7.1
Radicals and Rational Exponents
Motivation: diagonals of rectangles can be found by square roots
Def of
-- "square root
of a"
1. If
then b is a real
number such that b2 = a
2. b is the prinicpal root (i.e. the positive root [for even roots])
So the negative
square root of a is -![]()
Square roots of negative numbers are not defined (for real numbers)
-- so both a and b are positive only
Def of higher roots
Cube root of a: (real number b such that b3 = a) b pos or neg Ex ![]()
= -3
Fourth root of a: (real number b such that b4 = a) b and a are pos
only Ex
= 2
"nth root of a:" (real nr b such that bn = a);
General restriction: for even roots [i.e. n is even], both a and b can only be positive.
Negative radicands result in:
neg root if radicand is odd Ex: cube root (-125)
no real number if radicand is even Ex: fourth root (-16)
Terminology: radical expression, index, radical sign, radicand
For a rational (fractional) exponent,
Numerator = power
Denominator = root
Illustration: Solve x = 8 ^ (1/3) by cubing both sides; then solve x = 8 ^ (2/3)
If
n is odd, then ![]()
The same property holds whenever n is odd, even if x is negative.
If we assume all vars are nonnegative, then we can divide exponent by the index to simplify
(Restriction is necessary for even indices, not odd indices)
Ex: 5th root of (25x15y8z17w44)
If we rewrite this using rational exponents, and convert to mixed numbers (or divide)
Whole number is exponent outside the radical.
Numerator is exponent inside the radical.
or, if we divide:
Quotient is exponent outside the radical
Remainder is exponent inside the radical
Review properties of exponents -- these all apply to rational exponents, too.
1. (ar)(as) = ar+s 2. ar / as = ar-s so rule 3. 1/as = a-s = a-s/1 and 1/a-r = ar / 1 = ar
4. (ab)r = arbr 5. (a/b)r = ar / br
6. (ar)s = ars
Thm 5.1:
provided a, n positive
7.2
Simplifying Radical Expressions
Standards
1. No perfect squares under sq root symbol -- (so, factor out perfect squares)
Ex: ![]()
, ![]()
No perfect cubes under cube root symbol, etc -- (so, factor out perfect cubes, etc)
Ex: ![]()
So...no perfect nth powers under nth root symbol (so, factor out perfect nth powers)
Ex:
![]()
2. No fractions under a radical -- so, rationalize the denominator (as described below)
![]()
3. No radicals in a denominator -- so, rationalize the denominator (as described below)
![]()
Ex. for
class:
for n = 2, 3, 4, 5
Removing factors from the radicand:
Either:
(1) Factor out a perfect square (for sqrt), perfect cube (for cube roots), etc., or
(2) Write the prime factorization and divide each exponent by the index
The quotient is the exponent outside the radical
The remainder is the exponent inside the radical.
or, if it doesn't divide evenly, factor out just enough to get even division (factor out base remainder.)
Alternatively, convert to a rational exponent; write as a mixed number: integer part comes out of the radical, and the fraction part stays under the radical.
Rationalizing the denominator:
In general, multiply as necessary to make the radicand a perfect square (for sqrt), a perfect cube (for cube roots), a perfect 4th power (for 4th roots), etc.
If it's not obvious, do this:
1. Write the prime factorization of each factor in the radicand
2. Simplify the radical in the denominator. (Standard 1 above)
3. For each base, the power you need to multiply by is the index minus the remaining power
Ex: for
this simplifies to
Multiply by 23-2*x3-1
Applications:
Pythagorean theorem: a2 + b2 = c2 We need to take square roots to solve for any side
Example: Special triangles - It's useful to know the ratios of the sides
45 - 45 - 90 Ratios: 1 to 1 to sqrt(2) Why? Solve for the hypotenuse in the triangle.
30 - 60 - 90 Ratios: 1 to 2 to sqrt(3) Why? Solve for the altitude in an equilateral triangle.
Period of a pendulum: t = 2π * sqrt (L / 32), where t = period in seconds, L = length in feet.
Tasks: substitute for either t or L and solve for the otehr variable.
