unit #8 radicals and the quadratic formula 8 radicals...unit #8 radicals and the quadratic formula...
TRANSCRIPT
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Unit #8
Radicals and the Quadratic Formula
Lessons:
1 - Square Root Functions
2 - Solving Square Root Equations
3 - The Basic Exponent Properties
4 - Fractional Exponents Revisited
5 - More Exponent Practice
6 - The Quadratic Formula
7 - More Work with the Quadratic Formula
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Lesson 1: Square Root Functions and Graphs
Square roots are the natural inverses of squaring . In other words, to find the square root of an input, we must
find a number that when squared gives the input. Because of their important role in higher-level mathematics, it is
important to understand their graphs, as well as their domains and ranges.
Exercise #1: Consider the two functions and 3 2f x x g x x . (a) Graph π¦ = π(π₯) without the use of your
calculator on the grid shown. Label its equation.
(b) Using your calculator to generate a table of values, graph π¦ = π(π₯) on the same grid and
label its equation. Start your table at π₯ = β10 to
see certain x-values not in the domain of this
function.
(c) State the domain and range of each function below using set-builder notation.
π(π₯) = βπ₯ π(π₯) = βπ₯ + 3 β 2
Domain = Domain =
Range = Range =
Shifts in Graphs:
Weβve seen in previous chapter that horizontal movement (left or right) is controlled by any constants inside
with the x-term, and that vertical movement (up or down) is controlled by any constants grouped with the y-term
or alone.
Additionally, coefficients effect the graph by changing how wide or narrow the graph is, and can change the
direction that a graph goes in. For example, π¦ = βπ₯2 would open downwards because of the negative coefficient.
These major shifts hold true for many function graphs, including the square root
Exercise #2: Which of the following equations would represent the graph shown?
(1) π¦ = ββπ₯ + 4
(2) π¦ = 4β βπ₯
(3) π¦ = βπ₯ β 4
(4) π¦ = ββπ₯ β 4
y
x
y
x
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As we saw in the first exercise, the domains of square root functions are oftentimes limited due to the fact that
square roots of negative numbers do not exist in the Real Number System. We shall see in Unit #9 how these
square roots can be defined if a new type of number is introduced. For now, though, we are only working with real
numbers.
Recall: To determine the domain for these functions, we take the radicand (term(s) under the square root)
and set it greater than or equal to zero, and solve.
πππ πππππ β₯ π
Exercise #3: Which of the following values of x does not lie in the domain of the function π¦ = βπ₯ β 5?
Explain why it does not lie there.
(1) π₯ = 6 (3) π₯ = 5
(2) π₯ = 2 (4) π₯ = 7
Exercise #4: Determine the domain for each of the following functions. Show an inequality that justifies you work.
(a) π¦ = βπ₯ + 2 (b) π¦ = β3π₯ β 2 (c) π¦ = β8β 2π₯
Exercise #5: Consider the function π(π₯) = βπ₯2+ 4π₯ β 12.
(a) Use your calculator to sketch the function on the axes
given.
(b) Set up and solve a quadratic inequality that yields the
domain of π(π₯).
y
x
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Lesson 1 Homework: Square Root Functions and Graphs
1. Which of the following represents the domain and range of π¦ = βπ₯ β 5+ 7?
(1) Domain: [β5, β) (3) Domain: (β7, β)
Range: [7, β) Range: (5,β)
(2) Domain: [5, β) (4) Domain: [7, β)
Range: [7, β) Range: [5, β)
2. Which of the following values of x is not in the domain of π¦ = β1β 3π₯?
(1) π₯ =1
3 (3) π₯ = 0
(2) π₯ = β1 (4) π₯ = 4
3. Which of the following equations 4. Which equation below represents the
describes the graph shown below? graph shown below?
