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By definition, every real number has two square roots. For example, 64's
two square roots are 8 and -8 because 82 = 64 and (-8)2 = 64. However, the
principal square root of a real number is its nonnegative square root (8
would be the principal square root of 64). Examples:
1. Problem: SQRT(25)
Solution: 5
That is the answer because 5 * 5 = 25.
2. Problem: -SQRT(64)
Solution: -8
Note that the negative sign is outside the square root "symbol."
One special situation with square roots occurs when the number inside the
square root is squared. There is a special theorem that deals with this:
SQRT(a2) = |a|. Example:
3. Problem: SQRT((-16)2)
Solution: By the theorem above, the answer
is |-16|. The absolute
value of -16 is 16.
The cube root of 8 is 2 because 2 * 2 * 2 = 8. The cube root of -64 is -4
because (-4)3 = -64.
As you can see from the above examples, negative numbers have real cube
roots (unlike square roots). Example:
1. Problem: CBRT(27y3)
Solution: 3y
Although you usually see only square and cube roots, it is possible to see
and use 4th, 5th, and 6th roots of numbers. It is also mathematically
possible to use a one millionth root of a number. N-th roots, like square
roots have a special rule when the index is the same as the power the
radicand is raised to. Since the n-th root (n is any real number) can be any
number, a rule for each number would be not only cumbersome, but
impossible since numbers go on forever! Because of that slight problem,
the n-th Root Theorem was devised. It says that for any real numbers a and
k, the k-th root of ak is |a| when k is even and a when k is odd. Examples:
1. Problem: Find the 4th root of 16.
Solution: The problem asks us to find
a number that raised to the 4th
power equals 16.
Write 16 as a product of prime
factors.
2 * 2 * 2 * 2
To be able to remove something from
under the radical, there have to be
4 instances of it (because we're
taking a 4th root). There are four
instances of 2.
That leaves nothing under the radical,
so the answer is 2.
2. Problem: Find the 10th root of
5010.
Solution: By the n-th Root
Theorem, the 10th root of
5010 is
|50|, or 50.
For any nonnegative real numbers, a and b, SQRT(a) * SQRT(b) =
SQRT(ab). This theorem applies to any index. For example, the 4th root of a
times the 4th root of b equals the 4th root of ab. Example:
1. Multiply: SQRT(x + 2) * SQRT(x - 2)
Solution: Use the theorem above to put both
terms under the same radical.
SQRT((x + 2)(x - 2))
Multiply the binomials under the
radical out.
SQRT(x2 - 4)
Reversing the theorem stated above gives us a way to factor radical
expressions, thereby simplifying them. Example:
2. Problem: Simplify SQRT(20)
Solution: Factor the radicand as a
product of prime factors.
SQRT(2 * 2 * 5)
There are two instances of 2,
so by the definition of a square
root, you can take 2 out
from under the radicand. That gives
the following for an answer:
2(SQRT(5))
Taking the square root of fractions and dividing a radical by a radical are
discussed in this section.
When you have a problem like SQRT(27/y2), don't be scared of the fraction.
Just use the Roots of Fractions theorem, which says that nRT(a/b) =
(nRT(a))/(nRT(b)). Examples:
1. Simplify: CBRT(27/125)
Solution: Use the Roots of Fractions
theorem to rewrite the problem.
CBRT(27)
--------CBRT(125)
Take the cube root of both
the numerator and denominator
to get the final answer.
(3/5)
2. Simplify: SQRT(25/y2)
Solution: Use the Roots of Fractions
theorem to rewrite the problem.
SQRT(25)
-------SQRT(y2)
Take the square root of both
the numerator and denominator
to get the final answer.
(5/y)
Reverse the Roots of Fractions theorem when you are asked to divide a
radical by a radical of the same index when it can be simplified. Example:
3. Simplify: (SQRT(80))/(SQRT(5))
Solution: Use the converse of the Roots
of Fractions theorem and rewrite
it as one radical.
SQRT(80/5)
80/5 simplifies to 16.
SQRT(16) = 4
You add and subtract radicals the same way you would with polynomials, by
combining like terms. Example:
1. Problem: 6(SQRT(7)) + 4(SQRT(7))
Solution: Both terms are alike (like
radicals have the same index
and same radicand), so you
can add them.
