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Finding Real Zeros

  1. Mar 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the real zeros of the following quadratic functions. Every problem will have an f(x)= in front of it.
    1) x^2- 3x- 28
    2) x^2+12x+36
    3)x^2- 3
    4) 4- x^2
    2. Relevant equations


    3. The attempt at a solution
    1) Common factors are 7 and 4.
    So would it be (x+7)?(x-4)=0?
    Which means the zeros would be -7 and 4. If I'm doing it right that is.
     
    Last edited by a moderator: Mar 12, 2015
  2. jcsd
  3. Mar 12, 2015 #2

    RUber

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    I think you are on the right track.
    For the first one, you correctly identified the 4 and 7, but your signs were wrong when you put them into the polynomials.
    (x+7)(x-4) = x^2 +7x -4x -28 = x^2 +3x -28. Switch the signs, and you should be good to keep going in that direction to solve the rest.
     
  4. Mar 12, 2015 #3
    Wait so the zeros are 7 and -4? I saw somewhere that the factoring and zeros were opposite. Or am I confusing myself?
     
  5. Mar 12, 2015 #4

    RUber

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    You may be confusing the hint.
    Your factored form of (1) should be (x-7)(x+4) = x^2 -7x+4x-28 = x^2-3x-28 which is what you posted.
    The zeros of this are clearly when (x-7)=0 or when (x+4)=0. This happens when x = 7 or -4.
    Just saying signs are opposite sometimes leads to switching signs where you shouldn't just because you think you are supposed to change them somewhere.
     
  6. Mar 12, 2015 #5
    But are there cases where that happens because I get frustrated at that concept. Anyways
    2) The 2 zeros would both be positive 6, right?
    As for 3 and 4, I just need an example how to factor functions that look like that.
     
  7. Mar 12, 2015 #6

    RUber

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    for (2) your factors are (x+6)^2, is that zero when x=6? Finding zeros mean find an x such that the function equals zero at that x.
    For 3 and 4, I recommend you use the difference of squares formula:
    ## a^2 - b^2 = (a-b)(a+b) ##
    When in doubt, the quadratic equation will always work:
    For ##f(x) = ax^2 + bx +c ## zeros are at ## x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}##
     
  8. Mar 12, 2015 #7
    For 2 I skipped to the end so I can cover more ground at a quicker pace, Found the common factors and matched with positive 12.

    I'm still confused on solving 3 and 4. a^2 - b^2 seems simple enough but look at 3. a should be the first number but it's x^2.
     
  9. Mar 12, 2015 #8

    RUber

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    For (2) you found the right factors, x+6 and x+6. The zeros are not at +6.
    For (3) you have x^2 - 3. For this form, you want a^2 = x^2 and b^2 = 3.
     
  10. Mar 13, 2015 #9
    Sorry but I still don't understand. What am I doing wrong in 2? And how do I get those functions to equal x^2 and 3?
     
  11. Mar 13, 2015 #10

    RUber

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    If you are trying to find the zeros from a factored form, you need to make the factors zero.
    2 factors to (x+6)(x+6), what do you get for x=6? 6^2 + 12*6+ 36 = 144. You need to change the sign -- x=-6 gives (-6)^2 +12(-6)+ 36 = 72-72 = 0.
    This should make sense since (x+6) is zero when x=-6, and anything times zero is zero.

    For 3, let x = a and b =##\sqrt{3}.\quad a^2 = x^2, b^2 = 3.##
     
  12. Mar 13, 2015 #11

    Mentallic

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    Since you're obviously trying to learn this quickly, I'll give a short crash course on factorizing.

    We have the quadratic [itex]f(x)=x^2+12x+36[/itex].