(1) π¦ = βπ₯ + 4+ 1 (1) π¦ = βπ₯ β 2β 5
(2) π¦ = βπ₯ β 4β 1 (2) π¦ = ββπ₯ + 2+ 5
(3) π¦ = βπ₯ + 4β 1 (3) π¦ = ββπ₯ β 2+ 5
(4) π¦ = βπ₯ β 4+ 1 (4) π¦ = βπ₯ + 2+ 5
5. Determine the domains of each of the following functions. State your answers in set-builder notation.
(a) π¦ = βπ₯ + 10 (b) π¦ = β3π₯ β 5 (c) π¦ = β7β 2π₯
y
x
y
x
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6. Set up and algebraically solve a quadratic inequality that results in the domain of each of the following. Verify
your answers by graphing the function in a standard viewing window.
(a) π¦ = βπ₯2 β4π₯ β 5 (b) π¦ = β9β π₯2
7. Consider the function π(x) = ββπ₯ + 5+ 3.
(a) Graph the function y g x on the grid shown.
(b) Describe the transformations that have
occurred to the graph of y x to produce
the graph of y g x . Specify both the transformations and their order.
y
x
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Answers to Lesson 1 Homework
1) (2)
2) (4)
3) (1)
4) (3)
5) (a) {π₯ β₯ β10} (b) {π₯ β₯5
3} (c) {π₯ β€
7
2}
6) (a) (b) First the function is reflected in the x-axis. Then, the
function is shifted 5 units to the left and shifted
3 units up.
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Lesson 2: Solving Square Root Functions
Equations involving square roots arise in a variety of contexts, both applied and purely mathematical. As always,
the key to solving these equations lies in the applications of inverse operations. Recall that a square and square
root are inverses.
Get the square root alone
Square both sides
Finish solving
Check all answers for extraneous roots
Oftentimes, roots are introduced by various algebraic techniques that for one reason or another are not valid
solutions of the equations. These roots are known as extraneous and can always be found by calculator-checking
within the original equation.
Exercise #1: Solve each of the following square root equations. Check each equation for extraneous roots
(a) βπ₯ = 7 (b) βπ₯ β 3 = 5 (c) β2π₯ β 1 = 4
(d) 3βπ₯ β 4 = 20 (e) 2βπ₯ + 5+ 7 = 13 (f) 5β3π₯ β 2β 4 = 36
_____Exercise #2: Which of the following is the solution to 3βπ₯
2= 15?
(1) π₯ = 12.5 (2) π₯ = 50
(3) π₯ = 25 (4) π₯ = 4050
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Another scenario arises when a square root expression is equal to a linear expression. The next exercise will
illustrate both the graphical and algebraic issues involved.
Exercise #3: Consider the system of equations shown below.
π¦ = βπ₯ + 3 and π¦ = π₯ + 1
(a) Solve this system graphically using the grid to the
right.
(b) Solve this system algebraically for only the x-values
using substitution below.
(c) Why does your answer from part (a) contradict what you found in part (b)?
Exercise #4: Find the solution set of each of the following. Be sure to check your work and reject any extraneous
roots.
(a) β2π₯ β 3 = π₯ β 3 (b) 2π₯ = βπ₯ + 6 β 2
y
x
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Lesson 2 Homework: Solving Square Root Functions
1. Solve and check each of the following equations. As in the lesson, they are arranged from lesser to more
complex.
(a) βπ₯ = 5 (b) βπ₯ + 2 = 10 (c) β2π₯
3= 6
(d) 4βπ₯ = 24 (e) 2βπ₯ = 1 (f) β3π₯ = 4 = 8
(g) 1
2βπ₯ β 5 = 2 (h) β4π₯ β 1+ 3 = 4 (i) 5β1β 5π₯ β 3 = 27
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_____2. Which of the following values solves the equation β4π₯+19
2= 2?
(1) β9
2 (2) β
3
4
(3) 4
3 (4)
1
2
3. Solve and check each of the following equations for all values of x. Reject any extraneous roots.
(a) π₯ β 1 = βπ₯ + 11 (b) β4π₯ + 36 = 2π₯ β 6
.
