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Math Help - proof that sqrt(ab-ac)=a sqrt(b-c)

  1. #1
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    proof that sqrt(ab-ac)=a sqrt(b-c)

    Proof that: sqrt(ab-ac)=asqrt(b-c)
    without the use of substitution

    Thx
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  2. #2
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    Re: proof that sqrt(ab-ac)=a sqrt(b-c)

    It doesn't.

    \displaystyle \sqrt{ab - ac} = \sqrt{a(b - c)} = \sqrt{a}\sqrt{b - c}, not \displaystyle a\sqrt{b-c}.
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  3. #3
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    Re: proof that sqrt(ab-ac)=a sqrt(b-c)

    As Prove It said, you can't prove that- it's not true.

    Did you mean \sqrt{a^2b- a^2c}= a\sqrt{b- c}?
    (assuming that a is non-negative)


    I'm not sure what you mean by "without the use of substitution"- I see nothing to substitute there- but
    \sqrt{a^2b- a^2c}= \sqrt{a^2(b- c)}= \sqrt{a^2}\sqrt{b- c}= a\sqrt{b- c}

    ( \sqrt{a^2}= a if a is non-negative. \sqrt{a^2}= -a if a is negative.)
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  4. #4
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    Re: proof that sqrt(ab-ac)=a sqrt(b-c)

    Quote Originally Posted by HallsofIvy View Post
    As Prove It said, you can't prove that- it's not true.

    Did you mean \sqrt{a^2b- a^2c}= a\sqrt{b- c}?
    (assuming that a is non-negative)


    I'm not sure what you mean by "without the use of substitution"- I see nothing to substitute there- but
    \sqrt{a^2b- a^2c}= \sqrt{a^2(b- c)}= \sqrt{a^2}\sqrt{b- c}= a\sqrt{b- c}

    ( \sqrt{a^2}= a if a is non-negative. \sqrt{a^2}= -a if a is negative.)
    Alternatively, maybe the OP was being asked to prove \displaystyle \sqrt{ab - ac} = \sqrt{a}\sqrt{b-c}...
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