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Math Help - 2 problems involving the real numbers

  1. #1
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    2 problems involving the real numbers

    1. If a and b are real numbers, and a > b and b < 0, then which of the following inequalities must be true?
    A. a > 0
    B. a < 0
    C. a^2 > b^2
    D. a^2 < b^2
    E. b^2 > 0
    (Turns out it is E but not sure why)

    2. What is the smallest possible value for the product of two real numbers that differ by 6? (Without using derivatives and/or max/mins)

    Why is it -9??
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  2. #2
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    Any real number squared that isn't 0 is greater than 0
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  3. #3
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    Ahh yes of course, just by taking examples too But what about the second problem? Hmmm...
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  4. #4
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    2. Two numbers differ by \displaystyle 6. Call them \displaystyle x and \displaystyle x + 6.

    Their product is \displaystyle x(x + 6) = x^2 + 6x

    \displaystyle = x^2 + 6x + 3^2 - 3^2

    \displaystyle = (x+3)^2 - 9.


    This is a quadratic with minimum at \displaystyle (-3, -9). So the smallest possible value is \displaystyle -9.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    2. Two numbers differ by \displaystyle 6. Call them \displaystyle x and \displaystyle x + 6.

    Their product is \displaystyle x(x + 6) = x^2 + 6x

    \displaystyle = x^2 + 6x + 3^2 - 3^2

    \displaystyle = (x+3)^2 - 9.


    This is a quadratic with minimum at \displaystyle (-3, -9). So the smallest possible value is \displaystyle -9.
    I was wondering about that one. I took that same approach and wasn't sure what do to with the parabola.
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  6. #6
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    Hello, sarahh!

    2. What is the smallest possible value for the product of two real numbers
    that differ by 6? (Without using derivatives and/or max/mins)

    We have two numbers that differ by 6.

    Let \,x = larger number.
    and x-6 = smaller number.

    Their product is: . P \:=\:x(x-6) \:=\:x^2-6x

    The graph is an up-opening parabola: . \,\cup
    Its minimum point is at its vertex.
    The vertex is at: . x \:=\:\dfrac{\text{-}b}{2a} \:=\:\dfrac{\text{-}(\text{-}6)}{2(1)} \:=\:3

    Hence: . P \:=\:3^2 - 6(3) \:=\:-9

    The vertex of the parabola is (3,\,\text{-}9)

    Therefore, the minimum \,P is -9.

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