The document discusses various methods for solving quadratic equations, including factoring, square root method, completing the square, and the quadratic formula. It also defines terms related to quadratic equations like discriminant, complex numbers, and pure and complete quadratic equations. The objectives are for students to learn to solve different types of quadratic equations and problems using these various methods and concepts.
2. Week 3 Day 3
GENERAL OBJECTIVE
At the end of the chapter the students are expected to:
•
•
•
•
Solve quadratic equations using different methods,
Solve equations in quadratic form,
Solve equations leading to quadratic equation, and
Solve real-world problems that involve quadratic equation.
3. TODAY’S OBJECTIVE
Week 3 Day 3
At the end of the lesson the students are expected to:
• To distinguish between pure quadratic equation and complete
quadratic equation,
• To determine the number and type of solutions or roots of a
quadratic equation based on the discriminant,
• To define complex numbers, and
• To solve quadratic equations by factoring, square root
method, completing the square and quadratic formula.
4. Week 3 Day 3
DEFINITION
QUADRATIC EQUATION
A quadratic equation in x is an equation that can be written in
the standard form
2
ax bx c 0
where a, b, and c are real numbers and a 0 .
a represents the numerical coefficient of x2 ,
b represents the numerical coefficient of x, and
c represents the constant numerical term.
Example:
2x2 0
2x2 7x 0
2x2 50 0
5x 2 3x 3 0
5. Week 3 Day 3
• Pure Quadratic Equation
If b=0, then the quadratic equation is termed a "pure"
quadratic equation.
Example: 3x2 +6=0
• Complete Quadratic Equation
If the equation contains both an x and x2 term, then it
is a "complete" quadratic equation.
The numerical coefficient c may or may not be zero in
a complete quadratic equation.
Example: x2 +5x+6=0 and 2x2 - 5x = 0
6. DEFINITION
Week 3 Day 3
DISCRIMINANT OF A QUADRATIC EQUATION
The term inside the radical, b2 -4ac, is called the discriminant.
The discriminant gives important information about the
ax 2 bx c 0
corresponding solutions or roots of
where a, b, and c are real numbers and a 0 .
b2 -4ac
Positive
Zero
Negative
Solutions or Roots
Two distinct real roots
One real root (a double or repeated root)
Two complex roots(complex conjugates)
7. Week 3 Day 3
EXAMPLE
Determine the nature of roots of the following
quadratic equation.
1. x 2 4 x 5 0
2. x 2 2 x 5 0
3. 4 x 2 12 x 9 0
4. 9 x 2 12 x 4 0
5. x 2 6 x 3 0
8. Week 3 Day 3
DEFINITION
COMPLEX NUMBER
A complex number is an expression of the form a bi
where a and b are real numbers and i 1 i2 1
a is the real part and b is the imaginary part .
EXAMPLE
Real Part
Imaginary Part
3 4i
3
4
1 2
i
2 3
1
2
2
3
6i
0
6
-7
-7
0
9. SOLVING QUADRATIC EQUATIONS
Week 3 Day 3
There are four algebraic methods of solving quadratic equation
in one variable, namely:
• solution by factoring
• solution by square root method
• solution by completing the square
• solution by quadratic formula
10. Week 3 Day 3
SOLVING QUADRATIC EQUATIONS BY FACTORING
The Factoring Method applies the Zero Product Property which states
that if the product of two or more factors equals zero, then at least one
of the factors equals zero.
Thus if B·C=0, then B=0 or C=0 or both.
STEPS:
1. Write the equation in standard form ax2 + bx + c = 0.
2. Factor the left side completely.
3. Apply the Zero Product Property to find the solution set.
12. Week 3 Day 3
SOLVING QUADRATIC EQUATIONS BY SQUARE ROOT METHOD
The Square Root Property states that if an expression squared is equal
to a constant , then the expression is equal to the positive or negative
square root of the constant.
Thus, if x 2 P, then x P .
NOTE :
1. The variable squared must be isolated first ( coefficien t equal to 1)
2. If P 0 is a real number , the equation x 2 P
has real 2 distinct real solutions; x P and x P
2. If P 0, the equation x 2 P has a double root of zero
3. If P 0, the equation x 2 P has exactly two imaginary solutions
13. Week 3 Day 3
EXAMPLE
Solve the following equations.
