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Rational Expression Concepts -- Rational Equations

Module by: Kenny M. Felder. E-mail the author

Summary: This module introduces rational expressions in equations.

Rational Equations

A rational equation means that you are setting two rational expressions equal to each other. The goal is to solve for x; that is, find the x value(s) that make the equation true.

Suppose I told you that:

x 8 = 3 8 x 8 = 3 8 size 12{ { {x} over {8} } = { {3} over {8} } } {}
(1)

If you think about it, the x in this equation has to be a 3. That is to say, if x=3 then this equation is true; for any other x value, this equation is false.

This leads us to a very general rule.

A very general rule about rational equations

If you have a rational equation where the denominators are the same, then the numerators must be the same.

This in turn suggests a strategy: find a common denominator, and then set the numerators equal.

Table 1
Example: Rational Equation
3 x 2 + 12 x + 36 = 4x x 3 + 4x 2 − 12 x 3 x 2 + 12 x + 36 = 4x x 3 + 4x 2 − 12 x size 12{ { {3} over {x rSup { size 8{2} } +"12"x+"36"} } = { {4x} over {x rSup { size 8{3} } +4x rSup { size 8{2} } - "12"x} } } {} Same problem we worked before, but now we are equating these two fractions, instead of subtracting them.
3 ( x ) ( x − 2 ) ( x + 6 ) 2 ( x ) ( x − 2 ) = 4x ( x + 6 ) x ( x + 6 ) 2 ( x − 2 ) 3 ( x ) ( x − 2 ) ( x + 6 ) 2 ( x ) ( x − 2 ) = 4x ( x + 6 ) x ( x + 6 ) 2 ( x − 2 ) size 12{ { {3 \( x \) \( x - 2 \) } over { \( x+6 \) rSup { size 8{2} } \( x \) \( x - 2 \) } } = { {4x \( x+6 \) } over {x \( x+6 \) rSup { size 8{2} } \( x - 2 \) } } } {} Rewrite both fractions with the common denominator.
3 x ( x - 2 ) = 4 x ( x + 6 ) 3x(x-2)=4x(x+6) Based on the rule above—since the denominators are equal, we can now assume the numerators are equal.
3 x 2 – 6 x = 4 x 2 + 24 x 3 x 2 –6x=4 x 2 +24x Multiply it out
x 2 + 30 x = 0 x 2 +30x=0 What we’re dealing with, in this case, is a quadratic equation. As always, move everything to one side...
x ( x + 30 ) = 0 x(x+30)=0 ...and then factor. A common mistake in this kind of problem is to divide both sides by xx; this loses one of the two solutions.
x= 0 x=0 or x= -30 x=-30 Two solutions to the quadratic equation. However, in this case, x=0x=0 is not valid, since it was not in the domain of the original right-hand fraction. (Why?) So this problem actually has only one solution, x=–30x=–30.

As always, it is vital to remember what we have found here. We started with the equation 3x2+12x+36=4xx3+4x2−12x3x2+12x+36=4xx3+4x2−12x size 12{ { {3} over {x rSup { size 8{2} } +"12"x+"36"} } = { {4x} over {x rSup { size 8{3} } +4x rSup { size 8{2} } - "12"x} } } {}. We have concluded now that if you plug x=–30x=–30 into that equation, you will get a true equation (you can verify this on your calculator). For any other value, this equation will evaluate false.

To put it another way: if you graphed the functions 3x2+12x+363x2+12x+36 size 12{ { {3} over {x rSup { size 8{2} } +"12"x+"36"} } } {} and 4xx3+4x2−12x4xx3+4x2−12x size 12{ { {4x} over {x rSup { size 8{3} } +4x rSup { size 8{2} } - "12"x} } } {}, the two graphs would intersect at one point only: the point when x=–30x=–30.

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