Algebra II: Powers and Roots

On this page we hope to clear up problems you might have with powers (exponents) and roots (square, cube, n-th).  Square roots and exponents are used everywhere in math and can get quite complicated.  Read on or follow any of the links below to better your understanding of powers and roots.

Square roots
Cube roots
n-th roots
Multiplication and simplification
Division
Rationalizing the denominator
Roots and exponents
Quiz on powers and roots

Square Roots

By definition, every real number has two square roots.  For example, 64's two square roots are 8 and -8 becuase 8^2 = 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(a^2) = |a|.

Example:

```1.  Problem: SQRT((-16)^2)
Solution: By the theorem above, the answer is |-16|.  The absolute
value of -16 is 16.
```

Cube Roots

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(27y^3)
Solution:  3y
```

n-th Roots

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 a^k 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.

2.  Problem: Find the 10th root of 50^10.
Solution: By the n-th Root Theorem, the 10th root of 50^10
is |50|, or 50.
```

Multiplication and Simplification

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

SQRT((x + 2)(x - 2))

Multiply the binomials under the radical out.

SQRT(x^2 - 4)
```
Reversing the theorem stated above gives us a way to factor radical expressions, thereby simplifying them.

Example:

```1.  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/y^2), 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/y^2)
Solution: Use the Roots of Fractions Theorem to rewrite the
problem.

SQRT(25)
---------
SQRT(y^2)

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:

```1. Simplify: (SQRT(80))/(SQRT(5))
Solution: Use the converse of the Roots of Fractions Theore and

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

10(SQRT(7))
```
You have to look out for terms that do not look alike, but could be if factored.

Example:

```1.  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 sqaure 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:

```1.  Problem: CBRT(y) * (CBRT(y^2) + 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(y^2) + CBRT(y) * CBRT(2)

CBRT(y^3) + CBRT(2y)

Take the cube root of y^3.

y + CBRT(2y)
```

Rationalizing the Denominator

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 1 will
make the denomiator a perfect sqare (the conjugate is usually a
good choice).

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.

4(5) + 4(SQRT(2)) + 5(SQRT(2)) + (SQRT(2))^2
--------------------------------------------
5^2 + 5(SQRT(2)) - 5(SQRT(2)) - (SQRT(2))^2

Perform any indicated operations.

20 + 9(SQRT(2) + 2
------------------
25 - 2

Perform any indicated operations, and combine like terms, if you
can.

22 + 9(SQRT(2))
---------------
23
```

By definition, a^(1/2) * a^(1/2) = a^((1/2) + (1/2)) = a^1.  Thus, a^(1/2) = SQRT(a).  Similarly, a^(1/3) = 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, a^(m/k) equals the k-th root of a^m.

Example:

```1.  Problem: Simplify (27)^(2/3).
Solution: Rewrite using the definition stated above.

CBRT(27^2)
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 x^3. Solution: Rewrite by reversing the definition. x^(3/6) Use arithmetic to simplify the exponent. x^(1/2) By definition, x^(1/2) 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/(a^(m/n)).

Example:

```1.  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 4^(1/2) is the same as SQRT(4) and 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 a^n = b^n 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 = 7^2
x = 49
```

Take the quiz on powers and roots.  The quiz is very useful for either review or to see if you've really got teh topic down.

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