would be (h, k).)

    [tex]f(x) = (x + 1)^2 - 2[/tex] or
    [tex]f(x) = (x - (-1))^2 - 2[/tex],
    meaning that the vertex is (-1, 2).

    Now, if the quadratic was in standard form to begin with,
    [tex]f(x) = ax^2 + bx + c[/tex],
    then finding the vertex is more involved. If the vertex is (h, k), then
    [tex]h = -\frac{b}{2a}[/tex] and [tex]k = c - ah^2 = f(h)[/tex].

    I don't mean to offend, but what I've read so far seems to be overly complicated.

  12. Jul 25, 2009 #11


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    The standard form for the vertex of the parabola (at least in the eyes of a student first being exposed to these equations) looks complicated and just like another bit of random jiberish to remember. I personally prefer understanding the concepts and applying it through logic, rather than following some general form equation.

    This is what you find so overly complicated...?
  13. Jul 25, 2009 #12


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    Also, in case multiple solutions to the problem are possible, I am in favor of first following the solution which the OP attempted his/herself. In this case, from the quote
    I seems like the OP intended to use the quadratic formula, but was unable to identify a, b and c (which is an important skill in itself).
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