Transcript Chapter 1.4 - HCC Learning Web

Chapter 1.4 – Quadratic Equations and
Applications
What you should learn
1.
2.
3.
Solve quadratic
equations by factoring
Solve quadratic
equations by
extracting square roots
Solve quadratic
equations by
completing the square
4. Use Quadratic Formula
to solve quadratic
equations
5. Use quadratic
equations to model and
solve real-life problems
Chapter 1.4 – Definition
A Quadratic Equation in x
 is an equation that can be written in the
general form
ax2 + bx + c = 0
where a, b, and c are real numbers with a = 0
 is also called a second-degree polynomial
equation
Chapter 1.4 – Methods of solving
Quadratic Equations
1.
2.
3.
4.
Factoring
Extracting Square roots
Completing the square
Quadratic formula
Chapter 1.4 – Factoring
Factoring is based on Zero-Factor Property
If ab = 0 , then a=0 or b = 0
To use this property, write the left side of the
general form of a quadratic equation as a
product of tow linear factors. Then find the
solutions by setting each factor equal to 0.
Factoring

All terms must be collected on one side of
the equation before factoring.
Example 1:
Solve the quadratic equation by factoring :
p. 117
13. 6x2 + 3x = 0
18. 4x2 + 12x + 9 = 0
20. 2x2 = 19x + 33
Solution :
13. 6x2 + 3x = 0
3x ( 2x + 1) = 0
3x = 0 or 2x + 1 = 0
x = 0 or 2x = -1
x=-½
Solution :
18.
4x2 + 12x + 9 = 0
(2x +3 ) (2x +3 ) = 0
2x + 3 = 0
2x = -3
x = - 3/2
20. 2x2 = 19x + 33
2x2 – 19x – 33 = 0
(2x +3 ) ( x – 11) = 0
2x + 3 = 0 x – 11 = 0
x = -3/2
x = 11
Extracting the square root
The equation
u2 = d, where d > 0,
has exactly two solutions:
u = √d
and u = - √d
These solutions can also be written as
u = ± √d
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Example
Solve the equation by
extracting the square
roots. P. 117
28. 3x2 = 81
34. (x + 9)2 = 24
37. ( x – 7)2 = ( x + 3)2
Solution
28. 3x2 = 81
divide both sides by 3
x2 = 27
take the square root of
both sides
x = ± √27
x = ± 3√3
34. (x + 9)2 = 24
take the square root of
both sides
(x + 9) = ± √24
x + 9 = ± 2√6
solve for x, subtract 9
from both sides
x = 9 ± 2√6
Solution
37. ( x – 7)2 = ( x + 3)2
take the square root of both sides
( x – 7) = ± ( x + 3)
x – 7 = x + 3 ( not possible)
x – 7 = - ( x + 3)
x – 7 = -x – 3 , then solve for x
2x = 4,
x = 2, therefore the solution is
x=2
Completing the square

If the left side of the equation is not
factorable, and then you have to rewrite the
equation by completing the square so it can
be solved by extracting the square roots.
Completing the square

To complete the square for the expression
x2 + bx, add (b/2)2 , which is the square of
half of the coefficient of x. Consequently,
x2 + bx + (b/2)2 = ( x + b/2)2
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Example
Solve the quadratic equation by completing the
square. p. 117
42. x2 + 8x + 14 = 0
43. 9x2 - 18x = - 3
46. - x2 +x – 1 = 0
Solution
42.
x2 + 8x + 14 = 0
x2 + 8x
Subtract 14 from both sides
= -14
divide (8/2), then square, add to both sides
x2 + 8x + 16 = -14 + 16
Solution
x2 + 8x + 16 = -14 + 16
then factor the left side
(x + 4)2 = 2, then solve using
extracting the square root, take the square root
of both sides x + 4 = ± √2,
then solve for x
x = - 4 ± √2
Solution
43. 9x2 - 18x = - 3 divide each side by 9
x2 - 2x
= - 1/3
divide (-2/2), then square, add to both sides
x2 - 2x +1 = - 1/3 +1 then factor the left
side, simplify the right side
(x – 1)2 = 2/3
x = 1 ± √2/3
Solution
46. - x2 +x – 1 = 0
x2 - x + 1 = 0
x2 - x
multiply by -1
then follow the steps for
completing the square
= -1
divide -1 by 2, then square, then add to both sides
x2 - x + ¼ = -1 + ¼
(x + ½)2 = - ¾ no real solution, since the
square root of - ¾ will yield an imaginary number
The Quadratic Formula
The solutions of a quadratic equation in the
general form
ax2 + bx + c = 0 , a = 0
are given by the Quadratic Formula
x = -b ± √b2 – 4ac
2a
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Solutions of a Quadratic Equation
The quantity under the square root is called the
discriminant. It can be used to determine the
nature of the solutions of a quadratic
equation.
If the discriminant b2 – 4ac is
Positive
Equation has 2
distinct real
solutions
graph has 2 xintercepts
Zero
Equation has one graph has one xrepeated real
intercepts
solutions
Negative
Equation has no
real solution
graph has no xintercepts
Example:
Use the discriminant to determine the number
of real solutions of the quadratic equation
p.117, 118
67. 2x2 – x – 1 = 0
68. x2 – 4x + 4 = 0
94. (x + 6)2 = - 2x
Application - p. 118
#118. A rectangular classroom seats 72
students. If the seats were rearranged with
three more seats in each row, the classroom
would have two fewer rows. Find the original
number of seats in each row.
Solution
Let x = number of rows
y = number of seats
xy = 72 Original
y = 72 /x
(x – 2) ( y + 3) = 72 replace y by 72/x
(x – 2 ) (72/x + 3) = 72
Find the LCD, combine like terms, solve.
x = 8 original # rows, 72/8 = 9 seats per row
Practice:

Answer the following questions on pp. 117-119
# 22,19, 32, 36, 40, 41, 65, 91, 93, 113, 120
For the Position equation mentioned in #120, use
S = -16t2 + v0t + S0, where v0 is the initial velocity,
S is the height, S0 is the initial height, and t is the
time