Math in Our World
Section 6.6
Solving Quadratic Equations
Learning Objectives
 Identify the standard form of a quadratic
equation.
 Multiply binomials using FOIL.
 Factor trinomials.
 Solve quadratic equations using factoring.
 Solve quadratic equations using the quadratic
formula.
 Solve real-world problems using quadratic
equations.
Quadratic Equations
A quadratic equation is any equation that can be
written in the form ax2 + bx + c = 0,
where a, b, and c are real numbers, and a is not
zero. When written this way, we say that a
quadratic equation is in standard form.
To write a quadratic equation in standard form, we arrange
it so that zero is on the right side of the equation, the term
with exponent 2 comes first on the left side, followed by
the term with exponent 1 (if there is one) and then the
constant (numeric) term.
EXAMPLE 1
Writing a Quadratic Equation
in Standard Form
Write each equation in standard form and identify
a, b, and c.
(a) 7 + 9x2 = 3x
(b) 4x – 15 = 3x2
(c) 5x2 = 25
EXAMPLE 1
Writing a Quadratic Equation
in Standard Form
SOLUTION
(a)
Now we use the definition of standard form: a is the
coefficient of x2, b is the coefficient of x, and c is the constant
term, so we get a = 9, b = – 3, and c = 7.
(b)
Notice that if we wanted to, we could multiply both sides of
the equation in standard form by – 1, giving us the
equivalent equation 3x2 – 4x + 15 = 0. In this case, a = 3,
b = – 4, and c = 15.
EXAMPLE 1
SOLUTION
(c)
Writing a Quadratic Equation
in Standard Form
Binomials
A binomial is an algebraic expression with two
terms in which any variable has a whole number
exponent. Some examples of binomials are
Multiplying Binomials
FOIL Method:
F represents the product of the first terms of the binomial.
O represents the product of the outer terms.
I represents the product of the inner terms.
L represents the product of the last terms of the binomials.
EXAMPLE 2
Multiplying Binomials using
FOIL
Multiply (x – 8)(x + 3).
EXAMPLE 2
Multiplying Binomials using
FOIL
SOLUTION
Multiply the first terms: x · x = x2
Multiply the outer terms: x · 3 = 3x
Multiply the inner terms: – 8 · x = – 8x
Multiply the last terms: (– 8) · (3) = – 24
Combining like terms, the product is x2 – 5x – 24.
EXAMPLE 3
Multiplying Binomials using
FOIL
Multiply (2x – 5)(3x – 8).
SOLUTION
Factoring Trinomials
When a quadratic equation is written in standard
form, there are three terms on the left side if none
of the coefficients are zero. Three term expressions
like this are called trinomials. It’s important that
you’re really good at using FOIL, because one of
our methods for solving quadratic equations will
involve performing FOIL in reverse, a process we
call factoring.
Factoring Trinomials
Take another look at the result of Example 2:
If we read this equation from right to left, it’s an
example of factoring, because we start with the
trinomial x2 – 5x – 24 and write it as a product of
two factors: x – 8 and x + 3. The x2 comes from
multiplying the first terms of the two factors, x and
x. The constant term, 24, comes from multiplying
the last two numbers of the two factors, 8 and 3.
And the x term comes from combining the products
of the outsides and insides.
EXAMPLE 4
Factoring a Trinomial with
a=1
Factor x2 + 10x + 16.
SOLUTION
Step 1 The factored form will be the product of two
binomials; the first term of each is x (since the product of the
firsts must be x2). So x2 + 10x + 16 = (x
)(x
)
Step 2 Write all pairs of factors of the constant term. One of
these pairs must be the product of the last terms inside the
parentheses. In this case, they are 1 • 16, 2 • 8, and 4 • 4.
EXAMPLE 4
Factoring a Trinomial with
a=1
SOLUTION
Step 3 Set up the factorization with each possible pair of last
terms, found in Step 2. Then find the product of outside and
inside terms for each possibility.
1 • 16: (x 1)(x 16) Outsides: 16x Insides: x
2 • 8: (x 2)(x 8) Outsides: 8x Insides: 2x
4 • 4: (x 4)(x 4) Outsides: 4x Insides: 4x
Step 4 Find a sum or difference of the outsides and insides
that yields the x term in the trinomial. In this case, we need
10x, and 8x + 2x = 10x. That tells us that the two last terms
we need are +8 and +2, and the trinomial factors as
x2 + 10x + 16 = (x + 8)(x + 2)
EXAMPLE 5
Factoring a Trinomial with
a=1
Factor x2 – 7x + 12.
SOLUTION
Step 1 Write as a product of two binomials with first term x.
x2 – 7x + 12 = (x
)(x
)
Step 2 The pairs of factors of the constant term are 1 • 12,
2 • 6, and 3 • 4.
EXAMPLE 5
Factoring a Trinomial with
a=1
SOLUTION
Step 3 Possible factorizations.
(x 1)(x 12) Outsides: 12x
(x 2)(x 6) Outsides: 6x
(x 3)(x 4) Outsides: 3x
Insides: x
Insides: 2x
Insides: 4x
Step 4 The pair of outsides and insides that yields middle
term – 7x is 4x and 3x, if both are negative: – 4x – 3x = – 7x.
Also, (– 4)(– 3) = 12, which is the correct constant term. So
the factors are
x2 – 7x + 12 = (x – 3)(x – 4)
EXAMPLE 6
Factoring a Trinomial with
a=1
Factor x2 – 2x – 8.
