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MAT 150
Unit 2-2: Solving Quadratic
Equations
Objectives
 Solve
 Solve
quadratic equations using factoring
quadratic equations graphically using the
x-intercept method and the intersection method
 Solve quadratic equations by combining graphical and
factoring methods
 Solve quadratic equations using the square root
method
 Solve quadratic equations by completing the square
 Solve quadratic equations using the quadratic formula
 Solve quadratic equations having complex solutions
Factoring Methods
An equation that can be written in the form
ax2 + bx + c = 0, with a ≠ 0, is called a quadratic equation.
Zero Product Property
If the product of two real numbers is 0, then at least one
of them must be 0. That is, for real numbers a and b, if
the product ab = 0, then either a = 0 or b = 0 or
both a and b are equal to 0.
Solve with Factoring
A. 𝑥 2 +4𝑥 − 5 = 0
B.
3𝑥 2 + 7𝑥 = 6
C.
3𝑥 2 − 9𝑥 = 0
Example
The height above ground of a ball thrown upward at 64
feet per second from the top of an 80-foot-high building
is modeled by S(t) = 80 + 64t – 16t2 feet, where t is the
number of seconds after the ball is thrown. How long will
the ball be in the air?
Solution
Example
Consider the daily profit from the production and sale of x units of
a product, given by P(x) = –0.01x2 + 20x – 500 dollars.
a.
Use a graph to find the levels of production and sales that give
a daily profit of $1400.
b.
Is it possible for the profit to be greater than $1400?
Example
Consider the daily profit from the production and sale of x units of
a product, given by P(x) = –0.01x2 + 20x – 500 dollars.
Use a graph to find the levels of production and sales that give
a daily profit of $1400.
Solution
a.
Example (cont)
Consider the daily profit from the production and sale of x units of
a product, given by P(x) = –0.01x2 + 20x – 500 dollars.
b. Is it possible for the profit to be greater than $1400?
Solution
Combining Graphs and Factoring
Factor Theorem
The polynomial function f has a factor (x – a) if and only
if f(a) = 0. Thus, (x – a) is a factor of f(x) if and only if
x = a is a solution to f (x) = 0.
The Square Root Method
Square Root Method
The solutions of the quadratic equation x2 = C are
x =  C . Note that, when we take the square root of
both sides, we use a ± symbol because there are both a
positive and a negative value that, when squared, give
C.
Example
Solve the following equations using the square root
method.
a. 3x2 – 6 = 0
b. (x – 2)2 = 7
Solution
Quadratic Formula
The solutions of the quadratic equation ax2 + bx + c = 0
are given by the formula
2
b  b  4ac
x
2a
Note that a is the coefficient of x2, b is the coefficient of
x, and c is the constant term.
Example
Solve 5x2 – 8x = 3 using the quadratic formula.
Solution
The Discriminant
We can also determine the type of solutions a quadratic
equation has by looking at the expression b2  4ac,
which is called the discriminant of the quadratic
equation ax2 + bx + c = 0. The discriminant is the
expression inside the radical in the quadratic formula, so
it determines if the quantity inside the radical is positive,
zero, or negative.
• If b2  4ac > 0, there are two different real solutions.
• If b2  4ac = 0, there is one real solution.
• If b2  4ac < 0, there is no real solution.
Aids for Solving Quadratic Equations
Example
Solve the equations.
a. x2 = – 36
Solution
b. 3x2 + 36 = 0
Example
Solve the equations.
a. x2 – 3x + 5 = 0
Solution
a.
b. 3x2 + 4x = –3
Example (cont)
Solve the equations.
a. x2 – 3x + 5 = 0
Solution
b.
b. 3x2 + 4x = –3
Market Equilibrium
Suppose that the demand for artificial Christmas trees is given by the
function 𝑝 = 109.70 − 0.10𝑞 and that the supply of these trees is given by
𝑝 = 0.01 𝑞 2 + 5.91 where p is the price of a tree in dollars and q is the
quantity of trees that are demanded/supplied in hundreds. Find the price
that gives the market equilibrium price and the number of trees that will
be sold/bought at this price.