CHAPTER 7
QUADRATIC
EQUATIONS AND
FUNCTIONS
7-1
Completing the
Square
Completing the Square
 Transform the equation so that the
constant term c is alone on the right
side.
 If a, the coefficient of the seconddegree term, is not equal to 1, then
divide both sides by a.
 Add the square of half the coefficient
of the first-degree term, (b/2a)2, to
both sides (Completing the square)
Completing the Square
 Factor the left side as the
square of a binomial.
 Complete the solution
using the fact that (x +
q)2 = r is equivalent to
x + q = ±r
Solve:
2
a
– 5a + 3= 0
1. Move the 3 to the other side
2. a = 1
3. Complete the square, add
(5/2)2
4. Factor
5. Solve
Solve:
2
x
– 6x - 3= 0
1. Move the 3 to the other side
2. a = 1
3. Complete the square, add
(3)2
4. Factor
5. Solve
Solve:
2
2y
+ 2y + 5 = 0
1. move the 5 to the other side
2. divide both sides by 2 (a 
1)
3. Add (1/2)2 to both sides
4. Factor
5. Solve
Solve:
2
7x
– 8x + 3 = 0
1. move the 3 to the other side
2. divide both sides by 7 (a 
1)
3. Add (4/7)2 to both sides
4. Factor
5. Solve
7-2
Quadratic Formula
The Quadratic Formula
The solutions of the quadratic
equation
2
ax + bx + c = 0 (a  0)
are given by the formula:
Solve
2
• 3x + x – 1 = 0
• 5y2 = 6y – 3
2
• 2x
– 3x + 7 = 0
7-3
The Discriminant
Discriminant
The discriminant is used to
determine the nature of
the roots of a quadratic
equation and is equal to:
D=
2
b
– 4ac
Discriminant Cases
If D is positive, then the
roots are real and unequal.
If D is zero, then the roots
are real and equal (double
root) .
 negative, the roots are
imaginary.
Find the Discriminant
2
• x + 6x – 2 = 0
• 3x2 – 4x√3 + 4 = 0
2
•x
– 6x + 10 = 0
Discriminant
The discriminant also show
you whether a quadratic
equation with integral
coefficients has rational
roots.
2
D = b – 4ac
Test for Rational Roots
If a quadratic equation has
integral coefficients and its
discriminant is a perfect
square, then the equation
has rational roots.
Test for Rational Roots
 If the quadratic equation
can be transformed into an
equivalent equation that
meets this test, then it has
rational roots.
Find the Determinant and
Identify the Nature of the Roots
• 3x2 - 7x + 5 = 0
2
• 2x
- 13x + 15 = 0
• x2 + 6x + 10 = 0
7-4
Equations in
Quadratic Form
Quadratic Form
An equation in quadratic form
can be written as:
a[f(x)]2 + b[(f(x)] + c = 0
where a  0 and f(x) is some
function of x. It is helpful to
replace f(x) with a single
variable.
Example
(3x – 2)2 – 5(3x -2) – 6 = 0
Let z = 3x – 2, then
2
z
– 5z – 6 = 0
Solve for z and then solve for x
Solve Using Quadratic Form
 (x + 2)2 – 5(x + 2) – 14 = 0
 (3x +
4
x
+
2
4)
2
7x
+ 6(3x + 4) – 16 = 0
– 18 = 0
7-5
Graphing y – k = a(x-
2
h)
Parabola
Parabola is the set of all points
in the plane equidistant from a
given line and a given point not
on the line. Parabolas have an
axis of symmetry (mirror
image) either the x-axis or the
y-axis.
and
Parabola
The point where the parabola
crosses it axis is the vertex.
The graph is a smooth
curve.
Parabola
The graph of an equation
having the form
2
y – k = a(x - h)
has a vertex at (h, k) and its
axis is the line x = h.
Graph
y =
2
x
Use the form y – k = a(x –
2
h)
k= 0, h = 0, so the vertex is
(0,0) and x = 0
Table of Values
0
0
1
1
-1
1
2
4
-2
4
qx = x2
4
2
-5
5
-2
-4
Graph
y =
2
½x
Use the form y – k = a(x –
2
h)
k= 0, h = 0, so the vertex is
(0,0) and x = 0
fx =

1
2
x2
gx = x2
4
2
-5
5
-2
-4
Graph
y =
2
-½x
Use the form y – k = a(x –
2
h)
k= 0, h = 0, so the vertex is
(0,0) and x = 0
qx = x2
rx =
 
