Transcript Slide 1

Chapter 12:
Radicals and More
Connections to Geometry
By Marcy Schaeffer and Chris
Simmons
12.1 Functions Involving
Square Roots
Key Words
square root function: y=√x
domain: all the possible input values
of a function (all the numbers you
plug into the equation)
range: the collection of all output
values of a function (all the possible
answers after you plug a number into
an equation)
Function: y=x²
X
Y
(Domain) (Range)
Input
Output
1
1
2
4
3
9
4
16
12.2 Operations with Radical
Expressions
Key Words
simplest form of a radical
expression: form when all of the
possible square roots are no longer
part of the equation
The distributive property can be used
to simplify the expression when they
have the same radicand (i.e. if all
parts of the have a √3).
Examples!! Yay.
4√3 - √27= 4√3 - √(9 x 3)
= 4√3 - √9 x √3
= 4√3 - 3√3
= √3
Factor one of the numbers under the
square root only if one of the factors
is the same as another number
under a square root. Then distribute
the square root to the factors.
Finally, simplify.
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Examples (cont.)
√2 x √8= √16= 4
Multiply the numbers under the square
root as if there was no square root
sign. Then, put the square root sign
over the number and simplify.
 3/√5 = 3/√5 x (√5/√5)
If an expression has a square root sign
on the bottom, multiply the top and
the bottom by the number with the
square root sign.
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12.3 Solving Radical
Equations
Key Words
radical: the square root sign
extraneous solutions: solutions that
when plugged back into the original
equation do not work. These
solutions come from squaring both
sides. You can make sure you do not
have any of these solutions by
plugging the answers back into the
original equation.
Ejemplos
If a=b then a²=b².
 √x – 7= 0
√x= 7
(√x) ²= (7)²
x=49
First simplify the equation by moving
everything but the x to one side of
the equal sign. Then, square both
sides to get rid of the square root
sign over the x. By squaring both
sides, x is left by itself.
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Ejemplos (cont.)
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Solve x + 20 = x and check for extraneous solutions.
 Square each side of the equation. ( x + 2 0 ) ² = x²
Simplify the equation. x + 20 = x²
Write the equation in standard form. X² – x – 20 = 0
Factor the equation. (x – 5)(x + 4) = 0
Use the zero-product property to solve for x.
x = 5 or x = –4
CHECK Substitute 5 and –4 in the original equation.
CHECK X = 5 CHECK X = –4
x + 20 = x
x + 20 = x
5 + 20
?= 5 –4 + 20
?= –4
5 = 5 4 ≠ –4
ANSWER  The only solution is 5, because x = –4
does not satisfy the=original equation.
Solving Radical Equations
with Extraneous Solutions
First, square both sides of the equation
sign to get rid of the square root sign
above the x. Then, move everything
to one side of the equal side so it
equals 0. Solve for x by factoring or
by using the quadratic formula.
Finally, plug these solutions for x into
the original equation and see if the
equation works. If one of the
solutions you found for x does not
work, then that value of x is not a
solution for the equation.
12.4 Rational Exponents
Key Words
cube root of a: If b³ =a, then b is
called a cube root of a.
rational exponents: a^(m/n)=
(a^(1/n))^m
Properties of Rational
Exponents
a^m x a^n= a^(m+n)
If you are multiplying numbers of the same
base (i.e. “a”) with exponents, then add the
exponents of the numbers.
 (a^m)^n
If you are multiplying a number with an
exponent and just an exponent, then
multiply the exponents.
 (ab)^m
If you are multiplying numbers in parentheses
by an exponent, then distribute the
exponents to the numbers in the
parentheses.
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Examples
5^(1/3) x 5^(2/3)
Since the base of the exponents is the
same (the 5), then add the
exponents. 1/3+2/3=3/3=1. Thus the
answer is 5^1 or 5.
 (7^1/3)^6
Since you are multiplying a base with
an exponent and just an exponent,
simply multiply the exponents.
1/3 x 6=2 so the answer is 7² or 49.
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12.5 Completing the Square
Key Words:
completing the square: To
complete the square of x² + bx,
add the square of half of the
coefficient (the number in front
of) of x, that is, add (b/2)².
quadratic formula:
Methods for Solving Quadratic
Equations
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Finding square roots
Graphing
Using the Quadratic Formula
Factoring (if easy to do so)
Completing the Square (most
useful when number in front of
x² is 1 and the number in front of
the x is even)
Solving a Quadratic Equation
x² + 10x= 24
x² + 10x + 5² = 24 + 5²
(x + 5)²=49
x + 5= 7 or x + 5= -7
x= 2 or -12
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First, make sure all x’s are on one side of the
equal side, while all the numbers without
an x are on the other. Then, divide the
number in front of the x by 2. Square this
number and add it to both sides of the
equation. Then, find the factors of the
quadratic equation and solve it by factoring
or by using the quadratic equation.
Remember!
When completing the square,
always add the number divided
by two square to both sides of
the equation.
