Transcript 1.4: Quadratic Equations and Applications

1.4
QUADRATIC EQUATIONS AND APPLICATIONS
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What You Should Learn
• Solve quadratic equations by factoring.
• Solve quadratic equations by extracting square
roots.
• Solve quadratic equations by completing the
square.
• Use the Quadratic Formula to solve quadratic
equations.
• Use quadratic equations to model and solve
real-life problems.
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Factoring
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Example 1(a) – Solving a Quadratic Equation by Factoring
2x2 + 9x + 7 = 3
Original equation
2x2 + 9x + 4 = 0
Write in general form.
(2x + 1)(x + 4) = 0
2x + 1 = 0
x+4=0
The solutions are x =
original equation.
Factor.
x=
Set 1st factor equal to 0.
x = –4
Set 2nd factor equal to 0.
and x = –4. Check these in the
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Example 1(b) – Solving a Quadratic Equation by Factoring
6x2 – 3x = 0
Original equation
3x(2x – 1) = 0
3x = 0
2x – 1 = 0
cont’d
Factor.
x=0
Set 1st factor equal to 0.
x=
Set 2nd factor equal to 0.
The solutions are x = 0 and x = . Check these in the
original equation.
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Extracting Square Roots
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Example 2 – Extracting Square Roots
Solve each equation by extracting square roots.
a. 4x2 = 12
b. (x – 3)2 = 7
Solution:
a. 4x2 = 12
x2 = 3
x=
Write original equation.
Divide each side by 4.
Extract square roots.
When you take the square root of a variable expression,
you must account for both positive and negative
solutions.
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Example 2 – Solution
So, the solutions are x =
in the original equation.
b. (x – 3)2 = 7
x–3=
x=3
The solutions are x = 3 
equation.
cont’d
and x = –
. Check these
Write original equation.
Extract square roots.
Add 3 to each side.
. Check these in the original
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Completing the Square
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Example 3 – Completing the Square: Leading Coefficient Is 1
Solve x2 + 2x – 6 = 0 by completing the square.
Solution:
x2 + 2x – 6 = 0
Write original equation.
x2 + 2x = 6
Add 6 to each side.
x2 + 2x + 12 = 6 + 12
Add 12 to each side.
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Example 3 – Solution
(x + 1)2 = 7
x+1=
x = –1 
The solutions are x = –1 
equation as follows.
cont’d
Simplify.
Take square root of each side.
Subtract 1 from each side.
. Check these in the original
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Example 3 – Solution
cont’d
Check:
x2 + 2x – 6 = 0
(–1 +
)2 + 2 (–1 +
8–2
–2+2
)–6≟0
–6≟0
8–2–6=0
Write original equation.
Substitute –1 +
for x.
Multiply.
Solution checks.
Check the second solution in the original equation.
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The Quadratic Formula
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The Quadratic Formula
The Quadratic Formula is one of the most important
formulas in algebra. You should learn the verbal statement
of the Quadratic Formula:
“Negative b, plus or minus the square root of b
squared minus 4ac, all divided by 2a.”
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The Quadratic Formula
In the Quadratic Formula, the quantity under the radical
sign, b2 – 4ac, is called the discriminant of the quadratic
expression ax2 + bx + c. It can be used to determine the
nature of the solutions of a quadratic equation.
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Example 6 – The Quadratic Formula: Two Distinct Solutions
Use the Quadratic Formula to solve x2 + 3x = 9.
Solution:
The general form of the equation is x2 + 3x – 9 = 0. The
discriminant is b2 – 4ac = 9 + 36 = 45, which is positive. So,
the equation has two real solutions.
You can solve the equation as follows.
x2 + 3x – 9 = 0
Write in general form.
Quadratic Formula
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Example 6 – Solution
cont’d
Substitute a = 1, b = 3,
and c = –9.
Simplify.
Simplify.
The two solutions are:
Check these in the original equation.
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Applications
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Example 7 – Finding the Dimensions of a Room
A bedroom is 3 feet longer than it is wide (see Figure 1.20)
and has an area of 154 square feet. Find the dimensions of
the room.
Figure 1.20
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Example 7 – Solution
Verbal
Model:
Labels:
Width of room = w
Length of room = w + 3
Area of room = 154
Equation:
(feet)
(feet)
(square feet)
w(w + 3) = 154
w2 + 3w – 154 = 0
(w – 11)(w + 14) = 0
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Example 7 – Solution
w – 11 = 0
w = 11
w + 14 = 0
w = –14
cont’d
Choosing the positive value, you find that the width is
11 feet and the length is w + 3, or 14 feet.
You can check this solution by observing that the length is
3 feet longer than the width and that the product of the
length and width is 154 square feet.
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Applications
Another common application of quadratic equations
involves an object that is falling (or projected into the air).
The general equation that gives the height of such an
object is called a position equation, and on Earth’s
surface it has the form
s = –16t2 + v0t + s0.
In this equation, s represents the height of the object
(in feet), v0 represents the initial velocity of the object
(in feet per second), s0 represents the initial height of the
object (in feet), and t represents the time (in seconds).
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Applications
A third type of application of a quadratic equation is one in
which a quantity is changing over time t according to a
quadratic model.
A fourth type of application that often involves a quadratic
equation is one dealing with the hypotenuse of a right
triangle.
In these types of applications, the Pythagorean Theorem
is often used.
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Applications
The Pythagorean Theorem states that
a2 + b2 = c2
Pythagorean Theorem
where a and b are the legs of a right triangle and c is the
hypotenuse.
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