-1
2
x2
4
2
-5
5
-2
-4
Graph
2
ax
The graph of y =
opens
upward if a> 0 and
downward if a< 0. The
larger the absolute value of
a is, the “narrower” the
graph.
qx = x2
s x = 5x2
4
2
-5
5
-2
-4
Graph
y =
2
½(x-3)
Use the form y – k = a(x –
2
h)
k= 0, h = 3, so the vertex is
(3,0) and x = 3
fx =

1
2
x-32
4
2
-5
5
-2
-4
fx =


gx =
1
2
1
2
x-32
x+32
4
2
-5
5
-2
-4
Graph
To graph y = a(x – h)2, slide
2
the graph of y = ax
horizontally h units. If h > 0,
slide it to the right; if h < 0,
slide it to the left. The graph
has vertex (h, 0) and its axis
is the line x = h.
Graph
y – 3 =
2
½x
Use the form y – k = a(x –
2
h)
k= 3, h = 0, so the vertex is
(0,3) and x = 0
12
10
8
6
fx =

1
2
x2+3
4
2
Graph
y + 3 =
2
½x
Use the form y – k = a(x –
2
h)
k= -3, h = 0, so the vertex is
(0,-3) and x = 0
8
6
fx =


gx =
1
2
1
2
x2+3
x2-3
4
2
-5
5
-2
Graph
To graph y – k = ax2, slide
2
the graph of y = ax
vertically k units. If k > 0,
slide it upward; if k < 0, slide
it downward. The graph has
vertex (0, k) and its axis is
the line x = 0.
7-6
Quadratic Functions
Quadratic Functions
A function that can be written
in either of two forms.
General form:
2
f(x) = ax + bx + c
Completed square form:
a(x-h)2 + k
Graph
f(x) = 2(x –
y = 2(x –
2
3)
2
3)
+1
+1
Graph
It’s a parabola with vertex
(3,1) and axis x = 3
8
6
4
fx = 2x-32+1
2
-5
5
-2
10
Graph
f(x) =
y=
2
3x
2
3x
– 6x + 1
– 6x + 1
Graph
2
3x
y =
– 6x + 1
Rewrite the equation in the
2
form : y – k = a(x – h)
2
y -1 = 3(x – 2x)
2
y – 1 + 3 = 3(x – 2x + 1)
y + 2 = 3(x-1)2
fx = 3x2-6x+1
6
4
2
-5
5
-2
-4
-6
Quadratic Functions
2
ax
Let f(x) =
+ bx + c, a0
If a < 0, f has a maximum
value.
If a > 0, f has a minimum
value.
and
Quadratic Functions
The graph of f is a parabola.
This maximum or minimum
value of f is the y-coordinate
when x = - b/2a, at the vertex
of the graph.
Quadratic Functions
2
ax
Let f(x) =
+ bx + c, a0
This maximum or minimum
value of f is the y-coordinate
when
x = - b/2a, at the vertex of
the graph.
Example
2
1/2x
y=
+ 3x – 7/4
Find the maximum or
minimum value of f.
Find the vertex of the graph
of f
Example
2
1/2x
y=
+ 3x – 7/4
a = ½, a > 0, then f has a
minimum value.
Minimum occurs when
x = -b/2a
x = -3/2*1/2 = -3
Example
y = 1/2x2 + 3x – 7/4
y = ½(-3)2 + 3(-3) – 7/4
y = -25/4
So the minimum value of f is
-25/4 and the vertex is
(-3, -25/4)
7-7
Writing Quadratic
Equations and
Functions
Theorem
A quadratic equation with
roots r1 and r2 is
x2 – (r1 + r2)x + r1r2 = 0 or
2
a[x – (r1 + r2)x + r1r2 ] = 0
and
Theorem
The equation just given is
equivalent to
2
a[x – (sum of roots)x +
product of roots ] = 0
Example
Find a quadratic equation with
roots (2 + i)/3 and (2 – i)/3
Sum of roots =
(2 + i)/3 + (2 – i)/3 = 4/3
Product of roots =
(2 + i)/3 (2 – i)/3= 5/9
x2 – 4/3x + 5/9 = 9x2 -12x + 5
Theorem
If r1 and r2 are the roots of a
quadratic equation
ax2 + bx + c =0, then
r1 + r2 = sum of roots = -b/a
and
r1r2 = product of roots = c/a
Example
Find the roots of
2x2 + 9x + 5 = 0
r1 = -9 + 41 r2 = -9 - 41
4
4
Check
r1 + r2 = -9 + 41 + -9 - 41
4
4
= -18/4 = - b/a
r1
· r2 = -9 + 41 · -9 - 41
= 5/2 = c/a
4
4
END
END