1.
Classroom ex.1.3.4
pp.116
2x 32 0
2
Classroom ex. 1.3.5
2.
5a2 10 0
pp.117
4.
#30
pp.124
(4x - 1)2 16
#16
5.
16v 2 25 0
page 124
Classroom ex. 1.3.6
3.
(x - 3)2 25
pp.117
6. 2 y 2 4 y 72
2
2
7. 2m 5 m 3
from College Algebra by Exconde, Marquez and Sabino
exercise 5.4 numbers 9 and 10 page 112
14. Week 3 Day 3
SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE
STEPS:
1. Express the quadratic equation in the following form
x 2 bx c
2. Divide b by 2 and square the result, then add the
square to both sides.
2
b
b
x bx c
2
2
2
2
15. Week 3 Day 3
SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE
3. Write the left side of the equation as a perfect square
2
b
b
x c
2
2
2
4. Solve using the square root method.
16. Week 3 Day 3
EXAMPLE
Solve the following equations.
1.
Example #7 2
x 8x 3 0
pp.118
Example #2
2.
3x2 12x 13 0
pp.119
#55 x 2
1
3.
2x
pp.124 2
4
#56
4.
pp.124
t2 2t 5
0
3 3 6
What number should be added to complete the square of each expression .
#41
page 124
2
x2 x
5
#41
6.
page 124
x 2 12x
5.
17. Week 3 Day 3
SOLVING QUADRATIC EQUATIONS BY QUADRATIC FORMULA
THE QUADRATIC FORMULA
The roots of the quadratic equation ax2 + bx + c = 0, where a, b, and c
are constants and a 0 are given by:
b b 2 4ac
x
2a
Note : The quadratic equation must be in standard form
ax 2 bx c 0 in order to identify the parameters a, b, c.
18. DERIVATION OF QUADRATIC FORMULA BY COMPLETING THE
SQUARE
Consider the most general quadratic equation: ax2 bx c 0
Solve by completing the square:
WORDS
MATH
1. Divide the equation by
the leading coefficient a.
2. Subtract c from both sides.
a
b c
x2 x 0
a a
b
c
x x
a
a
2
3. Subtract half of b and add
the result b
2a
sides.
4.. Write the
2
a
to both
left side of the
equation as a perfect square
and the right side as a single
fraction.
2
2
b b b c
x2 x
a 2a 2a a
2
b b2 4ac
x
4a2
2a
19. DERIVATION OF QUADRATIC FORMULA BY COMPLETING THE
SQUARE
WORDS
5. Solve using the square root
method.
b
6. Subtract
from both sides
2a
and simplify the radical.
7. Write as a single fraction.
8. We have derived the quadratic
formula.
MATH
b
b2 4ac
x
2a
4a2
b
b2 4ac
x
2a
2a
b b2 4ac
x
2a
20. Week 3 Day 3
EXAMPLE
Solve the following equations using the quadratic formula.
1.
Your Turn 2
x 2 2x
pp.121
2.
Example #11
4x 2 4 x 1 0
pp.121
21. SUMMARY
Week 3 Day 3
The four methods for solving quadratic equations are:
1. factoring
3. completing the square
2. square root method
4. quadratic formula
Factoring and the square root method are the quickest
and easiest but cannot always be used.
Quadratic formula and completing the square work for all
quadratic equations and can yield three types of solutions:
1. two distinct real roots
2.one real root (repeated)
3.or two complex roots (conjugates of each other)
22. Week 4 Day 1
EQUATIONS IN QUADRATIC FORM
(OTHER TYPES)
23. Week 4 Day 1
CLASSWORK
Solve each quadratic equation using any method.
#87
2 2 4
1
1.
t t
page 125 3
3
5
#89
12
2.
x 7
page 125
x
#91
4x - 2 3
-3
3.
page 125
x3 x
x x - 3
#93
4.
x2 0.1x 0.12
page 125
24. TODAY’S OBJECTIVE
Week 4 Day 1
At the end of the lesson the students are expected to:
• To find the sum and product of roots of a quadratic equation.
• To find the quadratic equation given the roots.
• To transform a difficult equation into a simpler linear or quadratic
equation,
• To recognize the need to check solutions when the transformation
process may produce extraneous solutions,
• To solve radical equations.
