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Math Help - Problem #2 - Rationalizing

  1. #1
    Forum Admin topsquark's Avatar
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    Problem #2 - Rationalizing

    This is for the Pre-Calculus students.

    A common way to add fractions with radicals in the denominator is to "rationalize the denominator." For example, rationalize the denominator of
    \frac{2x + 3y}{\sqrt{x} - \sqrt{y} }

    To rationalize this we use (a - b)(a + b) = a^2 - b^2 and multiply the numerator and denominator by \sqrt{x} + \sqrt{y} giving

    \frac{2x + 3y}{\sqrt{x} - \sqrt{y} } \cdot \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x} + \sqrt{y}} = \frac{(2x + 3y) \cdot ( \sqrt{x} + \sqrt{y} ) }{x - y}

    How do you rationalize the denominator of this?
    \frac{2x + 3y}{\sqrt[3]{x} + \sqrt[3]{y}}

    Hint: How do you factor  a^3 + b^3?

    -Dan
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  2. #2
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    re: Problem #2 - Rationalizing

    Hello, topsquark!

    How do you rationalize the denominator of this?
    \frac{2x + 3y}{\sqrt[3]{x} + \sqrt[3]{y}}

    Hint: How do you factor  a^3 + b^3?

    Spoiler:

    We have: . \frac{2x+3y}{x^{\frac{1}{3}} + y^{\frac{1}{3}}}

    Multiply by \frac{x^{\frac{2}{3}} - x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}}}{x^{\frac{2}{3}} - x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}}}

    . . \frac{2x+3y}{x^{\frac{1}{3}} + y^{\frac{1}{3}} }\cdot\frac{x^{\frac{2}{3}} - x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}}}{x^{\frac{2}{3}} - x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}}} \;=\; \frac{(2x+3y)(x^{\frac{2}{3}} - x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}})}{x+y}

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    Forum Admin topsquark's Avatar
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    re: Problem #2 - Rationalizing

    And a big "nicely done" to Soroban for solving this week's challenge. His prize for submitting a correct entry is new car! (If I can still find those chocolate cars at Walgreen's.)

    -Dan
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