7.3 Addition and Subtraction of Radical Expressions
Def of similar radicals: have the same index and the same radicand
Technique: simplify first, then collect like terms (i.e. similar radicals)
We can factor out the common radical factor
Examples:
![]()
![]()
![]()
Modification of
previous problem
![]()
(But our text has
only square roots -- do this as square roots)
Special triangle: 30o-60o-90o
From an equilateral triangle: Label sides as 2x, use Pythagorean theorem to calculate height.

7.4
Multiplying and Dividing Rational
Expressions
Multiply factors outside the radicals and leave the results outside
Multiply factors inside the radicals and leave the results inside
Ex:
Simplify
when complete
![]()
![]()
![]()
Rationalizing denominator when the denominator has 2 terms, (where one or both are square roots)
Principle: When we multiply a binomial by the conjugate, the square root(s) disappear
Ex: ![]()
Ex:
![]()
![]()
Golden ratio problem: #110
To show that a given
rectangle is a golden rectangle, show that the length/width =
,
or
equivalently,
The text has us
show that these are the same.
Extension (not in our text): multiply as necessary to get a sum/difference of cubes
For
, multiply num and denom by
(the other factor if we factor x-8 as a difference of cubes)
7.5 Radical Equations
Principle: to clear radicals from radical equations:
1. Isolate one radical term on one side of the equation
2. Square both sides (for square roots) [For cube roots, cube both sides, etc.]
Remember: if the other side of the equation is x+3, the square is (x+3)2, not x2+32.
Caution: The solutions to the original equation will be included in the new solution set, but you may have introduced extraneous solutions as well. That is, the new equation may also include additional ("extraneous") solutions that are not valid in the original equation. Therefore, you must test all solutions obtained by this technique.
Ex: sqrt (3x + 7) = 5
Ex: cube root (3x + 7) = 5
Occasionally, it will be obvious that there is no solution:
Ex: sqrt (5x+1) = –8 All square roots have to be positive, so we can't possibly have –8.
Ex: sqrt(a+10) + 2 = a Poss sol –1 and 6; only 6 checks
Ex: sqrt(x2–1)=x+3
What do you do if there are two square roots? Go through the same "isolate and square" procedure twice
Ex: sqrt(2x+4) = sqrt(x+3) + 1 Poss sol –2 and 6; only 6 checks.
7.6 Complex Numbers
Definitions: we define i = sqrt(-1), or equivalently i2 = -1
Consequence: if 'a' is a positive number, sqrt(-a) = i*sqrt(a)
Tasks:
simplify radicals containing negative radicands: sqrt(-96)
simplify i2, i3, i4 using the principle i2 = -1
simplify expressions such as , i56, i18, i41, i39 using the principle i2 = -1
(quick key: remainder on div by 4)
Def: a complex number is any number that can be written in the form a+bi, where a and b are real.
"standard form"
equality: two complex numbers are equal if the real parts are equal and the imaginary parts are equal.
Ex: Solve sqrt(3) / 2 + 1/2 i = (2x-y) + y i
+, -, mult: straightforward: treat i as a variable, and clean up higher powers of i when complete.
For mult of 2 complex numbers, use FOIL.
Ex: [(3+2i)-(4-5i)](3-2i)2
Show that x=1-2i is a solution of x2-2x+5
division: multiply by the conjugate of the divisor/denominator
Divide 3+2i by 5-3i and write the result in standard form
Background for #79: electrical engineers use complex numbers to account for different "phase shifts" in alternating electric currents. For example, one voltage (or current) may be increasing while another is falling. Complex numbers make it possible to make calculations based on voltages and currents that are not "in phase" with each other.
Example (we haven't had the trig, so you'll have to take my word for that part - for now) Two voltage sources are at work in a circuit, 60 volts at a phase shift of 0 degrees, and 110 volts at a phase shift of 90 degrees. Picture these on a graph with the positive x-axis at 0 degrees. The positive x-axis is the "real" axis and the y-axis (at 90 degrees to it) becomes the "imaginary" axis. The voltages get converted to complex numbers: 60 + 0i and 0 + 110i. The sum is 60 + 110i. Converting back to the angle form, we get 125.3 volts (from the Pythagorean theorem) at 61.4 degrees (from tan θ = 110/60).
Return to: Merced College; Don Power Updated 11/13/07 by Don Power