# Loss of Roots

*Ref.*quadratic equations

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.