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Y = x^2-a(x+1)+3

  1. Jun 20, 2013 #1
    1. The problem statement, all variables and given/known data
    When the parabola y=x^2-a(x+1)+3 intersects the x-axis at one point then a=_ or =_.

    2. Relevant equations
    The equation of parabola y^2=4px,x^2=4py depending on the orientation

    3. The attempt at a solution
    Since the parabola meets the x axis once, I assume that to be the vertex.I try to rearrange the function into an appropriate form but I get stuck here.
    y=x^2-ax-a+3
    The equation already has two unknowns and hell breaks lose when I try to complete the square.
     
  2. jcsd
  3. Jun 20, 2013 #2
    What is the condition for a quadratic to have a single root?
     
  4. Jun 20, 2013 #3

    LCKurtz

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    How exactly does all hell break loose? What do you get?
     
  5. Jun 20, 2013 #4
    I get something to this effect;
    y+1/2a-3=(x-1/2a)^2
    I get stuck because the y part cannot be factorized but I am again thinking that the I could equate 1/2a-3=0 because apparently the left side should not have a constant since the vertex is along the x-axis.That could give me one of the values of a as 6.
     
  6. Jun 20, 2013 #5

    LCKurtz

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    When you write 1/2a do you mean a/2 or 1/(2a). You should use parentheses. In any case, show your steps on completing the square because what you have isn't correct.
     
  7. Jun 20, 2013 #6
    Oh my mistake,
    I got this y+(1/4)a^2-3=(x-(1/2)a)^2
    Is that what you have?
     
  8. Jun 20, 2013 #7

    LCKurtz

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    No, it isn't.
     
  9. Jun 20, 2013 #8

    verty

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    You may want to revisit parabolas and completing the square, you are missing some knowledge there.

    Some practice:
    1. Complete the square: 2x^2 + 5x - 4 = 0
    2. Complete the square: y = 2x^2 + 5x - 4
    3. Complete the square: y = ax^2 + bx + c

    Your answer for #3 should be ##y = a(x + \frac{b}{2a})^2 + \frac{4ac - b^2}{4a}##.
    Then...

    No, what I said after this was not correct. But this is good practice still.
     
    Last edited: Jun 20, 2013
  10. Jun 20, 2013 #9

    verty

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    If you let y = 0 and solve for x, you get the well-known formula for the roots of a parabola. But already in this form it is clear that something strange happens when ##4ac = b^2##.
     
  11. Jun 20, 2013 #10
    I redid it.
    I got this:
    y+(1/4)a^2+a-3=(x-(1/2)a)^2
    and solved a to be -6 or 2
     
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