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Page 269
When working with square roots, sometimes it is hard to tell when the expression is fully reduced and when
there is still work to be done. To help you recognize when (and what) to simplify, here is a list of rules to
which you can refer:
1.
To simplify a radical, you should identify any perfect squares and move them in front of the radical
according to this rule:
x2 = x
The absolute value is necessary because we always consider square roots of this form to be nonnegative numbers.
Also, if a number or variable within the radical has a factor that is a perfect square, the radical may be
simplified. For example, the number 20 is not a perfect square itself, but it is equal to 4 . 5 and 4 is a
perfect square. So
20 can be simplified:
20 = 4 × 5 = 2 2 × 5 = 2 5 .
Examples:
1)
52 = 5
2)
(-5)2 =
3)
49yz2 = 7 |z|
4)
36a5 =
5)
25 = 5
y
62 . a4 . a = 6a2 a
3 50 = 3 25 . 2 = 3 52 . 2 = 3 . 5 2 = 15 2
In the fourth example, notice that it is not necessary to enclose a2 in an absolute value sign because a2
is never negative.
2.
If there is a negative sign outside of the radical, don’t forget to transfer it to the answer. If there is a
negative inside of the radical, your answer is a complex number rather than a real number. The
portion of a complex number that involves a minus sign under a radical is called imaginary. For our
purposes, when a complex number results we say that there is no solution.
Examples:
- x2 = - | x |
- 25 = -5
- x 2 ® an imaginary number / no solution
- 25 ® no solution
3.
When multiplying two similar radicals (that is, two square roots, two cube roots, and so forth), you
can combine them by multiplying what is under the first radical by what is under the second radical
and writing this product under one radical. This new expression may need to be simplified. The rule
for square roots is
x × y = xy
Examples:
1.
5 × 2 = 5 × 2 = 10
2.
2 × 6 = 2×6 = 2×2×3 = 4 × 3 = 2 3
-4
3. However,
-9 ¹
36 because the original roots were not real numbers.
4. For x > 0 and h ³ 0, multiply and simplify :
(
(
)
x + h )( x + h ) + ( x + h )( x ) - ( x )(
x+h - x
)(
x+h + x =
(x + h ) - ( x ) = h
) ( x )( x ) =
x+h -
**When multiplying two conjugates together the middle terms will always cancel.
4.
Some teachers may require that an answer never have a radical in the denominator of a fraction. To
remove a radical from the denominator, multiply the fraction by 1, which is easily represented as a
fraction. (Remember that 2 and 6 are fractions equal to 1.) Choose the radical you are trying to
2
6
remove as the number over itself in the fraction. For example, if the radical in the denominator is 3 ,
3 . This is done because squaring a square root expression cancels out the
3
you should multiply by
x × x = x and
radical: For x > 0,
3 × 3 = 3.
Example:
10
3 2
5.
=
10
3 2
×
2
2
=
10 2
3 2× 2
=
10 2 10 2 5 2
=
=
3× 2
6
3
When a fraction has a radical in the numerator and a similar radical in the denominator, you are free
to combine the radicals into one. Whatever was originally in the numerator should be in the
numerator of the new fraction and whatever was originally in the denominator should remain in the
denominator. The rule for square roots is
x
=
y
x
y
Examples:
3
3
3
1
or
3
=
=
4
16
4
16
45
5
=
20
=
3
45
= 9 =3
5
20
3
, and this can be simplified as
20
3
×
3
3
=
20 × 3
=
3
4 × 5 × 3 2 5 × 3 2 15
=
=
3
3
3
6.
The expressions 2x and 4x can be added to obtain 6x because they are like terms. However, 3x and 5x2
cannot be added because they are not like terms. This rule is similar to the rule you must follow when
adding terms with square roots. To be added or subtracted, two terms must have exactly the same
expression under the radical.
a x + b x = ( a + b) x
Examples:
4 2 + 7 2 = 11 2
5 xt + 4 5 xt = 5 5 xt
2 8 - 7 8 = -5 8 = -5 2 × 4 = -5 × 2 2 = -10 2
2 6 + 3 2 cannot be further simplified
3 x +4 x
20 x
=
7 x
20 x
=
7
for x ¹ 0
20
2 3+ 3 3 3
=
= 3
3
3
Sometimes radicals can be simplified to obtain like terms:
3 50 + 6 8 = 3 2 . 25 + 6 4 . 2 = 3 . 5 2 + 2 . 6 2 = 15 2 + 12 2 = 27 2
Example: Certain problems in calculus require that radicals in the numerator of a fraction be removed. The
following problem is from the top of page 269 in your Calculus Concepts text. The
simplification below is more comprehensive than that shown in the text.
æ2 x+h -2 x ö
÷=
ç
÷
ç
h
ø
è
æ2 x+h -2 x ö æ2 x+h +2 x ö
÷×ç
ç
÷=
÷ ç2 x+h +2 x ÷
ç
h
ø è
è
ø
(2 x + h × 2 x + h ) + (2 x + h × 2 x ) + (-2 x × 2 x + h ) + (-2 x × 2 x )
h × (2 x + h + 2 x )
4( x + h) + (4 x + h × x ) + (-4 x + h × x ) - 4 x
h × (2 x + h + 2 x )
4h
h × (2 x + h + 2 x )
=
=
4( x + h ) - 4 x
h × (2 x + h + 2 x )
4
2 x+h +2 x
If h is set equal to zero, the above expression becomes:
=
=
4 x + 4h - 4 x
h × (2 x + h + 2 x )
=
4
2 x+0 +2 x
=
4
2 x +2 x
=
4
4 x
=
1
x