(c) 6π₯ = 2β2π₯ + 17 β 8 (d) β6π₯+4β1
4= π₯
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Answers to Lesson 2 Homework
1)
(a) x = 25 (b) x = 98 (c) x = 54 (d) x = 36 (e) π =π
π
(f) x = 20 (g) x = 196 (h) π =π
π (i) π = βπ
2) (2)
3) (π) π = π;π β βπ (ππππππππππ ππππ)
(b) π = π;π β π (ππππππππππ ππππ)
4) (π) π = {Β±π
π}
(b) π =π
π; π β β
π
π (ππππππππππ ππππ)
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Lesson 3: The Basic Exponent Properties
Exponents, which indicate repeated multiplication, are extremely important in higher-level mathematical study
because of their importance in numerous areas. The rules they play by, known as the exponent properties, are
critical to master. You should have seen these properties in previous math courses.
Recall Exponent Properties or Exponent Laws:
1) (ππ)(ππ)= ππ+π 2) ππ
ππ= ππβπ; πππππ π β π
3) (ππ)π= ππβπ 4) (π β π)π = ππ β ππ and (π
π)π
=ππ
ππ
5) πβπ=π
ππ and
π
πβπ= ππ 6) ππ = π
Exercise #1 (Property #1): Rewrite each of the following in simplest form:
(a) π₯10 β π₯3 = (b) (5π₯4)(6π₯3) = (c) π₯3π¦2π₯6π¦ =
Exercise #2 (Property #2): Rewrite each of the following in simplest form:
(a) π₯8
π₯2= (b)
6π₯10
12π₯4= (c)
2π₯6π¦3
8π₯π¦2=
Exercise #3 (Property #3): Simplify each of the following:
(a) (π₯2)3 = (b) (π¦4)6= (c) (π1
2)6
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(d) Which of the following expressions is not equivalent to 30x ?
(1) (π₯10)3 (3) π₯5 β π₯6
(2) (π₯6)5 (4) π₯10 β π₯20
Exercise #4 (Property #4): Rewrite each of the following as equivalent expressions:
(a) (2π₯2)3 = (b) (3
π₯2)4
= (c) (β2π₯2π¦5
3π§3)3
=
Exercise #5 (Property #5): Rewrite each of the following without the use of negative exponents:
(a) 32 (b) 4x (c)
1
π₯β3 (d)
3π₯4π¦10
15π₯6π¦9
(e) 5π₯4π¦β3
35π₯β2π¦8 (f) (
π₯10π¦β5
3π₯β2π¦)2
Exercise #6 (Property #6): Simplify each of the following:
(a) 05 (b) 03x (c) 4(2π₯)0=
Handle the negatives
Reduce the inside as
much as possible
Raise to the outside
power
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Lesson 3 Homework: The Basic Exponent Properties
1. Express each of the following expressions in "expanded" form, i.e., do all of the multiplication and/or division
possible and combine as many exponents as possible.
(a) π₯3 β π₯12 (b) 4π₯3 β 5π₯5 (c) (β3π₯2π¦)(5π₯7π¦3) (d) (4π₯3π¦6)(β7π₯4)
(e) π₯9
π₯3 (f)
5π₯3π¦7
15π₯π¦2 (g)
π₯3
π₯10 (h)
10π₯4π¦3
25π₯8
(i) (π₯5)8 (j) (10π₯3)0 (k) (β4π₯5)3 (l) (2π₯β2)4
_____2. Which of the following is not equal to 2β2? Do not use your calculator to do this problem. Show your
algebra.
(1) 1
4 (3) 0.25
(2) β4 (4) 1
22
_____3. If the expression 1
2π₯ was placed in the form ππ₯π where a and b are real numbers, then which of the following
is equal to π + π? Show how you arrived at your answer.
