Parametric equations are equations in which, alongside the unknown, one or more letters appear that represent unspecified values. These letters are called parameters.
For example:
\[ (a-1)x=2 \]
is a parametric equation in the unknown \(x\), with parameter \(a\).
The presence of a parameter fundamentally changes the way the equation is solved. Indeed, one does not seek a single numerical solution; instead, one studies how the solution set varies as the parameter changes.
In other words, a parametric equation does not merely ask:
"what is the value of the unknown?"
but also:
"for which values of the parameter does the equation have one solution, no solution, or infinitely many solutions?"
What a parameter is
A parameter is a letter that appears in an equation but is not treated as the main unknown.
In the equation:
\[ ax+1=0 \]
the unknown is \(x\), while \(a\) is a parameter.
This means that \(a\) can take various real values, and for each value of \(a\) a different equation is obtained.
For example:
if \(a=2\), the equation becomes:
\[ 2x+1=0 \]
if \(a=-1\), it becomes:
\[ -x+1=0 \]
if \(a=0\), it becomes:
\[ 1=0 \]
This last case immediately illustrates why parameters must be handled with care: certain values can completely alter the nature of the equation.
First-degree parametric equations
Consider the general form:
\[ A(a)x=B(a) \]
where \(A(a)\) and \(B(a)\) are expressions depending on the parameter \(a\).
The solution depends on the coefficient of the unknown \(x\), namely \(A(a)\).
If:
\[ A(a)\ne0 \]
then both sides may be divided by \(A(a)\), giving:
\[ x=\frac{B(a)}{A(a)} \]
If instead:
\[ A(a)=0 \]
division by \(A(a)\) is not permitted. In this case the value of the parameter must be substituted back into the equation to determine what remains.
The key point: never divide by an expression that may be zero
The most common mistake in parametric equations is dividing by an expression that depends on the parameter without first checking when it vanishes.
For example, from the equation:
\[ (a-1)x=2 \]
it would be incorrect to write immediately:
\[ x=\frac{2}{a-1} \]
without first observing that:
\[ a-1=0 \quad \Longleftrightarrow \quad a=1 \]
Indeed, if \(a=1\), the equation becomes:
\[ 0\cdot x=2 \]
that is:
\[ 0=2 \]
which is a contradiction.
Therefore the formula:
\[ x=\frac{2}{a-1} \]
is valid only for:
\[ a\ne1 \]
Case analysis
Solving a parametric equation typically requires a case analysis, that is, separating the values of the parameter into distinct cases.
The purpose of the case analysis is to determine:
- for which values of the parameter the equation has a unique solution;
- for which values it is inconsistent;
- for which values it is indeterminate.
For a first-degree equation:
\[ A(a)x=B(a) \]
there are three possibilities.
Case \(A(a)\ne0\)
The equation has a unique solution:
\[ x=\frac{B(a)}{A(a)} \]
Case \(A(a)=0\) and \(B(a)\ne0\)
The equation becomes:
\[ 0\cdot x=B(a) \]
with \(B(a)\ne0\). This yields a false equality:
\[ 0=B(a) \]
and the equation is inconsistent.
Case \(A(a)=0\) and \(B(a)=0\)
The equation becomes:
\[ 0\cdot x=0 \]
that is:
\[ 0=0 \]
This equality holds for every real number, so every real number is a solution.
In this case the equation is indeterminate:
\[ S=\mathbb{R} \]
First worked example
Solve and discuss the equation:
\[ ax=4 \]
The unknown is \(x\) and \(a\) is a real parameter.
The coefficient of \(x\) is \(a\). Two cases must be distinguished.
Case \(a\ne0\)
If \(a\ne0\), both sides may be divided by \(a\):
\[ x=\frac{4}{a} \]
Therefore, for \(a\ne0\), the equation has a unique solution:
\[ S=\left\{\frac{4}{a}\right\} \]
Case \(a=0\)
If \(a=0\), the equation becomes:
\[ 0\cdot x=4 \]
that is:
\[ 0=4 \]
This equality is false. The equation therefore has no solution:
\[ S=\varnothing \]
In summary:
\[ \begin{cases} a\ne0 & \Rightarrow S=\left\{\frac{4}{a}\right\} \\ a=0 & \Rightarrow S=\varnothing \end{cases} \]
Second worked example
Solve and discuss:
\[ (a-2)x=a-2 \]
The coefficient of the unknown is:
\[ a-2 \]
Before dividing by \(a-2\), one must determine when this coefficient vanishes:
\[ a-2=0 \quad \Longleftrightarrow \quad a=2 \]
Case \(a\ne2\)
If \(a\ne2\), then \(a-2\ne0\). Both sides may therefore be divided by \(a-2\):
\[ x=\frac{a-2}{a-2} \]
Since \(a-2\ne0\), the fraction equals:
\[ x=1 \]
Therefore:
\[ S=\{1\} \]
Case \(a=2\)
If \(a=2\), substituting into the original equation gives:
\[ (2-2)x=2-2 \]
that is:
\[ 0\cdot x=0 \]
hence:
\[ 0=0 \]
This equality holds for every value of \(x\). Therefore:
\[ S=\mathbb{R} \]
In summary:
\[ \begin{cases} a\ne2 & \Rightarrow S=\{1\} \\ a=2 & \Rightarrow S=\mathbb{R} \end{cases} \]
Third worked example
Solve and discuss:
\[ (a+1)x=a^2-1 \]
The coefficient of the unknown is:
\[ a+1 \]
The case in which this coefficient is nonzero must be distinguished from the case in which it vanishes.
