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More Equations
Disguised quadratic equations
Example 1
:
Solve for
x
:
=
+
x
4
3
4
x
2
.
Solution
:
=
+
x
4
3
4
x
2
 (i).
is equivalent to the equation
=

+
x
4
4
x
2
3
0.
Letting
=
u
x
2
we obtain a quadratic equation for
u
.
=

+
u
2
4
u
3
0.
The lefthand side can be factored to give
=
(
)

u
1 (
)

u
3
0.
This gives
=
u
1
or
=
u
3.
Writing these two equations in terms of
x
gives:
=
x
2
1
or
=
x
2
3.
Solving each of these equations for
x
gives
=
x

1
or
=
x
1
or
=
x

3
or
=
x
3.
Hence the solution set of equation (i) is
{
}
,
, ,

3

1 1
3 .
_______
Note
:
The solutions of equation (i) give the
x
coordinates of the points of intersection of the two graphs
=
y
+
x
4
3
and
=
y
4
x
2
.
The solutions can also be visualised as the
x
intercepts of the graph of
=
y

+
x
4
4
x
2
3.
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View Full Document Example 2
:
Solve for
x
:
=

+
2
x
9
x
4
0.
Solution
:
=

+
2
x
9
x
4
0
 (ii).
Letting
=
r
x
we obtain a quadratic equation for
r
.
=

+
2
r
2
9
r
4
0.
The lefthand side can be factored to give
=
(
)

2
r
1 (
)

r
4
0.
This gives
=
r
1
2
or
=
r
4.
Writing these two equations in terms of
x
gives:
=
x
1
2
or
=
x
4.
Solving each of these equations for
x
gives
=
x
1
4
or
=
x
16.
Hence the solution set of equation (i) is
{
}
,
1
4
16 .
____
Notes
:
(1) The solutions can be checked in the original equation (ii).
When
=
x
1
4
, the lefthand side of (ii) is:
=

+
2
1
4
9
1
4
4

+
1
2
9
2
4
=
0.
When
=
x
16, the lefthand side of (ii) is:
=

+
2 (
)
16
9
16
4

+
32
36
4
=
0.
(2) Since (ii) is equivalent to the equation
=
+
2
x
4
9
x
,
the solutions of (ii) can also be visualised as the
x
coordinates of the points of interection of the line
=
y
+
2
x
4
with the curve
=
y
9
x
.
(3)
The solutions can also be visualised as the
x
intercepts of the graph of
=
y

+
2
x
9
x
4.
(4) The equation (ii) can be solved in another way.
Isolating the term involving the square root on the right side of the equation gives:
=
+
2
x
4
9
x
.
Squaring both sides of this equation gives:
=
+
+
4
x
2
16
x
16
81
x
.
This is a quadratic equation for
x
and it can be written in the form:
=

+
4
x
2
65
x
16
0.
The lefthand side can be factored to give the equivalent equation:
=
(
)

4
x
1 (
)

x
16
0,
so gives:
=
x
1
4
or
=
x
16,
as before.
The process of squaring an equation can introduce
extraneous solutions
, but we this has not happened here since we know that both of
the values
=
x
1
4
and
=
x
16
check in the original equation.
Equations involving rational expressions
Example 3
:
Solve for
t
:
=

+
t
1
3

t
1
2
1.
Solution
:
=

+
t
1
3

t
1
2
1
<=>
=

2 (
)
+
t
1
3 (
)

t
1
6
( multiplying both sides of the first equation by 6 )
<=>
=
+

+
2
t
2
3
t
3
6
<=>
=

t
1
<=>
=
t

1.
_____
Example 4
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This note was uploaded on 11/12/2011 for the course MATH 111 taught by Professor Uri during the Spring '08 term at Rutgers.
 Spring '08
 Uri
 Equations

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