(1) 1 (3) 1
2
(2) 3
2 (4) β
1
2
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_____4. If π(π₯) = 5π₯0 +4π₯β3 then π(π) =
(1) 12π β 5 (3) 1
4π3+5
(2) 5+4
π3 (4) β12π + 1
_____5. Which of the following is equivalent to (4π₯8)
3
(6π₯5)2 for all π₯ β 0? Show the work that leads to your final answer.
(1) 16
9π₯14 (3)
2
3π₯14
(2) 16
9π₯4 (4)
2
3π₯4
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Answers to Lesson 3 Homework
1)
(a) π₯15 (b) 20π₯8 (c) β15π₯9π¦4 (d) β28π₯7π¦6
(e) π₯6 (f) 1
3π₯2π¦5 (g)
1
π₯7 (h)
2π¦3
5π₯4
(i) π₯40 (j) 1 (k) β64π₯15 (l) 16
π₯8
2) (2) Algebra required
3) (4) Algebra required
4) (2)
5) (1)
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Lesson 4: Fractional Exponents Revisited
Recall that in Unit #4 we introduced the concept that roots could be represented by rational or fractional
exponents.
Recall we can also combine integer powers with roots with the following:
Exercise #1: Rewrite each expression in the form bax where a and b are both rational numbers.
(a) 5βπ₯ (b) βπ₯5
4 (c)
7
βπ₯3 (d)
5
3 βπ₯10
Exercise #2: Rewrite each of the following power/root combinations as a rational exponent in simplest form.
(a) βπ₯7 (b) βπ₯64
(c) (βπ₯)6 (d) (βπ₯
3 )10
_____Exercise #3: If π(π₯) = 10π₯32 β β 24π₯β1, then which of the following represents the value of π(4)? Find the
value without the use of a calculator. Show the steps in your calculation.
(1) 36 (3) 54
(2) 48 (4) 74
_____ Exercise #4: Which of the following is not equivalent to π₯β7
3β ?
(1) 1
π₯73β (3)
1
βπ₯73
(2) 1
βπ₯37 (4) β
1
π₯7
3
UNIT FRACTION EXPONENTS
For n given as a positive integer: π1πβ = βπ
π
πππ€ππ
ππππ‘
βpower over rootβ
RATIONAL EXPONENT CONNECTION TO ROOTS
For the rational number π
π, π
ππβ is equivalent to: βππ
π or (βπ
π )π
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Variables are perfect
roots if the exponent is
divisible by the index
Fractional exponents play by the same rules (properties) as all other exponents. It is, in fact, these properties that
can justify many standard manipulations with square roots (and others). For example, simplifying roots.
Exercise #5: Simplify each of the following roots. Show manipulations. Be thoughtful about the index being used!
It may be helpful to write or calculate a list of perfect squares, cubes, etc
(a) β28 (b) βπ₯6π¦11
(c) β18π₯4 (d) β200π₯5π¦3 (e) β147π₯9π¦4
(f) β163 (g) β108
3 (h) β2503
(i) β12π₯83
(j) β1624 (k) β16π₯8
4
(l) β48π₯10π¦54 (m) β64π₯12π¦155
SIMPLIFYING RADICALS
βπ β π = βπ β βπ or βπ β π π = βπ
π β βππ
; try to keep π as the largest perfect square divisor
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Lesson 4 Homework: Fractional Exponents Revisited
_____1. Which of the following is equivalent to π₯52β ?
(1) 5π₯
2 (2)
2π₯
5 (3) βπ₯5 (4) βπ₯2
5
_____2. If the expression 1
βπ₯ was placed in π₯π form, then which of the following would be the value of a?
