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{{ printedBook.courseTrack.name }} {{ printedBook.name }} When you solve equations algebraically, there are the occasional rare cases when the given solutions are extraneous or one more more solutions are lost.

If both sides of an equation is squared or if both sides is multiplied with an expression that contains a variable, it is necessary to check the solutions.

**Squaring:**If both sides of an equation are squared, for example, $x=4$, the same operation is done to both sides. This means that equality is maintained. However, there might still appear solutions which do not solve the original equation.

$xx_{2} =4=16⇔x=±4 $
Two solutions appear, $x=-4$ and $x=4,$ but only **one** solves the equation above. As a result, there is a risk of adding solutions that do not solve the original equation when squaring both sides.

**Multiplication by a variable:**If an equation has a variable in the denominator, an extraneous solution can be introduced by multiplying both sides with $x.$ An example is the equation $x2x =0$, which lacks solutions. However, if both sides are multiplied by $x$ it can look like $x=0$ is a solution.

$x2x 2x =0=0⇔x=0 $

If $x=0$ is tested in the original equation, the left-hand side will be undefined due to division by $0$. It is, in fact, already understood that $x$ cannot equal $0$ because $x$ stands by itself in the denominator. When the equation is multiplied with $x,$ in order to move it to the numerator, this condition disappears.

For some equations, the variable is present in all terms, like for example in $3x_{2}=3x$. In cases like these, it can be very tempting to divide by $x$. $3x_{2}3x =3x=3⇔x=1 $

However, $x=0$ is also a solution for the original equation. Why did it disappear? When dividing by $x$, it is assumed that $x$ does not equal $0$ because division by zero is not allowed. Therefore, it is necessary to be careful when dividing both sides by an expression that contains variables. When dividing by a variable, it is important to investigate which values that make the expression equal to $0$. It must be tested whether these are actual solutions to the original equation.