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Radical Expressions
A radical expression is an
expression that contains a square root - like
√xy
The radicand is the expression
under the radical sign, (XY in the above example). The
radicand is in simplest form if it contains no perfect
square factors other than 1.
Product property of square roots -
for any numbers c and d, where c
≥ 0, d
≥ 0,
√cd
=
√c
·
√d
Quotient property of square roots -
for any numbers c and d, where c
≥ 0, d
≥ 0,
√c/d
=
√c
/
√d
Rationalizing the denominator of a
radical expression means eliminating radicals from the
denominator of a fraction. Multiply the numerator and the
denominator by the radical in the denominator.
To simplify
(√11)
/ (√8
)
= (√11)
/ (√4
·
√2)
=
(√11)
/ 2
·
√2)
Multiply numerator and denominator by the
radical in the denominator:
(√11·
√2
) /
(
2
·
√2
·
√2)
=
(√11·
√2
) /
(
2
·
2) =
(√11·
√2
) / 4 =
√22
/ 4
Using a conjugate to simplify a radical
expression:
Conjugates are binomials in the form:
a√b
+ c√d
and a√b
- c√d
Their product is always a rational number with
no radicals (if a,b,c and d are rational numbers).
Rationalize:
(3)
/ (2
-
√7
)
(3)
/ (2
-
√7
) =
(3)
·
(2
+
√7
) / (2
-
√7
)
·
(2
+
√7
) =
(6
+ 3√7
) / (4
-
√7
·
√7) = (6
+ 3√7
) / (4
-
√49
) = (6
+ 3√7
) / 4 - 7=
(6
+ 3√7
) / -3
When is a radical expression in its
simplest form?
-
No radicands have perfect square
factors other than 1
-
No radicands contain factors.
-
No radicals appear in the denominator
of a fraction.
Add and subtract radical
expressions:
9√8
+ 4√8
- 7√8
= (9 + 4 - 7)
√8
= 6√8
Example 2:
3√45
+ 2√20
+ 6√625
=
3√9
· 5
+ 2√4
· 5
+ 6√25
· 25
=
3
·
3√5
+ 2
· 2√5
+ 6
· 25
=
9√5
+
4√5
+
150 = 13√5
+
150
Radical equations - are
equations that have variables in the radicand, like a =
√2b
/ 4
To solve radical equations: isolate the
radical on one side of the equation, then square each side
of the equation to eliminate the radical. Squaring each side
of the equation may introduce extraneous solutions,
that are not solutions to the original (before squaring)
equation. Ignore these extraneous solutions.
Example:
x - 3 =
√2x
+ 2
(x - 3)2 = 2x
+ 2
x2 - 6x + 9 = 2x + 2
x2 - 8x + 7 = 0
(x - 7) (x - 1) = 0
x = 7 and x = 1
Substitute x = 7 in the original
equation:
7 - 3 =
√2
·7 + 2 =
√14
+ 2 =
√16
4 = 4
7 is a valid solution.
Substitute x = 1 in the original
equation:
1 - 3 =
√2
·1 + 2 =
√2
+ 2 =
√4
- 2 ≠
2
1 is not a valid solution.
The e-Learning Research Foundation 2008 |