(1) -2 (2) 2 (3) 1
2 (4) β
1
2
_____3. Which of the following is not equivalent to βπ₯9 ?
(1) π₯3 (2) (βπ₯)9 (3) π₯
92β (4) π₯4βπ₯
_____4. The radical expression β50π₯5π¦3 can be rewritten equivalently as
(1) 25π₯π¦β2π₯π¦ (3) 5π₯2π¦β2π₯π¦
(2) 5π₯π¦βπ₯π¦ (4) 10π₯2π¦β5π₯π¦
_____5. If the function π¦ = 12βπ₯3 was placed in the form π¦ = πππ₯ then which of the following is the value of π β π?
(1) β36 (3) 36
(2) β4 (4) 4
6. Rewrite each of the following expressions without roots by using fractional exponents.
(a) βπ₯5 (b) βπ₯3 (c) βπ₯
7 (d) βπ₯6
(e) βπ₯113
(f) 1
βπ₯4 (g)
1
βπ₯23 (h)
1
βπ₯9
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7. Rewrite each of the following without the use of fractional or negative exponents by using radicals.
(a) π₯16β (b) π₯
110β (c) π₯
β13β (d) π₯
β15β
(e) π₯35β (f) π₯
β72β (g) π₯
94β (h) π₯
β211β
8. Simplify each of the following square roots that contain variables in the radicand.
(a) β8π₯9 (b) β75π₯16π¦11 (c) 2π₯β18π₯7 (d) 3π₯2π¦β98π₯5π¦8
9. Express each of the following roots in simplest radical form.
(a) β16π₯83
(b)β108π₯5π¦103 (c) β64π₯12π¦143 (d) β375π₯7π¦113
10. Mikayla was trying to rewrite the expression 25π₯12β in an equivalent form that is more convenient to use. She
incorrectly rewrote it as 5βπ₯. Explain Mikalya's error.
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Answers to Lesson 4 Homework
1) (3)
2) (4)
3) (1)
4) (3)
5) (4)
6) (a) πππβ (b) π
ππβ (c) π
ππβ (d) π
ππβ
(e) πππ
πβ (f) πβπ
πβ (g) πβπ
πβ (h) πβπ
πβ
7) (a) βππ (b) βπ
ππ (c) π
βππ (d)
π
βππ
(e) βπππ
(f) π
βππ (g) βππ
π (h)
π
βππππ
8) (a) πππβππ (b) πππππβππ (c) πππβππ (d) ππππππβππ
9) (a) πππ βππππ
(b) ππππ βπππππ (c) πππππ βπππ (d) πππππ βππππ π
10) Mikalya raised both the x and the 25 to the π
π power. However, only the x should be raised to the
π
π
power. The correct answer should be ππβπ. She would be correct if the original expression contained
parentheses: (πππ)π
π.
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Lesson 5: Exponent Practice
For further study in mathematics, especially Calculus,
it is important to be able to manipulate expressions
involving exponents, whether those exponents are
positive, negative, or fractional. The basic laws of
exponents, which you should have learned in Algebra
1 and have used previously in this course, are shown
to the right. They apply regardless of the nature of
the exponent (i.e. positive, negative, or fractional).
Make each step carefully & thoughtfully, keeping in mind the order in which you are working.
Exercise #1: Simplify each of the following expressions. Leave no negative exponents in your answers.
(a) π₯3βπ₯4
(π₯5)2 (b)
(π₯2π¦)4
π₯5π¦7
(c) π₯β3π¦4
π₯β6π¦ (d)
(π₯β3π¦β4)2
(π₯π¦3)β4
In the last exercise, all of the powers were integers. In the next exercise, we introduce fractional powers.
Remember, though, that they will still follow the exponent rules above. If needed, use your calculator to help add
and subtract the powers.
Exercise #2: Simplify each of the following expressions. Write each without the use of negative exponents.
