The misinterpretation of the role of the unknown that appears to the left and right of the equal sign often leads to root loss, which is a common error when solving quadratic equations. This error arises due to a need for more understanding regarding the proper interpretation of the unknown values within the equation. It is crucial to approach the problem with a clear understanding of the roles and relationships of the variables involved to avoid such mistakes and achieve accurate results.

For example, try to solve the equation:

\[x(2x-5) = x\]

Solving equations can be tricky, and it’s common to make mistakes, especially if you’re new to it. It’s important to remember that simplifying both sides of the equation by removing x might seem like the right approach, but it can lead to confusion and errors. In this case, someone could be tempted to do the following simplification:

\[\cancel{x}(2x-5) = \cancel{x}\]

This simplification leads to obtaining a first-degree equation in x that has a unique solution.

\[2x-5=0\] \[x= \frac{5}{2}\]

This error generates an incorrect solution and leads to the loss of a second equation solution. In more general terms, the variable \(x\) should always be retained and never cancelled when solving equations to avoid the loss of solutions.


Correct procedure

It is essential to follow the correct procedure when dealing with second-degree equations. First, the equation should be returned to its standard form, simplifying it to the form \(ax^2+bx+c\).

We have:

\begin{align} 2x^2-5x-x &= 0 \\[0.5em] 2x^2-6x &= 0 \\[0.5em] 2x(x-3) &= 0 \end{align}

The equation is reduced to an incomplete quadratic equation of the second degree with the coefficient \(c\) being null. The equations admits two distinct solutions, \(x1 = 0\) and \(x2 = 3\).

The solution to the equation is: \[x_1 = 0 \quad x_2 = 3\]

The correct procedure demonstrates that there are two solutions, and neither is equal to the value found in the case of the wrong procedure.