SOLUTION
Step 1 Write as a product of two binomials with first term x.
x2 – 2x – 8 = (x
)(x
)
Step 2 The pairs of factors of the constant term are 1 • 8 and
2 • 4.
EXAMPLE 6
Factoring a Trinomial with
a=1
SOLUTION
Step 3 Possible factorizations.
(x 1)(x 8) Outsides: 8x Insides: x
(x 2)(x 4) Outsides: 4x Insides: 2x
Step 4 The pair of outsides and insides that yields middle
term – 2x is – 4x and 2x; – 4x + 2x = – 2x.
Also, (– 4)(2) = – 8, which is the correct constant term. So
the factors are
x2 – 2x – 8 = (x + 2)(x – 4)
EXAMPLE 7
Factoring a Trinomial with
a≠1
Factor 3x2 – 4x – 15.
SOLUTION
Step 1 Write as a product of two binomials; the product of
the first two terms has to be 3x2, so we will use 3x and x.
3x2 – 4x – 15 = (3x
)(x
)
Step 2 The pairs of factors of the constant term are 1 • 15
and 3 • 5.
EXAMPLE 7
Factoring a Trinomial with
a≠1
SOLUTION
Step 3 Possible factorizations (important note: this time the
order matters, so we have to write the possible combinations
in both orders):
(3x 1)(x 15) Outsides: 45x
Insides: x
(3x 15)(x 1) Outsides: 3x
Insides: 15x
(3x 3)(x 5) Outsides: 15x
Insides: 3x
(3x 5)(x 3) Outsides: 9x
Insides: 5x
Step 4 The pair of outsides and insides that yields middle
term – 4x is – 9x and 5x; – 9x + 5x = – 4x. This comes from
the factor 5 and –3. So the factors are
3x2 – 4x – 15 = (3x + 5)(x – 3)
EXAMPLE 8
Factoring a Trinomial with
a≠1
Factor 14x2 – 33x + 10.
SOLUTION
Step 1 Write as a product of two binomials; the product of
the first two terms has to be 14x2, so we can use 14x and x
or 7x and 2x.
14x2 – 33x + 10 = (14x
)(x
) or (7x
)(2x
)
Step 2 The pairs of factors of the constant term are 1 • 10
and 2 • 5.
EXAMPLE 8
Factoring a Trinomial with
a≠1
SOLUTION
Step 3 We’ll start trying the possible combinations starting
with 14x and x.
(14x 10)(x 1) Outsides: 14x
Insides: 10x
(14x 1)(x 10) Outsides: 140x Insides: x
(14x 2)(x 5) Outsides: 28x
Insides: 5x
Step 4 We haven’t written all the combinations, but at this
point we can stop. With factors (14x – 5) and (x – 2), we’ll get
outsides – 28x and insides – 5x, which gives us the – 33x we
need.
14x2 – 33x + 10 = (14x – 5)(x – 2)
Solving Quadratic Equations
Solving Quadratic Equations by Factoring
Step 1 Write the quadratic equation in standard
form.
Step 2 Factor the left side.
Step 3 Set both factors equal to zero.
Step 4 Solve each equation for x.
This works since the only way a product can equal zero is
if one or both of the factors are equal to zero.
EXAMPLE 9
Solving a Quadratic Equation
Using Factoring
Solve x2 – 13x = – 36.
SOLUTION
The solution set is {9, 4}. They can be checked by
substituting the values into the original equation.
EXAMPLE 10
Solving a Quadratic Equation
Using Factoring
Solve 6x2 – 6 = – 5x.
SOLUTION
Solving Quadratic Equations
Solving Using the Quadratic Formula
Step 1 Write the equation in standard form.
Step 2 Identify a, b, and c.
Step 3 Substitute the values into the formula.
If ax2 + bx + c = 0, then
b  b  4ac
x
2a
2
EXAMPLE 11
Using the Quadratic Formula
Solve 2x2 – x – 8 = 0 using the quadratic formula.
SOLUTION
The equation is in standard form, so a = 2, b = –1 and c = –8.
EXAMPLE 11
Using the Quadratic Formula
SOLUTION
Substitute a = 2, b = –1 and c = –8 into the formula.
EXAMPLE 12
Using the Quadratic Formula
Solve 5 = 5y2 + 8y using the quadratic formula.
SOLUTION
Before identifying a, b, and c, we need to rewrite the
equation in standard form by subtracting 5y2 and 8y from
both sides.
– 5y2 – 8y + 5 = 0
So a = – 5, b = –8 and c = 5.
EXAMPLE 12
Using the Quadratic Formula
SOLUTION
Substitute a = – 5, b = –8 and c = 5 into the formula.
EXAMPLE 13
Applying Quadratic Equations
to Woodworking
The plans for a make-it-yourself picnic table call for
the length to be 2 feet more than the width. If you
want a table with 15 square feet of area, what
dimensions should you choose?
SOLUTION
Let x = the width of the table. The length is 2 feet more, so x
2 is the length. The area of a rectangle is length times width,
so we can set up an equation:
Length times Width is Area
(x + 2) •
x = 15
EXAMPLE 13
Applying Quadratic Equations
to Woodworking
SOLUTION
Now we solve.
Even though we got two solutions, only one makes sense in
the problem – you can’t build a table that is – 5 feet wide. So
the width is 3 feet and the length is 5 feet.