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Thread: a and b

  1. #1
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    a and b

    If a and b are positive numbers, explain how the value of
    (sqrt{a}*sqrt{b})^2 compares with the value of a + b.

    I used FOIL and ended up with a + 2(sqrt{ab}) + b, which does not equal a + b.
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  2. #2
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    And you would be 100% correct
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  3. #3
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    Quote Originally Posted by magentarita View Post
    If a and b are positive numbers, explain how the value of
    (sqrt{a}*sqrt{b})^2 compares with the value of a + b.

    I used FOIL and ended up with a + 2(sqrt{ab}) + b, which does not equal a + b.
    $\displaystyle (\sqrt{a}\sqrt{b})^2 = ab$.

    Your statement will only be true if

    $\displaystyle ab = a + b$

    $\displaystyle 0 = a - ab + b$

    $\displaystyle 0 = a(1 - b) + b$

    $\displaystyle -b = a(1 - b)$

    $\displaystyle -\frac{b}{1 - b} = a$

    $\displaystyle \frac{b}{b - 1} = a$.


    So your situation will only be true for positive values of $\displaystyle a, b$ so that $\displaystyle a = \frac{b}{b - 1}$.
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  4. #4
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    sorry...but

    Quote Originally Posted by Prove It View Post
    $\displaystyle (\sqrt{a}\sqrt{b})^2 = ab$.

    Your statement will only be true if

    $\displaystyle ab = a + b$

    $\displaystyle 0 = a - ab + b$

    $\displaystyle 0 = a(1 - b) + b$

    $\displaystyle -b = a(1 - b)$

    $\displaystyle -\frac{b}{1 - b} = a$

    $\displaystyle \frac{b}{b - 1} = a$.


    So your situation will only be true for positive values of $\displaystyle a, b$ so that $\displaystyle a = \frac{b}{b - 1}$.
    I am sorry. I meant to type (sqrt{a} + sqrt{b})^2 in the original question. Can you show me now?
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    $\displaystyle (\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{a}\sqrt{b} + b$

    $\displaystyle = a + 2\sqrt{ab} + b$.


    The statement will be true if

    $\displaystyle a + 2\sqrt{ab} + b = a + b$

    $\displaystyle 2\sqrt{ab} = 0$

    $\displaystyle \sqrt{ab} = 0$

    $\displaystyle ab = 0$.


    This will only happen if one or both of a and b is 0.


    So your equation only holds true for if one or both of a and b is 0.
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  6. #6
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    Quote Originally Posted by magentarita View Post
    If a and b are positive numbers, explain how the value of
    (sqrt{a}*sqrt{b})^2 compares with the value of a + b.

    I used FOIL and ended up with a + 2(sqrt{ab}) + b, which does not equal a + b.
    You used "foil" on what? Of course, a+ 2sqrt{ab}+ b is not equal to a+ b, why should it be? Were you asked to about (sqrt{a}*sqrt{b})^2 or (sqrt{a}+ sqrt{a})^2? And you weren't asked to show these were equal, only to "compare" them. Which is larger, (sqrt{a}+ sqrt{b})^2 or (a+ b)^2?
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  7. #7
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    the book

    Quote Originally Posted by Prove It View Post
    $\displaystyle (\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{a}\sqrt{b} + b$

    $\displaystyle = a + 2\sqrt{ab} + b$.


    The statement will be true if

    $\displaystyle a + 2\sqrt{ab} + b = a + b$

    $\displaystyle 2\sqrt{ab} = 0$

    $\displaystyle \sqrt{ab} = 0$

    $\displaystyle ab = 0$.


    This will only happen if one or both of a and b is 0.


    So your equation only holds true for if one or both of a and b is 0.
    The question is asking to compare. After using FOIL, I saw that a + 2(sqrt{ab}) + b is > than a + b, which is the book's answer.
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