    When we talk about f(x), we are actually talking about [itex]x^2+12x+36[/itex] in this case.
    f(1) will then represent f(x) when x=1, hence
    [tex]f(x)=x^2+12x+36[/tex]
    and
    [tex]f(1)=(1)^2+12(1)+36=1+12+36=49[/tex]

    Similarly, [itex]f(6)=(6)^2+12(6)+36=12+72+36=144[/itex] and [itex]f(-6)=(-6)^2+12(-6)+36=36-72+36=0[/itex]

    Notice f(-6)=0, hence x = -6 is a zero of the quadratic f(x). But of course, it's not easy to figure out the values of x this way. We need to turn to factorizing.
    Since you already know how to factorize, we'll move on.

    So [itex]f(x)=x^2+12x+36=(x+6)^2[/itex].

    Now, f(1)=49 that we saw earlier, but with this factorized form, we can find that out a lot more easily. [itex]f(x)=(x+6)^2[/itex] hence [itex]f(1)=(1+6)^2=7^2=49[/itex].
    Similarly,
    [tex]f(6)=(6+6)^2=12^2=144[/tex]
    and
    [tex]f(-6)=(-6+6)^2=0^2=0[/tex]

    It was easy to see that x=-6 gave us f(x)=0 because if k represents any number then k2=0 only when k=0. So if we think of x+6 as being k, hence [itex](x+6)^2=k^2[/itex] then we can only get [itex](x+6)^2=0[/itex] by letting x+6=0, solving gives us x=-6.
     
  13. Mar 13, 2015 #12

    HallsofIvy

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    iYou need to stop memorizing formulas and think about what you are doing! To "find a zero of f(x)" means to solve the equation f(x)= 0. So to find a zero 0 of [itex]f(x)= x^2- 3[/itex] means to solve the equation [itex]x^2- 3= 0[/itex]. I presume you know that you can add the same thing to both sides of the equation so that is the same as [itex]x^2= 3[/itex]. There are two numbers that make that true.

    You should also know that if ab= 0 then either a= 0 or b= 0 or both. So to solve [itex]x^2-3x- 28= (x- 7)(x+ 4)= 0 you must have x- 7= 0 and x+ 4= 0.
     
    Last edited: Mar 13, 2015
  14. Mar 13, 2015 #13

    Ray Vickson

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    To not confuse the OP even more, that should say: to solve ##x^2 - 3x - 28 = (x-7)(x+4) = 0## you must have x-7 = 0 OR x+4 = 0.
     
  15. Mar 13, 2015 #14

    HallsofIvy

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    Yes, thank you.
     
  16. Mar 15, 2015 #15
    I think I understand how you got to the x^2=3 part but is the way you get the zero by squaring the 3. You said there were two numbers and.. Ugh
     
  17. Mar 15, 2015 #16

    Ray Vickson

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    No, you do NOT square the 3---you do the exact opposite! Of course, there are two roots; if x = r is one root, so is x = -r, because ##(-r)^2 = (+r)^2##, so both -r and +r have the same "square".
     
  18. Mar 15, 2015 #17
    Crap, you're right, I meant square root. So the answer is 1.73?
     
  19. Mar 15, 2015 #18

    Ray Vickson

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    No, because ##\sqrt{3} \neq 1.73##. Your 1.73 is an approximation, not the exact value. In fact, you cannot write down any finite decimal number that gives the answer, because ##\sqrt{3}## is a so-called irrational number. A better approximation to the answer is ##\sqrt{3} \doteq 1.732050808##; an even better approximation is ##\sqrt{3} \doteq 1.7320508075688772935274463415058723669428052538104##, but that is still not exact. (Note that I used "##\doteq##" instead of "##=##", to indicate these are approximations.)

    Of course, in practice you might use a calculator to compute an answer, in which case you would make do with a decimal approximation such as 1.73 or 1.7321 or whatever meets the accuracy of your analysis. However, I am trying to get you to realize that you would be dealing with an approximate answer, and somewhere you should indicate that you understand that---either by saying "approximately", or by using a symbol such as "≈" instead "=".
     
    Last edited: Mar 15, 2015
  20. Mar 16, 2015 #19
    Got it, thanks. Can you help me understand 2 a little better?
     
  21. Mar 16, 2015 #20

    Mentallic

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    Was my post at #11 not sufficient? What don't you understand?
     
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