10(SQRT(7))
You have to look out for terms that do not look alike, but could be if
factored. Example:
2. Problem: 3(SQRT(8)) - 5(SQRT(2))
Solution: Factor 8.
3(SQRT(4 * 2)) - 5(SQRT(2))
Factor SQRT(4 * 2) into two
radicals.
3(SQRT(4))(SQRT(2)) - 5(SQRT(2))
Take the square root of 4.
3 * 2(SQRT(2)) - 5(SQRT(2))
6(SQRT(2)) - 5(SQRT(2))
Combine like terms.
SQRT(2)
To multiply radical expressions that have factors which contain more than
one term, use the same procedure you would when multiplying
polynomials. Example:
3. Problem: CBRT(y) * (CBRT(y2) + CBRT(2))
Solution: Use the distributive law of
multiplication, which says that
a(b + c) = ab + ac to
multiply the expression out.
CBRT(y) * CBRT(y2) + CBRT(y) * CBRT(2)
Multiply the radicals.
CBRT(y3) + CBRT(2y)
Take the cube root of
y3.
y + CBRT(2y)
When dealing with radicals and fractions, you will, on many occasions, get
an answer with a radical in the denominator. Usually, an answer is not
considered simplified until there are no radicals in the denominator. The
process of removing radicals from the denominator is called rationalizing
the denominator. An important thing to remember when rationalizing
denominators is that anything divided by itself is 1. For example, 67/67 is
the same as 1. Example:
1. Problem: Rationalize the denominator.
4 + SQRT(2)
----------5 - SQRT(2)
Solution: Multiply by 1 (make sure
the fraction you choose to
use as one will make the
denominator a perfect
square — the conjugate is
usually a good number).
4 + SQRT(2)
5 + SQRT(2)
----------- * ----------5 - SQRT(2)
5 + SQRT(2)
Multiply the problem as you
would multiply any fractions.
Also, the FOIL method of
multiplying binomials will
come in handy.
20 + 4(SQRT(2)) + 5(SQRT(2)) + (SQRT(2))2
----------------------------------------25 + 5(SQRT(2)) - 5(SQRT(2)) - (SQRT(2))2
Perform any indicated operations.
20 + 4(SQRT(2)) + 5(SQRT(2)) + 2
-------------------------------25 + 5(SQRT(2)) - 5(SQRT(2)) - 2
Perform any indicated operations,
and combine like terms, if you
can.
22 + 9(SQRT(2))
--------------23
By definition, a1/2 * a1/2 = a(1/2) + (1/2) = a1. Thus, a1/2 = SQRT(a). Similarly,
a1/3 = CBRT(a), and so on.
Those examples show us that fractions can be used for exponents.
However, this could become confusing when we see a number raised to the
four-fifths power. Actually, it's not that bad. There is a definition that
states, for any natural numbers, m and k, and any nonnegative number a,
am/k equals the k-th root of am. Example:
1. Problem: Simplify (27)2/3
Solution: Rewrite using the definition
stated above.
CBRT(272)
CBRT(729)
9
By reversing the definition used in the example problem, you can simplify
problems that use abnormally large indexes and powers. Example:
2. Problem: Simplify the 6th root of x3
Solution: Rewrite by reversing the
definition.
x3/6
Use arithmetic to simplify the
exponent.
x1/2
By definition, x1/2 is
the same as SQRT(x).
Also, exponents can be written as negative numbers. Since having negative
exponents in an answer is not considered simplified, you need to know that
a-(m/n) = 1/am/n. Example:
3. Problem:
Simplify 4-(1/2)
Solution: Use the definition of negative
exponents and rewrite the
expression with positive
exponents for a simplified expression.
1
----4(1/2)
Since 41/2 is the same
as SQRT(4) and the square root
of 4 is 2, the answer
is 1/2.
Radical equations are equations that contain variables under one or more
radicals. The Principle of Powers theorem helps us get past such problems.
It says, For any real number n, if an equation a = b is true then the equation
an = bn is true. Example:
1. Problem: SQRT(x) - 3 = 4
Solution: Add 3 to each side.
SQRT(x) = 7
Using the Principle of Powers
theorem, square both sides.
x = 72
x = 49