Multiplying by the Conjugate - University of Washington
Multiplying by the Conjugate
Sometimes it is useful to eliminate square roots from a fractional expression.
A way to do this is to utilize the fact that (A + B)(A - B) = A2 - B2 in order to eliminate square roots via squaring.
For instance, consider the expression
x
+
x2
x-2 .
Suppose we want to eliminate the square root from the numerator (this is sometimes called rationalizing the numerator).
What we can do it multiply the entire expression by
x
-
x2
.
x - x2
Since this is essentially equal to 1 (that is, it is 1 unless x = 1 or x = 0, in which case it is
undefined), our resulting expression will be essentially equivalent but it will have a different
form:
x + x2 x-2
=
x + x2 x-2
x - x2 x - x2
=
(x
-
x -x4 2)( x -
x2)
The equality if true as long as x = 1 or 0.
Note that we have moved the root from the numerator to the denominator; that's what this technique does.
Here's another example:
x- x = x- x
s
s
x x
+ +
x
x
=
x2 -x s(x + x)
which is true provided x = 0. The place students see this often is when working with the difference quotient expression
f (x + h) - f (x) . h
If f (x) is a square root function, then multiplication by the conjugate can be used to simplify this expression (in particular, to eliminate the h from the denominator).
Here's an example of this.
Suppose f (x) = 2x - 1.
Then
f (x
+
h) h
-
f (x)
=
=
2(x + h) - 1 - 2x - 1
h 2(x + h) - 1 - 2x - 1
h
2(x + h) - 1 + 2x - 1
2(x + h) - 1 + 2x - 1
=
2(x + h) - 1 - (2x - 1)
h( 2(x + h) - 1 + 2x - 1)
=
2h
h( 2(x + h) - 1 + 2x - 1)
=
2
( 2(x + h) - 1 + 2x - 1)
provided h = 0.
We can, as well, move radicals from the denominator to the numerator:
x1+ 1 = x1+ 1
x x
- -
1 1
=
x-1 x-1
provided x = 1, since x - 1 = 0 when x = 1.
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