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Conics Help Please

  1. Apr 21, 2008 #1
    Conics!!! Help Please!!!!

    okay i think i have solved this correctly...still a little unsure though....
    y=x^2-6x+3----find x intercept
    first i used complete the square---- y-3=(x-3)^2
    then i solved for x---- 3+(y-3)^(1/2)=x
    then i plugged 0 in for y and got 3+(3^(1/2)) and 3-(3^(1/2))
    is that correct??
     
  2. jcsd
  3. Apr 21, 2008 #2

    exk

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    x intercept is where y=0.
     
  4. Apr 21, 2008 #3
    yes i know that is why i made the equation equal to x and plugged 0 in for y
     
  5. Apr 21, 2008 #4

    exk

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    why not just 0=x^2-6x+3?
     
  6. Apr 21, 2008 #5
    you get the same answer i was just taught to solve for x first.....so is my answer right or wrong??
     
  7. Apr 21, 2008 #6

    exk

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    Try applying the quadratic formula to x^2-6x+3=0 and see if you get the same answer. I certainly don't.

    Following your work:
    then i solved for x---- 3+(y-3)^(1/2)=x
    then i plugged 0 in for y and got 3+(3^(1/2)) and 3-(3^(1/2))

    if you plug in 0 for y in 3+(y-3)^(1/2)=x you most certainly don't get 3+(3)^(1/2)=x, but 3+(-3)^(1/2)=x, which is not the answer.

    The reason is that you didn't complete the square correctly: y=x^2-6x+3 --> y=x^2-6x+3+6-6 --> y=(x-3)^2-6 --> y+6=(x-3)^2
     
  8. Apr 21, 2008 #7
    okay i see where i made my mistake..so the correct answer is x=3+(6^(1/2)) right???
     
  9. Apr 21, 2008 #8

    exk

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    well yes and x=3-6^.5 also (remember there are 2 roots for a quadratic)
     
  10. Apr 21, 2008 #9
    okay thank you for your help :)
     
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