Rational Equation with a Parameter and a Hidden Domain Restriction
A tutoring group is analyzing the rational equation (x^2 − (a + 2)x + 2a)/(x − 1) = 0, where a is a real parameter. A student substituted a = 4, canceled the denominator immediately, and reported the solutions x = 1 and x = 2. The tutor points out that cancellation is allowed only after recording x ≠ 1, and that the parameter can change the number of valid solutions. The group must decide for which values of a the equation has exactly one real solution, and must distinguish a repeated root from a root excluded by the domain. They also need to explain what happens at a = 1 and at a = 2, because those values are easy to mishandle when the numerator is factored. A second exercise asks them to solve the equation completely when a = 4 and to verify each candidate in the original equation rather than in the canceled form. The tutor requires a written answer that names the domain restriction, gives the factorization, and states the final count for every relevant parameter value.
1. State the domain restriction, factor the numerator, and determine the number of distinct real solutions for every real value of a. Justify the cases a = 1 and a = 2.
Read the worked answer
Domain. The denominator is zero at x = 1, so every candidate must satisfy x ≠ 1.
Factor the numerator. x^2 − (a + 2)x + 2a = (x − 2)(x − a). The equation is therefore (x − 2)(x − a)/(x − 1) = 0 with x ≠ 1.
The candidates are x = 2 and x = a.
- If a ≠ 1 and a ≠ 2, both candidates are valid and distinct, so there are two distinct solutions.
- If a = 1, the candidates are x = 2 and x = 1. The value x = 1 is excluded by the domain, leaving only x = 2. So there is one distinct solution.
- If a = 2, the numerator is (x − 2)^2. The only candidate is x = 2, which is valid. The root has multiplicity 2, but there is still one distinct real solution.
Conclusion: the equation has exactly one distinct real solution when a = 1 or a = 2; otherwise it has two. It never has zero real solutions.
2. Solve the equation completely for a = 4, and verify the solutions in the original equation.
Read the worked answer
For a = 4, the numerator factors as (x − 2)(x − 4). The equation is (x − 2)(x − 4)/(x − 1) = 0, with x ≠ 1.
The candidates are x = 2 and x = 4. Neither equals 1, so both are valid.
Check in the original equation:
- At x = 2, the numerator is 4 − 24 + 8 = −12? Wait: compute directly: x^2 − 6x + 8 = 4 − 12 + 8 = 0, denominator 1, so the value is 0.
- At x = 4, the numerator is 16 − 24 + 8 = 0, denominator 3, so the value is 0.
Thus the solution set is x = 2 or x = 4. The canceled form would have suggested x = 1 as a possible root, but the original expression is undefined there.
3. A classmate claims that the equation always has two solutions because the numerator is quadratic. Identify the flaw in that reasoning and connect it to the parameter values found above.
Read the worked answer
The numerator being quadratic guarantees at most two real roots of the numerator, but it does not guarantee two valid solutions of the original rational equation. Some numerator roots may be excluded by the denominator, and a quadratic may have a repeated root that gives only one distinct value.
Here the numerator roots are x = 2 and x = a. If a = 1, the root x = 1 is excluded because the denominator is zero there. If a = 2, the two numerator roots coincide at x = 2, so there is only one distinct candidate. In both cases the count of valid solutions drops from two to one. The correct count must always be made after applying the domain restriction x ≠ 1.