What is the value of k?

What is the value of k, when parabola y=X^2 + (K+2) +2K +1 touches the x-axis?

Update:

ATTENTION!!!! correct equation

y = x² + (k + 2)x + 2k + 1

3 Answers

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  • 7 years ago
    Favorite Answer

    Hello,

    It bears repeating: you should have applied the well-known following formulas: "Hello", "Please" and "Thanks".

    Anyway...

    = = = = = = = = = = = = = = = = = =

    y = x² + (k + 2)x + 2k + 1

    If the curve touch the x-axis, then y must be nil:

    y = 0

    x² + (k + 2)x + 2k + 1 = 0

    [x + (k + 2)/2]² – (k + 2)²/4 + 2k + 1 = 0 →→→ Completing the square.

    [x + (k + 2)/2]² – (k² + 4k + 4)/4 + (8k + 4)/4 = 0

    [x + (k + 2)/2]² – (k² + 4k + 4 – 8k – 4)/4 = 0

    [x + (k + 2)/2]² – (k² – 4k)/4 = 0

    Now three cases are possible:

    ♠ If 0<k<4

    Then (k²–4k)<0

    And

       [x + (k + 2)/2]² – (k² – 4k)/4

    Being the sum of a perfect square and a positive value will NEVER be nil.

    ♠ If k=0 or k=4,

    Then (k²–4k)=0

    And

       [x + (k + 2)/2]²

    Being a perfect square will be nil when x=-(k+2)/2.

    So if k=0 (resp. k=4) the curve will touch the x-axis at only one point (-1;0) (resp. (-3;0)).

    ♠ If k<0 or k>4,

    Then (k²–4k)>0

    And

       [x + (k + 2)/2]² – (k² – 4k)/4 = [x + (k + 2)/2]² – [√(k² – 4k)/2]²

    Being the difference od two squares will be nil at two points of the x-axis.

    So if k<0 or k>4, the curve will touch the x-axis at two points:

    (-[k + 2 + √(k² – 4k)]/2;0) and (-[k + 2 – √(k² – 4k)]/2;0).

    = = = = = = = = = = = = = = = = = =

    Since you want the curve to "touch" the x-axis, the only values of k that result in that is:

    k=0 and k=4.

    Regards,

    Dragon.Jade :-)

  • DWRead
    Lv 7
    7 years ago

    The question says "touches," not "crosses," so the vertex lies on the x-axis.

    y = x² + (k+2)x + (2k+1)

     = x² + (k+2)x + ((k+2)/2)² - ((k+2)/2)² + (2k+1)

     = (x + (k+2)/2)² - (k² + 4k + 4)/4 + (2k+1)

     = (x + (k+2)/2)² + (-k² - 4k - 4)/4 + (8k+4)/4

     = (x + (k+2)/2)² + (-k² + 4k)/4

    vertex ((k+2)/2, (-k² + 4k)/4)

    (-k² + 4k)/4 = 0

    k² - 4k = 0

    k - 4 = 0

    k = 4

  • Anonymous
    7 years ago

    You know what else it touches?! My heart!

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