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Homework Help: Is this question possible without having learnt derivatives?

  1. Dec 31, 2009 #1
    nvm got it
     
    Last edited: Dec 31, 2009
  2. jcsd
  3. Dec 31, 2009 #2

    rock.freak667

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    Homework Helper


    Right, well when you reach here, you will get

    [tex]3x^2+6x-(1+b)=0[/tex]

    Now if we solve for x using the quadratic equation we'll get

    [tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

    If the line intersects the curve, you'll get two real values for x right? (in the form a±b)

    Now for a tangent, the line intersects the curve once. So what condition should you put so that in the quadratic formula above, to get only value for x?
     
  4. Dec 31, 2009 #3
    Suggestion (without working it all out): Look for a value of b such that the line intersects the parabola at one point.

    From where you were:

    0=3x^2+6x-1-b

    Isolate b:

    b = 3x^2+6x-1

    Solve the quadratic equation for b. Does that work?
     
  5. Dec 31, 2009 #4

    HallsofIvy

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    Science Advisor

    No, tha'ts not a quadratic equation "for b"- and it is already solved for b.

    rockfreak667 suggested the right method: the equation [itex]ax^2+ bx+ c= 0[/itex] has a single solution if and only if the "discriminant", [itex]b^2- 4ac= 0[/math]. For this equation that would be 36+ 12(b+1)= 0. Solve that equation for b.
     
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