Quadratic equations and quadratic functions
Introduction
.
A
quadratic function
is a function f of the form
=
(
)
f
x
+
+
a x
2
b x
c
 (i),
where
≠
a
0.
A
quadratic equation
is an equation of the form
=
+
+
a x
2
b x
c
0
 (ii),
where
≠
a
0.
The real number solutions of equation (ii), if any,
provide the
x
intercepts of the graph of the function f.
Any real number
x
in the domain
of a general function g, which has real number values, such that
=
(
)
g
x
0 is called a
zero
of g.
Hence any solution of the quadratic
equation (ii) will be a zero of the quadratic function f defined by equation (i).
Example 1
:
Solve the quadratic equation
=


2
x
2
x
3
0,
and illustrate the solutions as the
x
intercepts of the graph of
=
y


2
x
2
x
3.
Solution
:
By factoring the left side of the equation
=


2
x
2
x
3
0,
we see that it is equivalent to
=
(
)

2
x
3
(
)
+
x
1
0.
This gives
=
x
3
2
or
=
x

1.
_______
To sketch the graph of
=
y


2
x
2
x
3 we make use of the technique of
completing the square
.
Given a pair of terms
+
x
2
2
p
,
we can add
p
2
, which is "half the coefficient of
x
squared" to form the
perfect square
=
+
+
x
2
2
p
p
2
(
)
+
x
p
2
.
=
y


2
x
2
x
3
is equivalent to
=
y

2

x
2
x
2
3
 (iii).
Adding
=

1
4
2
1
16
inside the bracket

x
2
x
4
will form the perfect square
=

+
x
2
x
2
1
16

x
1
4
2
.
It is useful to incorporate this perfect square in a rearrangement of the right side of equation (iii), but we must make an adjustment to the
constant

3
in the right side of (iii).
By adding
1
16
inside the bracket we have effectively added
=
2
1
16
1
8
to the right side of
equation (iii) because of the 2 factor to the left of the bracket. Thus we shall have an equation equivalent to (iii) if we also subtract
1
8
"outside the bracket" to give
=
y


2

+
x
2
x
2
1
16
1
8
3.
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Alternatively, introduce
+

1
16
1
16
inside the bracket so that equation (iii) becomes:
=
y

2

+

x
2
x
2
1
16
1
16
3 ,
which gives:
=
y


2

+
x
2
x
2
1
16
1
8
3.
This equation can be written in the form
=
y

2

x
1
4
2
25
8
 (iv).
Using ideas about transformations of graphs of functions, the equation (iv) shows that the graph of
=
y


2
x
2
x
3 has the same shape as
the parbabola
=
y
2
x
2
, because it can be obtained by first translating the curve
=
y
2
x
2
to the right by
1
4
unit to give the curve
=
y
2

x
1
4
2
,
and then shifted
=
25
8
3
1
8
units down to form the curve
=
y

2

x
1
4
2
25
8
.
In particular, the lowest point or
vertex
of
the parabola
=
y
2
x
2
moves from the origin to the point
,
1
4

25
8
.
The parabola
=
y
2
x
2
is obtained by scaling the
y
coordinates of the points on the curve
=
y
x
2
with the factor 2 so that they all move
vertically to lie at twice the distance from the
x
axis, which stretches the graph
=
y
x
2
parallel to the
y
axis.
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 Spring '08
 Uri
 Equations, Quadratic equation, Q7., Peter Stone

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