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Math Help - Help rationalize this numerator so that...

  1. #1
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    Help rationalize this numerator so that...

    rationalize this numerator so that

    [(√2(x+delta x)-3) - (√2x-3)]/delta x = 2/[(√2(x+delta x)-3) + √2x-3)]
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  2. #2
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    Hello, yess!

    Is there a typo on the problem?


    \text{Rationalize this numerator so that:}

    . \displaystyle \frac{\sqrt{2}(x+ \Delta x)-3) - (\sqrt{2}x-3)}{\Delta x} \;=\; \frac{2}{[\sqrt{2}(x+\Delta x)-3] + [\sqrt{2}x-3]}

    This is not true . . .
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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, yess!

    Is there a typo on the problem?



    This is not true . . .
    ohh oops the square root should be over the whole expression so square root of 2(x+delta x) -3 minus the square root of 2x-3 all over delta x equals 2 over the square root 2(x+delta x)-3 + square root of 2x-3.
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  4. #4
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    if you mean
    \frac{\sqrt{2(x+\Delta x)- 3}- \sqrt{2x- 3}}{\Delta x}

    you "rationalize the numerator" by multiplying by
    \frac{\sqrt{2(x+\Delta x)- 3}+ \sqrt{2x- 3}}{\sqrt{2(x+\Delta x)- 3}+ \sqrt{2x- 3}}
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