Chapter 7 Rational Functions - College of the Redwoods

718 Chapter 7 Rational Functions
Cancel.
Simplify each side.
x(x − 4)
[ 1
x(x − 4)
x + 1
[ 1
x]
+ x(x − 4)
]
= [2] x(x − 4)
x − 4
[ ] 1
= [2] x(x − 4)
x − 4
[ [ ]
1 1
x(x − 4) + x(x − 4) = [2] x(x − 4)
x]
x − 4
(x − 4) + x = 2x(x − 4)
2x − 4 = 2x 2 − 8x
This last equation is nonlinear, so we make one side zero by subtracting 2x and adding
4 to both sides of the equation.
0 = 2x 2 − 8x − 2x + 4
0 = 2x 2 − 10x + 4
Note that each coefficient on the right-hand side of this last equation is divisible by
2. Let’s divide both sides of the equation by 2, distributing the division through each
term on the right-hand side of the equation.
0 = x 2 − 5x + 2
The trinomial on the right is a quadratic with ac = (1)(2) = 2. There are no integer
pairs having product 2 and sum −5, so this trinomial doesn’t factor. We will use the
quadratic formula instead, with a = 1, b = −5 and c = 2.
x = −b ± √ b 2 − 4ac
2a
= −(−5) ± √ (−5) 2 − 4(1)(2)
2(1)
= 5 ± √ 17
2
It remains to compare these with the graphical solutions found in part (c). So, enter the
solution (5- √ (17))/(2) in your calculator screen, as shown in Figure 8(a). Enter
(5+ √ (17))/(2), as shown in Figure 8(b). Thus,
5 − √ 17
2
≈ 0.4384471872
and
5 + √ 17
2
≈ 4.561552813.
Note the close agreement with the approximations found in part (c).
(a)
Figure 8.
(b)
Approximating the exact solutions.
Version: Fall 2007