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Conjugates / Dividing by Square Roots (page 4 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(3) + sqrt(5)) (sqrt(3) - sqrt(6))

    I do the multiplication:

      multiplication

    Then I complete the calculations by simplifying:

      (sqrt(3) + sqrt(5)) (sqrt(3) - sqrt(6)) = 3 + sqrt(15) - 3sqrt(2) - sqrt(30)

  • Simplify:  (sqrt(3) + sqrt(5)) (sqrt(3) - sqrt(5))

    I do the multiplication:

      multiplication

    Then I simplify:

      (sqrt(3) + sqrt(5)) (sqrt(3) - sqrt(5)) = 3 - 5 = -2


Note in the last example above how I ended up with all whole numbers. (Okay, technically they're integers, but the point is that the terms do not include any radicals.) I multiplied two radical "binomials" together and got an answer that contained no radicals. You may also have noticed that the two "binomials" were the same except for the sign in the middle: one had a "plus" and the other had a "minus". This pair of factors, with the second factor differing only in the one sign in the middle, is very important; in fact, this "same except for the sign in the middle" second factor has its own name:

    Given the radical expression sqrt(a) + sqrt(b), the "conjugate" is the expression sqrt(a) - sqrt(b).

The conjugate (KAHN-juh-ghitt) has the same numbers but the opposite sign in the middle. So not only is sqrt(a) - sqrt(b) the conjugate of sqrt(a) + sqrt(b), but sqrt(a) + sqrt(b) is the conjugate of sqrt(a) - sqrt(b).

When you multiply conjugates, you are doing something similar to what happens with a difference of squares:

    a2b2 = (a + b)(ab)   Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved

 

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When you multiply the factors a + b and ab, the middle "ab" terms cancel out:

    (a + b)(a - b) = a^2 + ab - ab - b^2 = a^2 - b^2

The same thing happens when you multiply conjugates:

    (sqrt(a) + sqrt(b)) (sqrt(a) - sqrt(b)) = a - b

We will see shortly why this matters. To get to that point, let's take a look at fractions containing radicals in their denominators.


Dividing by Square Roots

Just as you can swap between the multiplication of radicals and a radical containing a multiplication, so also you can swap between the division of roots and one root containing a division.

  • Simplify:  sqrt[ 8 / 2 ]
  • I can simplify this by working inside, and then taking the square root:

      sqrt[ 8 / 2 ] = sqrt[4] = 2

    ...or else by splitting the division into two radicals, simplifying, and cancelling:

      sqrt[8/2] = sqrt[8] / sqrt[2] = (2 sqrt[2]) / sqrt[2] = 2

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Cite this article as:

Stapel, Elizabeth. "Conjugates / Dividing by Square Roots." Purplemath. Available from
    http://www.purplemath.com/modules/radicals4.htm. Accessed
 

 

 

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