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Math Help - Square root problem

  1. #1
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    Square root problem

    If  \sqrt{a + b} = \sqrt{a} + \sqrt{b}, which of the following must be true?

    A. a = b
    B. a = 0 and b = 0
    C. a = 0 or b = 0
    D. a + b > 1
    E. a = 1 and b > 0

    I mean clearly C works but why don't the other choices work?
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  2. #2
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    Quote Originally Posted by sarahh View Post
    If  \sqrt{a + b} = \sqrt{a} + \sqrt{b}, which of the following must be true?

    A. a = b
    B. a = 0 and b = 0
    C. a = 0 or b = 0
    D. a + b > 1
    E. a = 1 and b > 0

    I mean clearly C works but why don't the other choices work?
    For each of A, B, D, E, let a= 1, b= 0
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  3. #3
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    Hmm so wait Ivy--would you say that for this question all the answers are correct but choice C is more correct?? (Because of your values)
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  4. #4
    Senior Member BAdhi's Avatar
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    no he says only C is correct :-)
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  5. #5
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    Quote Originally Posted by sarahh View Post
    If  \sqrt{a + b} = \sqrt{a} + \sqrt{b}, which of the following must be true?

    A. a = b
    B. a = 0 and b = 0
    C. a = 0 or b = 0
    D. a + b > 1
    E. a = 1 and b > 0

    I mean clearly C works but why don't the other choices work?
    \displaystyle \sqrt{a+b} = \sqrt{a} + \sqrt{b}

    \displaystyle (\sqrt{a+b})^2 = (\sqrt{a} + \sqrt{b})^2

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

    \displaystyle 0 = 2\sqrt{a}\sqrt{b}.


    For this to be true, \displaystyle a = 0, b = 0 or \displaystyle a=b=0.

    The one that must be true is C.
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  6. #6
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    Thanks Prove It! That's quite ingenius and beautiful actually
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  7. #7
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    Quote Originally Posted by sarahh View Post
    If  \sqrt{a + b} = \sqrt{a} + \sqrt{b}, which of the following must be true?

    A. a = b

    B. a = 0 and b = 0

    C. a = 0 or b = 0

    D. a + b > 1

    E. a = 1 and b > 0
    Or...

    \sqrt{a+b}=\sqrt{a}+\sqrt{b}\Rightarrow\sqrt{a+b}\  left[\sqrt{a}-\sqrt{b}\right]=\left[\sqrt{a}+\sqrt{b}\right]\left[\sqrt{a}-\sqrt{b}\right]

    \Rightarrow\sqrt{a+b}\sqrt{a}-\sqrt{a+b}\sqrt{b}=a-\sqrt{a}\sqrt{b}+\sqrt{a}\sqrt{b}-b

    \Rightarrow\sqrt{a^2+ab}-\sqrt{ab+b^2}=a-b\Rightarrow\ ab=0


    I mean clearly C works but why don't the other choices work?
    A.

    a=b

    \sqrt{2a}=\sqrt{2}\sqrt{a}

    \sqrt{a}+\sqrt{a}=2\sqrt{a}

    Not equal.

    B.

    \sqrt{0+0}=\sqrt{0}+\sqrt{0}

    however the equation is true for either one of the two values being non-zero.

    Hence it is unnecessary that both be zero.

    D.

    a+b>1

    a=1,\;\;\ b=0\Rightarrow\sqrt{1+0}=\sqrt{1}+\sqrt{0}

    It's unnecessary for a+b>1

    E.

    a=1,\;\;b=0 works, hence the condition is not required.
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