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Rationalizing Denominators (page 5 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


  • Simplify:  sqrt[ 25 / 3 ]
    • sqrt[25/3] = sqrt[25] / sqrt[3] = 5 / sqrt[3]

This looks very similar to the previous exercise, but this is the "wrong" answer. Why? Because the denominator contains a radical. The denominator must contain no radicals, or else it's "wrong". (Why "wrong" in quotes? Because this issue may matter to your instructor right now, but it probably won't later on. It's like when you were in elementary school and improper fractions were "wrong" and you had to convert everything to mixed numbers instead. But now that you're in algebra, improper fractions are fine, even preferred. Once you get to calculus or beyond, they won't be so uptight about where the radicals are.)

    To get the "right" answer, I must "rationalize" the denominator. That is, I must find some way to convert the fraction into a form where the denominator has only "rational" (fractional or whole number) values. But what can I do with that radical-three? I can't take the 3 out, because I don't have a pair of threes.

Thinking back to those elementary-school fractions, you couldn't add them unless they had the same denominators. To create these "common" denominators, you would multiply, top and bottom, by whatever the denominator needed. Anything divided by itself is just 1, and multiplying by 1 doesn't change the value of whatever you're multiplying by the 1. But multiplying that "whatever" by a strategic form of 1 could make the necessary computations possible, such as:

    2/5 + 3/7 = (2/5)(7/7) +(3/7)(5/5) = 14/35 + 15/35 = 29/35

We can use the same technique to rationalize radical denominators.

    I could take a 3 out of the denominator if I had two factors of 3 inside the radical. I can create this pair of 3's by multiplying by another copy of root-three. If I multiply top and bottom by root-three, then I will have multiplied the fraction by a strategic form of 1. I won't have changed the value, but simplification will now be possible:

      5/sqrt[3] = (5/sqrt[3])(sqrt[3]/sqrt[3]) = (5sqrt[3]) / (sqrt[3]sqrt[3]) = (5 sqrt[3]) / 3

This last form, "five, root-three, divided by three", is the "right" answer they're looking for.

  • Simplify:  (6 sqrt[2]) / sqrt[3]
    • (6 sqrt[2]) / sqrt[3] = ((6sqrt[2])/sqrt[3])(srqt[3]/sqrt[3]) = (6 sqrt[2*3])/(sqrt[3*3]) = (6 sqrt[6])/(3) = 2 sqrt[6]

Don't stop once you've rationalized the denominator. As the above demonstrates, you should always check to see if something remains to be simplified.

  • Simplify:  3 / (2 + sqrt[2])
  • This expression is in the "wrong" form, due to the radical in the denominator. But if I try to multiply through by root-two, I won't get anything useful:

      (3/(2+sqrt[2]))(sqrt[2]/sqrt[2]) = (3sqrt[2]) / (2sqrt[2] + 2)

    Multiplying through by another copy of the whole denominator won't help, either:

      (3/(2+sqrt[2]))((2+sqrt[2])/(2+sqrt[2])) = (6 + 3sqrt[2]) / (6 + 4sqrt[2])

    But look what happens when I multiply by the same numbers, but with the opposite sign in the middle:   Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved

      (2 + sqrt[2])(2 - sqrt[2]) = 4 - 2sqrt[2] + 2sqrt[2] - sqrt[2]sqrt[2] = 4 + 0 - 2 = 2

    This multiplication made the radical terms cancel out, which is exactly what I want. This "same numbers but the opposite sign in the middle" thing is the "conjugate" of the original expression. By using the conjugate, I can do the necessary rationalization.

      (3/(2 + sqrt[2]))((2 - sqrt[2])/(2 - sqrt[2])) = (3(2 - sqrt[2])/(4 - 2) = (6 - 3sqrt[2])/2

Do not try to reach inside the numerator and rip out the 6 for "cancellation". The only thing that factors out of the numerator is a 3, but that won't cancel with the 2 in the denominator. Nothing cancels!

  • Simplify:  (1 + sqrt[7]) / (2 - sqrt[7])
  • I'll multiply by the conjugate in order to "simplify" this expression. The denominator's multiplication results in a whole number (okay, a negative, but the point is that there aren't any radicals):

      (2 - sqrt[7])/(2 + sqrt[7]) = 4 - 2sqrt[7] + 2sqrt[7] - sqrt[7]sqrt[7] = 4 + 0 - 7 = -3

    The numerator's multiplication looks like this:

      (1 + sqrt[7])(2 + sqrt[7]) = 2 + 2sqrt[7] + 1sqrt[7] + sqrt[7]sqrt[7] = 2 + 2sqrt[7] + 1sqrt[7] + 7 = 9 + 3sqrt[7]

    Then the simplified (rationalized) form is:

      ((1+sqrt[7])/(2-sqrt[7]))((2+sqrt[7])/(2+sqrt[7])) = (9 + 3sqrt[7])/(-3) = (3 + sqrt[7])/(-1) = -(3+sqrt[7]) = -3 - sqrt[7]

It can be helpful to do the multiplications separately, as shown above. Don't try to do too much at once, and make sure to check for any simplifications when you're done with the rationalization.

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Cite this article as:

Stapel, Elizabeth. "Rationalizing the Denominators." Purplemath. Available from
    http://www.purplemath.com/modules/radicals5.htm. Accessed
 

 

 

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