http://www.ck12.org Chapter 2. Polynomials and Rational Functions
x(x+ 3 )− 5 = 12 (x+ 3 )
x^2 + 3 x− 5 − 12 x− 36 = 0
x^2 − 9 x− 41 = 0
x=−(−^9 )±√(− 9 ) (^2) − 4 · 1 ·(− 41 )
2 · 1
x=^9 ±^7
√ 5
2
The only potential extraneous solution would have been -3, so both answers are possible.
Example B
Solve the following rational equation
x^3 +x 4 −x+^12 =−x^2 +^26 x+ 8
Solution:Multiply each part of the equation by the common denominator ofx^2 + 6 x+ 8 = (x+ 2 )(x+ 4 ).
(x+ 2 )(x+ 4 )[ 3 x
x+ 4 −1
x+ 2]
=
[ − 2
(x+ 2 )(x+ 4 )]
(x+ 2 )(x+ 4 )
3 x(x+ 2 )−(x+ 4 ) =− 2
3 x^2 + 6 x−x− 2 = 0
3 x^2 + 5 x− 2 = 0
( 3 x− 1 )(x+ 2 ) = 0
x=^13 ,− 2Note that -2 is an extraneous solution. The only actual solution isx=^13.
Example C
Solve the following rational equation fory.
x= 2 + 2 +^1 y+ (^11)
Solution: This question can be done multiple ways. You can use the clearing fractions technique twice.
(
2 +y+^11
)
x=[
2 + 2 +^11
y+ 1](
2 +y+^11)
2 x+y+x 1 = 2(
2 +y+^11)
+ 1
2 x+y+x 1 = 4 +y+^21 + 1(y+ 1 )[
2 x+y+x 1]
=
[
5 +y+^21]
(y+ 1 )
2 x(y+ 1 )+x= 5 (y+ 1 )+ 2
2 xy+ 2 x+x= 5 y+ 5 + 2Now just get theyvariable to one side of the equation and everything else to the other side.