A rational expression is one that has the form
$\frac{p\left(x\right)}{q\left(x\right)}$p(x)q(x),
where $p\left(x\right)$p(x) and $q\left(x\right)$q(x) are polynomials and $q\left(x\right)\ne0$q(x)≠0.
A rational equation is then an equation containing polynomials and at least one rational expression.
So, for example, the following are all rational equations:
| $\frac{x}{x+1}=0$xx+1=0 | $4=\frac{p^2-9}{2p-1}$4=p2−92p−1 | $5x^3-1=\frac{2x^3-3}{2x}$5x3−1=2x3−32x |
Remember that a rational expression is undefined whenever the denominator of the expression is zero.
So an expression such as $\frac{1}{x}$1x is undefined for $x=0$x=0 and defined for every other value of $x$x.
More complicated expression such as $\frac{1}{x^2-3x+2}$1x2−3x+2 are also undefined for certain values of $x$x, but these values are less obvious. To find the values of $x$x for which the expression is undefined, we can construct an equation by setting the denominator to equal zero.
| $x^2-3x+2$x2−3x+2 | $=$= | $0$0 | (Setting the denominator to zero) |
| $\left(x-2\right)\left(x-1\right)$(x−2)(x−1) | $=$= | $0$0 | (Factoring the expression) |
| $x$x | $=$= | $1,2$1,2 | (Solving for each factor) |
So we can conclude that the expression $\frac{1}{x^2-3x+2}$1x2−3x+2 is undefined for $x=1$x=1 and $x=2$x=2.
A rational equation is then undefined whenever any of the expressions in the equation are undefined.
A rational expression is undefined whenever the denominator of the expression is zero.
A rational equation is then undefined whenever any expression in the equation is undefined.
Worked example
For what values of $x$x is the following rational equation undefined?
$\frac{x-3}{\left(x-3\right)\left(x+7\right)}=0$x−3(x−3)(x+7)=0
Think: A rational equation is undefined when the expressions of the equation are undefined. In this case, that happens when the denominator $\left(x-3\right)\left(x+7\right)$(x−3)(x+7) is zero.
Do: To determine when the denominator is zero, we set it equal to zero and solve for $x$x:
| $\left(x-3\right)\left(x+7\right)$(x−3)(x+7) | $=$= | $0$0 | (Setting the denominator to zero) |
| $x-3$x−3 | $=$= | $0$0 | (Solving when one factor is zero) |
| $x$x | $=$= | $3$3 | (Adding $3$3 to both sides) |
| $x+7$x+7 | $=$= | $0$0 | (Solving for the other factor) |
| $x$x | $=$= | $-7$−7 | (Subtracting $7$7 from both sides) |
| $x$x | $=$= | $-7,3$−7,3 | (Combining both solutions) |
So we can conclude that the equation $\frac{x-3}{\left(x-3\right)\left(x+7\right)}=0$x−3(x−3)(x+7)=0 is undefined for $x=-7$x=−7 and $x=3$x=3.
Reflect: The expression $\frac{x-3}{\left(x-3\right)\left(x+7\right)}$x−3(x−3)(x+7) has a common factor of $x-3$x−3 between the numerator and denominator.
If we were to cancel this common factor, we would end up with the expression $\frac{1}{x+7}$1x+7. Notice that this expression is only undefined when $x=-7$x=−7, and is defined when $x=3$x=3.
So the expressions $\frac{x-3}{\left(x-3\right)\left(x+7\right)}$x−3(x−3)(x+7) and $\frac{1}{x+7}$1x+7 are different, since the second one is defined for $x=3$x=3.
Now we can simplify a rational expression by cancelling common factors, but only over values for which it is defined.
That is, $\frac{x-3}{\left(x-3\right)\left(x+7\right)}=\frac{1}{x+7}$x−3(x−3)(x+7)=1x+7 for all real values of $x$x except for $x=-7$x=−7 and $x=3$x=3.
An algebraic expression is undefined whenever the denominator is zero before any factors have been cancelled.
Practice questions
Question 1
For what value of $w$w is the following rational equation undefined?
$\frac{4}{w}=0$4w=0
Question 2
For what values of $b$b is the following rational equation undefined?
$\frac{6}{b-2}-\frac{7b}{\left(b-5\right)\left(b-8\right)}=0$6b−2−7b(b−5)(b−8)=0
Write each answer on the same line, separated by commas.
question 3
For what values of $j$j is the following rational equation undefined?
$-\frac{4}{j}+\frac{7}{j+5}=\frac{24}{j^2+j}$−4j+7j+5=24j2+j
Write each answer on the same line, separated by commas.