Solving:
\[ a+1=0 \quad \Longleftrightarrow \quad a=-1 \]
Case \(a\ne-1\)
If \(a\ne-1\), then \(a+1\ne0\). Both sides may be divided by \(a+1\):
\[ x=\frac{a^2-1}{a+1} \]
Factoring the numerator:
\[ a^2-1=(a-1)(a+1) \]
Therefore:
\[ x=\frac{(a-1)(a+1)}{a+1} \]
Since we are working in the case \(a\ne-1\), we have \(a+1\ne0\), so the common factor cancels:
\[ x=a-1 \]
Hence:
\[ S=\{a-1\} \]
Case \(a=-1\)
If \(a=-1\), substituting into the original equation gives:
\[ (-1+1)x=(-1)^2-1 \]
that is:
\[ 0\cdot x=1-1 \]
hence:
\[ 0=0 \]
The equation holds for every real value of \(x\). Therefore:
\[ S=\mathbb{R} \]
In summary:
\[ \begin{cases} a\ne-1 & \Rightarrow S=\{a-1\} \\ a=-1 & \Rightarrow S=\mathbb{R} \end{cases} \]
Parametric equations with the parameter in the denominator
In some equations the parameter appears in the denominator. In such cases the first step is not to solve the equation, but to establish for which values of the parameter the equation is defined.
Consider:
\[ \frac{x}{a-1}=3 \]
The denominator cannot be zero, so one must require:
\[ a-1\ne0 \]
that is:
\[ a\ne1 \]
The equation is defined only for \(a\ne1\). In that case, multiplying both sides by \(a-1\):
\[ x=3(a-1) \]
hence:
\[ x=3a-3 \]
In summary:
\[ a\ne1 \quad \Rightarrow \quad S=\{3a-3\} \]
For:
\[ a=1 \]
the equation is undefined, since the denominator would be zero.
Second-degree parametric equations
Parametric equations may also be of second degree. In this case the parameter can affect the discriminant and hence the number of real solutions.
Consider the general form:
\[ Ax^2+Bx+C=0 \]
where at least one of \(A\), \(B\), \(C\) depends on a parameter.
If \(A\ne0\), the equation is quadratic and the discriminant is examined:
\[ \Delta=B^2-4AC \]
Depending on the sign of \(\Delta\), three cases arise:
- if \(\Delta>0\), the equation has two distinct real solutions;
- if \(\Delta=0\), the equation has a repeated real solution;
- if \(\Delta<0\), the equation has no real solutions.
If instead \(A=0\), the equation is no longer quadratic and must be treated as a first-degree equation.
Example of a second-degree parametric equation
Discuss the equation:
\[ x^2-2x+a=0 \]
Here the parameter \(a\) appears in the constant term.
The coefficient of \(x^2\) is \(1\), so the equation is quadratic for all values of \(a\).
Computing the discriminant:
\[ \Delta=(-2)^2-4\cdot1\cdot a \]
that is:
\[ \Delta=4-4a \]
Factoring out \(4\):
\[ \Delta=4(1-a) \]
The number of real solutions depends on the sign of \(1-a\).
Case \(\Delta>0\)
We need:
\[ 4(1-a)>0 \]
Since \(4>0\), the sign is determined by \(1-a\):
\[ 1-a>0 \]
hence:
\[ a<1 \]
For \(a<1\), the equation has two distinct real solutions.
Case \(\Delta=0\)
We need:
\[ 4(1-a)=0 \]
that is:
\[ 1-a=0 \]
hence:
\[ a=1 \]
For \(a=1\), the equation has a repeated real solution.
Case \(\Delta<0\)
We need:
\[ 4(1-a)<0 \]
hence:
\[ 1-a<0 \]
from which:
\[ a>1 \]
For \(a>1\), the equation has no real solutions.
In summary:
\[ \begin{cases} a<1 & \text{two distinct real solutions} \\ a=1 & \text{repeated real solution} \\ a>1 & \text{no real solutions} \end{cases} \]
Closing remarks
Parametric equations require a more careful mode of reasoning than purely numerical equations. A parameter is not a mere decorative symbol: it can change the degree of the equation, make coefficients vanish, make division impossible, or alter the number of solutions.
For this reason, the correct approach does not consist in solving mechanically, but in carrying out a case analysis.
In particular, whenever a quantity depends on the parameter, one must ask whether it can vanish. Only after this check is it valid to divide, cancel common factors, or apply solution formulas correctly.
Understanding parametric equations therefore means learning to read an equation not as a single problem, but as a family of problems — one for each value of the parameter.