(a) π₯13β βπ₯
12β
π₯16β
(b) (π₯
12β )
5
π₯32β βπ₯3
(c) (4π₯
23β )
3
32π₯8
EXPONENT LAWS
1. π₯π β π₯π = π₯π+π 5. π₯βπ =1
π₯π and
1
π₯βπ= π₯π
2. π₯π
π₯π= π₯πβπ 6. (π₯π)π = π₯πβπ
3. (π₯ β π¦)π = π₯π β π¦π and (π₯
π¦)π
=π₯π
π¦π
4. π₯π
π = βπ₯ππ 7. π₯0 = 1
(For integers m and n)
-
To be fully simplified, an expression should not contain negative exponents and should not contain fraction
exponents.
Exercise #3: Rewrite each expression below in its simplest form
(a) π₯53β (b)
π₯52β
π₯43β (c)
1
π₯β3
2β
(d) π₯3
βπ₯ (e) (8π₯5)
23β (f)
(27π₯)13β
6βπ₯
Consider the real number component and the variable component individually. Be thoughtful about what is
actually happening to which piece:
_____Exercise #4: Which of the following is equivalent to β8π₯73
?
(1) 8π₯73β (3) 2π₯
37β
(2) 2π₯73β (4) 8π₯
37β
_____Exercise #5: The expression 1
β4π₯ is the same as
(1) 1
2π₯β1
2β (3) 4π₯12β
(2) 2π₯β1
2β (4) 1
2π₯12β
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Lesson 5 Homework: Exponent Practice
1. Rewrite each of the following expressions in simplest form and without negative exponents.
(a) π₯3π₯7
(π₯2)3 (b)
5π₯4
25π₯10 (c)
(π₯3π¦4)2
(π₯3π¦)3 (d)
(2π₯3)5
8π₯β4
_____2. Which of the following represents the value of πβ4
πβ2 when π = 3 πππ π = 2?
(1) 4
9 (2)
4
81
(3) 1
36 (4)
1
3
3. Simplify each expression below so that it contains no negative exponents. Do not write the expressions using
radicals.
(a) π₯72β π¦
12β
π₯34β π¦2
(b) (π₯
13β )
4
π₯β2
3β (c) (5π₯
23β π¦
β12β )(2π₯2π¦β3)
_____4. Which of the following represents the expression 24 π₯
β12β
6π₯52β
written in simplest form?
(1) 4
π₯3 (3)
π₯2
4
(2) 4π₯3 (4) 4π₯2
-
5. Rewrite each of the following expressions using radicals. Express your answers in simplest form.
(a) (4π₯)32β (b) π₯
β23β (c) (π₯4)
35β
34 5x
(d) βπ₯3
βπ₯ (e)
βπ₯βπ₯2
π₯53β
(f) (2βπ₯)
3
24π₯
_____6. Which of the following is equivalent to 5βπ₯
20π₯3?
(1) 1
4βπ₯3 (3)
1
4 βπ₯25
(2) 4
βπ₯5 (4)
1
4βπ₯5
_____7. When written in terms of a fractional exponent the expression βπ₯βπ₯
π₯β2 is
(1) π₯72β (3) π₯
β12β
(2) π₯52β (4) π₯β
32β
_____8. Expressed as a radical expression, the fraction π₯13β π₯
12β
π₯β1 is
(1) 1
βπ₯6 (3) βπ₯
611
(2) 1
βπ₯611 (4) βπ₯
116
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Answers to Lesson 5 Homework
1) (a) ππ (b) π
πππ (c)
ππ
ππ (d) ππππ
2) (2)
3) (a) πππ
πβ
πππβ
(b) ππ (c) πππ
ππβ
πππβ
4) (1)
5) (a) ππβπ (b) π
βπππ (c) π
πβπππ
(d) π
βππ (e) βπ
ππ (f) βπ
π
6) (4)
7) (1)
8) (4)
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Lesson 6: The Quadratic Formula
There are three main paths to solving for the roots (solutions, zeros, x-intercepts) of a trinomial equation: Factoring (although not all are factorable!) Completing the square Quadratic Formula
Exercise #1: Solve the following quadratic using the given method.
Express your answers in simplest radical form.