25. Week 4 Day 1
RECALL
The four methods for solving quadratic equations are:
1. factoring
3. completing the square
2. square root method
4. quadratic formula
Factoring and the square root method are the quickest
and easiest but cannot always be used.
Quadratic formula and completing the square work for all
quadratic equations and can yield three types of solutions:
1. two distinct real roots
2.one real root (repeated)
3.or two complex roots (conjugates of each other)
26. SUM AND PRODUCT OF ROOTS
Recall from the quadratic formula that when
ax 2 bx c 0
b b2 4ac
x
2a
b
r
Let the roots be r and s
b
s
b 2 4ac
2a
b 2 4ac
2a
Week 4 Day 1
27. SUM OF ROOTS
Sum of roots = r + s
b b 2 4ac b b 2 4ac
rs
2a
2a
b b 2 4ac b b 2 4ac
2a
2b
rs
2a
b
rs
a
Week 4 Day 1
28. Week 4 Day 1
PRODUCT OF ROOTS
Product of roots = (r) (s)
b b 2 4ac b b 2 4ac
(r)( s )
*
2a
2a
( b)
2
b
2
4ac
4a 2
b 2 b 2 4ac
(r )(s)
4a 2
c
(r)(s)
a
2
29. Week 4 Day 1
EXAMPLE
Determine the value of k that satisfies the given condition
1. kx 2 5x 4 0;
sum of roots is 20.
2. (3k 2)x 2 2x k - 1 0; product of roots is - 1.
3. 2k 1x 2 - 10x k 2 5k 6 0; one of the roots is 0.
4. 2k - 1x 2 k - 5 x - 6 0;
the roots numerically equal but with
opposite signs
30. Week 4 Day 1
FINDING THE QUADRATIC EQUATION GIVEN THE ROOTS
Let the roots be r and s, the quadratic equation is
x - r x s 0
Example: Find the quadratic equations with the given roots.
1.
3
1
and
4
2
2. 5 2 and 7 2
3. 1 2i and 1 2i
31. RADICAL EQUATIONS
Week 4 Day 1
Radical Equations are equations in which the variable is
inside a radical (that is square root, cube root, or higher
root).
x 3 2,
2x 3 x,
x 2 7x 2 6
32. RADICAL EQUATIONS
Week 4 Day 1
Steps in solving radical equations:
1. Isolate the term with a radical on one side.
2. Raise both (entire)sides of the equation to the power that will
eliminate this radical and simplify the equation.
3. If a radical remains, repeat steps 1 and 2.
4. Solve the resulting linear or quadratic equation.
5. Check the solutions and eliminate any extraneous solutions.
Note: When both sides of the equations are squared extraneous
solutions can arise , thus checking is part of the solution.
33. EXAMPLE
Week 4 Day 1
Solve the following equations.
1.
Example 1
pp.128
x -3 2
Classroom ex. 1.4.1 a.
2.
pp.128
b.
Your Turn
3.
pp.128
#29
6.
page 133
2x 6 x 3
2 x x
Example 3
pp.129
x 2 7x 2 6
#28
5.
page 133
4.
x 5 1 x 2
#24
page 133
2x2 8x 1 x 3
2-x 3
2 - x 3
7.
34. Week 4 Day 1
CATCH THE MISTAKE
#83
8.
Explain the mistake that is made.
page 134
Solve the equation
3t 1 4
Solution :
3t 1 16
3t 15
t 5
This is incorrect.What mistake was made?
35. Week 4 Day 1
SUMMARY
.
b
r
Let the roots be r and s
b
s
Sum of roots :
rs
b
a
b 2 4ac
2a
b 2 4ac
2a
Productof roots :
c
(r)(s)
a
Steps in solving radical equations:
1. Isolate the term with a radical on one side.
2. Raise both (entire)sides of the equation to the power that
will eliminate this radical and simplify the equation.
3. If a radical remains, repeat steps 1 and 2.
4. Solve the resulting linear or quadratic equation.
5. Check the solutions and eliminate any extraneous solutions.
36. TODAY’S OBJECTIVE
Week 4 Day 2
At the end of the lesson the students are expected to:
• To solve equations that are quadratic in form,
• To realize that not all polynomial equations are factorable.
