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Math Help - [SOLVED] Write x^2 + 10x + 38 in the form (x+b)^2+c...

  1. #1
    Member looi76's Avatar
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    [SOLVED] Write x^2 + 10x + 38 in the form (x+b)^2+c...

    Question:
    Write x^2 + 10x + 38 in the form (x+b)^2+c where the values of b and c are to be found.
    (a) State the minimum value of x^2 + 10x + 38 and the value of x for which this occurs.
    (b) Determine the values of x for which x^2 + 10x + 38 \geq 22.


    Attempt:

    x^2 + 10x + 38

    = (x + \frac{10}{2})^2 - (\frac{10}{2})^2 + 38

    = (x + 5)^2 - 25 + 38

    What to do next?

    = (x + 5)^2  + 13
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  2. #2
    Moo
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    Hello,

    The beginning is correct.

    To continue, you can remember that something squared is always > 0.

    So (x+5)+13 has a minimum when (x+5) = 0, because 0 is the minimal value for a square.

    For b), (x+5)+13>22 <=> (x+5)>9=3

    -> |x+5|>3
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  3. #3
    o_O
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    For a parabola in the form of y = a(x - b)^{2} + c (note the negative sign in front of b), the vertex is given by (b, c).

    x^{2} + 10x + 38 \geq 22
    x^{2} + 10x + 16 \geq 0

    Now solve the inequality by finding the roots and finding the regions in which the inequality holds.
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