(a) completing the square. (b) the quadratic formula.
π₯2β 6π₯ + 1 = 0 π₯2 β 6π₯ + 1 = 0
_____Exercise #2: Which of the following represents the solutions to the equation π₯2 β10π₯ + 20 = 0?
(1) π₯ = {5Β±β10} (3) π₯ = {β5Β±β10}
(2) π₯ = {β5Β±β5} (4) π₯ = {5Β±β5}
Quadratic Formula: π₯ =βπΒ±βπ2β4ππ
2π
Notice that factoring
would not work in this
case. There are no
numbers which multiply
to +1 and add to -6!
Notice that factoring
would not work in this
case. There are no
numbers which multiply
to +20 and add to -10!
Make a conclusion: When a trinomial is NOT factorable, the roots will contain:
-
Exercise #3: Solve the following quadratic using the given method.
(a) factoring (b) the quadratic formula
2π₯2+ 11π₯ β 6 = 0 2π₯2+ 11π₯ β 6 = 0
Exercise #4: Solve each of the following quadratics by using the quadratic formula. Place all answers in simplest
form.
(a) 3π₯2+ 5π₯ + 2 = 0 (b) π₯2 β8π₯ + 13 = 0
(c) 2π₯2 β2π₯ β 5 = 0 (d) 5π₯2+ 8π₯ β 4 = 0
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Lesson 6 Homework: The Quadratic Formula
1. Solve each of the following quadratic equations using the quadratic formula.
Express all answers in simplest form.
(a) π₯2+7π₯ β 18 = 0 (b) π₯2 β2π₯ β 1 = 0
(c) π₯2 +8π₯ + 13 = 0 (d) 3π₯2β 2π₯ β 3 = 0
(e) 6π₯2β 7π₯ + 2 = 0 (f) 5π₯2 + 3π₯ β 4 = 0
Remember the conclusion made at the end of exercise two above. Circle a big green circle around any
trinomial that must have been factorable, based on your solutions.
π₯ = βπΒ± βπ2β 4ππ
2π
-
_____2. Which of the following represents all solutions of π₯2 β4π₯ β 1 = 0?
(1) {2Β±β5} (3) {2 Β±β10}
(2) {2Β±β5} (4) {β2Β±β12}
_____3. Which of the following is the solution set of the equation 4π₯2 β12π₯ β 19 = 0?
(1) {5
2Β±β3} (3) {
3
2Β±β7}
(2) {β2
3Β±β2} (4) {β
7
3Β±β6}
_____4. Rounded to the nearest hundredth the larger root of π₯2 β22π₯ + 108 = 0 is
(1) 18.21 (3) 6.74
(2) 13.25 (4) 14.61
-
5. Algebraically find the x-intercepts of the function π¦ = π₯2β4π₯ β 6. Express your answers in simplest radical
form.
6. A missile is fired such that its height above the ground is given by β = β9.8π‘2+38.2π‘ + 6.5, where t represents
the number of seconds since the rocket was fired. Using the quadratic formula, determine, to the nearest tenth
of a second, when the rocket will hit the ground.
-
Answers to Lesson 6 Homework
1) (a) π = {βπ,π} (b) π = {πΒ± βπ} (c) π = {βπΒ±βπ}
(d) π = {πΒ±βππ
π} (e) π = {
π
π,π
π} (f) π = {
βπΒ±βππ
ππ}
(You should have circled two questions)
2) (1)
3) (3)
4) (4)
5) π = {πΒ± βππ}
6) π = {βππ.πΒ±βπππ.ππ
βππ.π}
π = {βπ.ππππ,π.ππππ}
Since the time must be positive, we reject the negative time.
It will take approximately 4.1 seconds for the rocket to hit the ground.
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Lesson 7: More Quadratic Formula
The quadratic formula (shown above) is extremely useful because it allows us to solve quadratic equations,
whether they are prime or factorable. In this lesson, we will get more practice using this formula.