• To solve equations that are factorable.
37. Week 4 Day 2
EQUATIONS QUADRATIC IN FORM: u-SUBSTITUTION
Equations that are higher order or that have fractional powers often
can be transformed into quadratic equation by introducing a usubstitution, thus the equation is in quadratic form.
Example:
Original Equation
Substitution
New Equation
x 3x 4 0
ux
2
u 3u 4 0
1
t3
u2 2u 1 0
4
2
2
t3
1
2t 3
1 0
u
2
38. Week 4 Day 2
EQUATIONS QUADRATIC IN FORM: u-SUBSTITUTION
Steps in solving equations quadratic in form:
1. Identify the substitution.
2. Transform the equation into a quadratic equation.
3. Apply the substitution to rewrite the solution in terms the original
variable.
4. Solve the resulting equation.
5. Check the solution in the original equation.
39. EXAMPLE
Week 4 Day 2
Solve the following equations.
1.
Example 4 -2 1
x x 12 0
pp.131
Classroom ex. 1.4.4
2x 12 102x 11 9 0
2.
pp.131
3.
Classroom ex. 1.4.5
pp.131
2
z5
1
8z 5
8
5
16 0
2
Classroom ex. 1.4.5 3
4.
2x x 3 x 3 0
pp.132
41. SUMMARY
Week 4 Day 2
Radical equations, equations quadratic in form, and factorable
equations can often be solved by transforming them into simpler
linear or quadratic equations.
Radical Equations: Isolate the term containing a radical and raise
it to the appropriate power that will eliminate the radical. If there
is more than one radical, it does not matter which radical is
isolated first. Raising radical equations to powers may cause
extraneous solutions, so check each solutions.
Equations quadratic in form: Identify the u-substitution that
transforms the equation into a quadratic equation. Solve the
quadratic equation and then remember to transform back to the
original equation.
Factorable equations: Look for a factor common to all terms or
factor by grouping.
43. Start
Week 4 Day 3
RECALL
A
Read and analyze
the problem
Make a diagram or
sketch if possible
Solve the equation
Determine the
unknown quantity.
Check the solution
Set up an equation,
assign variables to
represent what you
are asked to find.
no
Did you set up
the equation?
no
yes
A
Is the unknown
solved?
yes
End
44. APPLICATION PROBLEMS
Week 4 Day 3
1. If a person drops a water balloon off the rooftop of an 81 foot
building, the height of the water balloon is given by the
equation h 16t2 81 where t is in seconds. When will the
water balloon hit the ground?
(Classroom example 1.3.12 page 122)
2. You have a rectangular box in which you can place a 7 foot
long fishing rod perfectly on the diagonal. If the length of the
box is 6 feet, how wide is that box?
(Classroom example 1.3.13 page 123)
3. A base ball diamond is a square. The distance from base to
base is 90 feet. What is the distance from the home plate to
the second base?
(#108 page 125)
45. Week 4 Day 3
4. Lindsay and Kimmie, working together, can balance the financials for
the Kappa Kappa Gama sorority in 6days. Lindsay by herself can
complete the job in 5days less than Kimmie. How long will it take
Lindsay to complete the job by herself? (# 113 page 125)
5.A rectangular piece of cardboard whose length is twice its width is
used to construct an open box. Cutting a I foot by 1 foot square off of
each corner and folding up the edges will yield an open box. If the
desired volume is 12 cubic feet, what are the dimensions of the original
piece of cardboard? (# 110 page 125)
6.Aspeed boat takes 1 hour longer to go 24 miles up a river than to
return. If the boat cruises at 10mph in still water, what is the rate of
the current? (#140 page 126)
46. Week 4 Day 3
7. Cost for health insurance with a private policy is given by C 10 a
where C is the cost per day and a is the insured’s age in years. Health
insurance for a six year old, a=6, is $4 a day (or $1,460 per year). At
what age would someone be paying $9 a day (or $3,285 per year).
(#73 page 134)
8. The period (T) of a pendulum is related to the length (L) of the
pendulum and acceleration due to gravity (g) by the formula
L
T 2
. If the gravity is 32ft / s2 and the period is 1 second find the
g
approximate length of the pendulum. Round to the nearest inch.
(#80 page 134)