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A Special Case of Rationalizing / Radicals & Exponents / Radicals & Domains (page 7 of 7)

Sections: Square roots, More simplification / Multiplication, Adding (and subtracting) square roots, Conjugates / Dividing by square roots, Rationalizing denominators, Higher-Index Roots, A special case of rationalizing / Radicals & exponents / Radicals & domains

A Special Case of Rationalizing

If your class has covered the formulas for factoring the sums and differences of cubes, then you might encounter a special case of rationalizing denominators. The reasoning and methodology are similar to the "difference of squares" conjugate process for square roots.

  • Simplify:  2 / (1 + cbrt[4])
  • I would like to get rid of the cube root, but multiplying by the conjugate won't help much:

      (1 + cbrt[4])(1 - cbrt[4]) = 1 - (cbrt[4])^2 = 1 - cbrt[16] = 1 - 2 cbrt[2]

    But I can "create" a sum of cubes, just as using the conjugate allowed me to create a difference of squares earlier. Using the fact that a3 + b3 = (a + b)(a2 ab + b2), and letting a = 1 and b equal the cube root of 4, I get:

      (1 + cbrt[4])(1 - cbrt[4] + (cbrt[4])^2) = 1 + (cbrt[4])^3 = 1 + 4 = 5

    If I multiply, top and bottom, by the second factor in the sum-of-cubes formula, then the denominator will simplify with no radicals:

      (2/(1 + cbrt[4]))*((1 - cbrt[4] + cbrt[16])/(1 - cbrt[4] + cbrt[16])) = (2 - 2cbrt[4] + 4cbrt[2]) / 5

Naturally, if the sign in the middle of the original denominator had been a "minus", I'd have applied the "difference of cubes" formula to do the rationalization. This sort of "rationalize the denominator" exercise almost never comes up. But if you see this in your homework, expect one of these on your next test.

Radicals Expressed With Exponents

Radicals can be expressed as fractional exponents. Whatever is the index of the radical becomes the denominator of the fractional power. For instance:

    sqrt[9] = 2nd-rt[9] = 9^(1/2) = 3

The second root became a one-half power. A cube root would be a one-third power, a fourth root would be a one-fourth power, and so forth.This conversion process will matter a lot more once you get to calculus. For now, it allows you to simplify some expressions that you might otherwise not have been able to.   Copyright Elizabeth Stapel 1999-2011 All Rights Reserved

  • Express  cbrt[2] * 4th-rt[2]  as a single radical term.
  • I will convert the radicals to exponential expressions, and then apply exponent rules to combine the factors:

      cbrt[2] 4th-rt[2] = 2^(1/3) 2^(1/4) = 2^(1/3 + 1/4) = 2^(7/12) = 12th-rt[2^7]

  • Simplify:  cbrt[5] / sqrt[5]
    • 5^(1/3) / 5^(1/2) = 5^(1/3 - 1/2) = 5^(-1/6) = 1/5^(1/6) = 1/(6th-rt[5]) = (1/(6th-rt[5]))*(6th-rt[5^5]/6th-rt[5^5]) = (6th-rt[5^5]) / 5

A Few Other Considerations

Usually, we cannot have a negative inside a square root. (The exception is for "imaginary" numbers. If you haven't done the number "i" yet, then you haven't done imaginaries.) So, for instance, sqrt(-4) is not possible. Do not try to say something like " sqrt(-4) = -2 ", because it's not true: (-2)^2 = +4, which does not equal -4 . You must have a positive inside the square root. This can be important for defining and graphing functions.

  • Find the domain of the following:
    • y = sqrt(x - 2)

    The fact that I have the expression x 2 inside a square root requires that x 2 be zero or greater, so I must have x 2 > 0. Solving, I get:

      domain:  x > 2

On the other hand, you CAN have a negative inside a cube root (or any other odd root). For instance:

    cbrt(-8) = -2

...because (2)3 = 8.

  • Find the domain of the following:
    • y = cbrt(x - 2)

    For cbrt(x - 2), there is NO RESTRICTION on the value of x, because x 2 is welcome to be negative inside a cube root. Then the domain is:

      domain:  all x

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Cite this article as:

Stapel, Elizabeth. "A Special Case of Rationalizing / Radicals & Exponents  / Radicals &
    Domains." Purplemath. Available from


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