Exercise #1: Consider the quadratic function π(π₯) = π₯2 β4π₯ β 36.
(a) Algebraically determine this functionβs x-intercepts using
the quadratic formula. Write the answers in simplest
radical form AND to the nearest hundredth
(b) Find the vertex of this parabola
(c) Sketch a graph of the quadratic on the axes given. Use the ZERO command on your calculator to verify your
answers from part (b). Label the zeros on the graph.
20
50
y
x
Quadratic Formula: π =βπΒ±βππβπππ
ππ
-
_____ Exercise #2: Which of the following sets represents the x-intercepts of π¦ = 3π₯2 β19π₯ + 6?
(1) {1
2,7
3}
(2) {2ββ5,2 +β5}
(3) {1
6ββ
17
2,1
6+ β
17
2}
(4) {1
3, 6}
Exercise #3: (Revisiting the Crazy Carmel Corn Company) β Recall that the Crazy Carmel Corn company modeled
the percent of popcorn kernels that would pop, P, as a function of the oil temperature, T, in degrees Fahrenheit
using the equation
π = β1
250π2+ 2.8π β 394
The company would like to find the range of temperatures that ensures that at least 50% of the kernels will pop.
Write an inequality whose result is the temperature range the company would like to find. Solve this inequality
with the help of the quadratic formula. Round all temperatures to the nearest tenth of a degree.
-
Exercise #4: Find the intersection points of the linear-quadratic system shown below algebraically using the
quadratic formula. Then, use you calculator to help produce a sketch of the system. Label the intersection points
you found on your graph.
π¦ = 4π₯2β 6π₯ + 2 and π¦ = 6π₯ β 3
Note: The fact that the solutions to this system were rational numbers indicates that the quadratic equation in
Exercise #4 could have been solved using factoring and the Zero Product Law
5
15
y
x
-
Lesson 7: More Quadratic Formula
_____1. Which of the following represents the solutions to π₯2 β4π₯ + 12 = 6π₯ β 2?
(1) π₯ = {4Β±β7} (3) π₯ = {5Β±β22}
(2) π₯ = {5Β±β11} (4) π₯ = {4Β±β13}
_____2. The smaller root, to the nearest hundredth, of 2π₯2β 3π₯ β 1 = 0 is
(1) 0.28 (3) 1.78
(2) 0.50 (4) 3.47
_____3. The x-intercepts of π¦ = 2π₯2 +7π₯ β 30 are
(1) π₯ = {β7Β±β191
2} (3) π₯ = {β6,
5
2}
(2) π₯ = {β3,5} (4) π₯ = {β3Β±β131}
Solve the following equation for all values of x. Express your answers in simplest radical form.
4) 4π₯2 β4π₯ β 5 = 8π₯ + 6 5) 9π₯2 = 6π₯ + 4
π₯ = βπΒ±βπ2β 4ππ
2π
-
6. Algebraically solve the system of equations shown below. Note that you can use either factoring
or the quadratic formula to find the x-coordinates, but the quadratic formula is probably easier.
π¦ = 6π₯2+ 19π₯ β 15 πππ π¦ = β12π₯ + 15
7. The Celsius temperature, C, of a chemical reaction increases and then decreases over time
according to the formula πΆ(π‘) = β1
2π‘2 +8π‘ + 93, where t represents the time in minutes. Use
the Quadratic Formula to help determine the amount of time, to the nearest tenth of a minute, it
takes for the reaction to reach 110 degrees Celsius.
-
Answers to Lesson 7 Homework
1) (2)
2) (1)
3) (3)
4) π₯ = {3Β±2β5
2}
5) π₯ = {1Β±β5
3}
6) {(β6,87), (5
6, 5)}
7) π‘ = {β8Β±β30
β1}
π‘ β {2.5, 13.5}
It will take approximately 2.5 minutes for the chemical reaction to reach 110Β° Celcius.