12.6 The Pythagorean
Theorem and Its Converse
Key Words:
theorem: a statement that can be proven true
Pythagorean theorem: If a triangle is a right
triangle, then the sum of the squares of the
lengths of legs a and b equals the square
of the length of the hypotenuse c. a² + b²=
c²
hypotenuse: the side of a triangle across
from the right angle, the longest side
legs: the other two sides of a triangle that are
not the hypotenuse
converse: a statement in which the
hypothesis and conclusion are
interchanged
Using the Pythagorean
Theorem
Usually the two sides or legs of a right
triangle are labeled a or b, and the
hypotenuse is labeled c.
Example…..who knew?
By using the Pythagorean theorem, just
determine the given values of a, b, and c.
Remember, a and b are the legs of the
triangle and c is the hypotenuse. Then,
plug the values into the equation and solve
for the unknown value.
A harder example
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A right triangle has one leg that is 7 inches longer than the other
leg. The hypotenuse is 17 inches. Find the unknown lengths.
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SOLUTION
a² + b² = c²
Write Pythagorean theorem.
x² + (x + 7) ² = 172
Substitute for a, b, and c.
x² + x² + 14x + 49 = 289
Simplify.
2x² + 14x – 240 = 0
Write in standard form.
X² + 7x – 120 = 0
Divide each side by 2.
(x – 8)(x + 15) = 0
Factor.
x = 8 or x = –15
Zero-product property
ANSWER  Length is positive, so the solution x = –15 is
extraneous. The
sides have lengths of 8 inches and 8 + 7 = 15 inches.
CHECK Substitute 8 and 15 for the lengths of the legs in the
Pythagorean theorem.
a² + b² = c²
82 + 152
?= 172
64 + 225 = 289
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Converse of the Pythagorean
Theorem
If a triangle has side lengths a, b, and c
such that a² + b² = c², then the
triangle is a right triangle.
For example, by plugging side lengths
15, 20, and 25 into the Pythagorean
theorem, it shows that the triangle is
a right triangle; the numbers on both
sides of the equal sign are the same.
Remember that the longest side is
the hypotenuse, or c.
12.7 The Distance Formula
Key Words:
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distance formula: Given the
two points (x1, y1) and (x2, y2),
the distance between these
points is given by the formula:
Distance Formula.
Distance Formula Example
Use the distance formula to find
the distance between points (37,
122) and (34, 117)
Tough Distance Formula
Example
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A player kicks a soccer ball from a position that is 8 yards
from a sideline and 10 yards from a goal line. The ball
lands at a position that is 45 yards from the same sideline
and 38 yards from the same goal line. How far was the
ball kicked?
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SOLUTION
Assuming the kicker is left of the goalie, the ball is
kicked from the point (8, 10) and lands at the point (45,
38). Use the distance formula.
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d = (x2 – x1)² + (y2 – y1) ²
= (45 – 8)² + (38 – 10) ²
= 372 + 282
= 1369 + 784
= 2153
≈ 46.4
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ANSWER  The ball was kicked about 46.4 yards.
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Write the formula.
Substitute.
Simplify.
Evaluate powers.
Add.
Find the square root.
X-pla-nay-shun
To find the distance between two
points, first determine which
point is the “first” point and
which point is the “second” point
(which point is “first” or “second”
does not matter). Then, plug in
the x and y values into the
distance formula. Simplify and
solve.
12.9 The Midpoint Formula
Key Words:
midpoint: the point on a segment
that is the same distance from
both the endpoints
midpoint formula:
Midpoint Example
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Given the points ( 1, -2 ) and ( -3, 5 ), find
the midpoint of the line segment joining
them.
Using the midpoint formula, label the points
as x1 = -1, y1 = -2, x2 = -3, and y2 = 5. To
find the midpoint M of the line segment
joining the points, use the midpoint
formula:
Overview
To find the midpoint of a line,
determine the endpoints of the
segment. Of these endpoints,
determine the “first” and
“second” points. Plug these
values of x and y into the
equation.
12.9 Logical Reasoning: Proof
Key Words:
postulate/axiom: properties that
mathematicians accept without proof
counterexample: used to show that a
general statement is false
indirect proof: proof by assuming the
statement is false
conjecture: a statement that is thought
to be true but has not been proven
true
The Basic Axioms of Algebra
Let a, b, and c be real numbers.
Axioms of Addition and Multiplication
Closure: a + b is a real number
Commutative: a + b = b + a
Associative: (a + b) + c= a + (b + c)
Identity: a + 0 = a, 0 + a= a
Inverse: a + (-a) = 0
Axiom Relating Addition and Multiplication
Distributive: a(b + c) = ab + ac
Axioms of Equality
Addition: If a = b, then a + c = b + c
Multiplication: If a = b, then ac = bc
Substitution: If a = b, then a can be
substituted for b.
The End
Get excited!
You just learned about equations
with the square root sign and
with exponents, the
Pythagorean theorem, the
distance formula, and the
midpoint formula. How fun.