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Completing the square (How to)

  1. Jul 23, 2007 #1
    Here's the problem I'm currently working on:
    x[tex]^{2}[/tex] + 6x + 1 = 0
    Solve.

    Now I'm supposed to be learning completing the square here and the text book I have to work with doesn't quite explain it in much detail - so I don't know that I've even got the whole concept on how to complete the square.

    This is what I've done, which will end up giving me the wrong answer.
    [tex]x^{2} + 6x + 1 = 0 [/tex]
    [tex]x^{2} + 6x + 1 + 8 = 8 [/tex]
    [tex](x + 3)^{2}= 8 [/tex]
    [tex]\sqrt{(x+ 3)^{2}}[/tex] = [tex]+ or -\sqrt{8}[/tex]
    [tex] x = \sqrt{8} - or + 3 [/tex]

    I'm really just following an example given from the textbook, to tell the truth I have no idea if the concept actually applies to this situation or not. :grumpy:

    The answer is supposed to be
    [tex]x = -3 + or - 2 \sqrt{2}[/tex]
     
  2. jcsd
  3. Jul 23, 2007 #2
    when you have

    x+3 = +/- sqrt(8),

    the next step is x = -3 +/- sqrt(8)

    so I don't understand how you end up with sqrt(8) +/- 3

    By the way sqrt(8) = 2sqrt(2).
     
  4. Jul 23, 2007 #3

    Kurdt

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    Completing the square comes from:

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

    [tex] (x+a)^2 - a^2 = x^2+2ax[/tex]

    I'm sure the text probably introduced it in that manner. In that case what are you stuck on specifically?
     
  5. Jul 23, 2007 #4

    VietDao29

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    This line is wrong. You don't square root it. Square root something will only return you a non-negative value.

    Say, if a2 = 16, then [tex]\sqrt{a ^ 2} = |a| = 4[/tex]

    What you should do is just leave out the square sign, like this:

    [tex](x + 3) ^ 2 = 8[/tex]
    [tex]\Rightarrow x + 3 = \pm \sqrt{8}[/tex]

    i.e, (x + 3) can have 2 different values (i.e, [tex]\pm \sqrt{8}[/tex]), both of which if you square, you'll get 8.

    Can you get it? :)
     
  6. Jul 23, 2007 #5
    standard form is

    ax^2 + bx + c

    which term do you use in "completing the square," ax^2 or bx ? what would your next step me?
     
  7. Jul 23, 2007 #6

    VietDao29

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    Well, he did show his work in the first post. =.=" :rolleyes: He just asked where he had gone wrong. :)
     
  8. Jul 23, 2007 #7
    i don't see why you kept working with him, it's wrong from the 2nd line?
     
  9. Jul 23, 2007 #8

    VietDao29

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    Err, sorry, but where is the mistake in the second line? I cannot seem to find one. :frown:
     
  10. Jul 23, 2007 #9
    x^2 + 6x + 1

    (6/2)^2
     
  11. Jul 23, 2007 #10

    VietDao29

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    1 + 8 = (6/2)2 = 9, isn't that correct? He adds 8 to both sides, and comes up with the second line, to get 1 + 8 = 9, as required. Well, I see nothing wrong there. :)
     
  12. Jul 23, 2007 #11
    i PM'ed u so that OP doesn't see my solution.
     
    Last edited: Jul 23, 2007
  13. Jul 23, 2007 #12

    VietDao29

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    Just read what the problem asks, does it ask you to find the vertex of the parabola? >"<

    Btw, doesn't your third line, and his second line, look exactly the same? They are twins, can't you see? :)
     
  14. Jul 23, 2007 #13
    i know they are, i was typing out all my steps (obviously)

    ok by reading his again, i thought he had "added" 8 to both sides, my bad

    btw, i was "confirming" my answer, get over it.

    anyways OP, post 4 is good. i got confused with your 2nd line, oops.
     
    Last edited: Jul 23, 2007
  15. Jul 24, 2007 #14

    cristo

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    But he has added 8 to both sides. There are many ways to do this, all equivalent, but some people prefer other methods. The OP has added 8 to both sides, whereas others would perhaps add 0(=8-8) to the LHS.

    There's no need to take that tone.
     
  16. Jul 24, 2007 #15
    oh yes, and since i'm human i took this a little offensive as in implying i'm an idiot not even realizing my lines were the "twins." it's not like i cussed him out. he threw something small at me, so i threw something equivalent, no harm :-]

    like i said, get over it. come on now, be realistic. don't make it something bigger than it is.
     
    Last edited: Jul 24, 2007
  17. Jul 24, 2007 #16
    Well that certainly helped, I had no idea why it was being done in the original example in my text - now it makes sense!

    Ok so if [tex]\sqrt{8}[/tex] is the same as [tex]2\sqrt{2}[/tex] then my
    answer was basically right except, well except where it was wrong.

    So it would look more like:

    [tex]x^{2} + 6x + 1 = 0[/tex]
    [tex]x^{2} + 6x + 1 + 8 = 8[/tex]
    [tex](x + 3)^{2} = 8[/tex]
    [tex](x + 3) = +/- \sqrt {8}[/tex]
    [tex]x = +/- \sqrt {8} - 3 [/tex]

    Would that be correct?
     
    Last edited: Jul 24, 2007
  18. Jul 24, 2007 #17

    Kurdt

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    Thats correct. Of course you can simplify [itex]\sqrt{8} = 2 \sqrt{2} [/itex] to get the answer from your text. Have you seen the method I posted before? Sometimes thats easier to use I find if you're not comfortable with the method presented in your book.
     
    Last edited: Jul 24, 2007
  19. Jul 24, 2007 #18
    Thanks! Your efforts are definately appreciated!! :)

    And I did take a look at the other method too, I've got it written down here I'm gonna try it out on some more practice problems... once I get beck to my work... I'm in the midst of procrastinating
     
    Last edited: Jul 24, 2007
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