Extraneous solutions

Suppose that x is the variable to be solved for in some equation. If we increase the degree of x in the equation (for example, by multiplying through by x − a, clearing the denominator, cross-multiplying, squaring both sides of the equation, etc.) then we run the risk of introducing extraneous solutions. These are false solutions that are created in the algebra process but that do not satisfy the original equation.

For example, the equation
x − 3 = 0
obviously has one solution, x = 3. But, if (for some reason) we multiply both sides of the equation by x then we get the equation
x (x − 3) = 0,
which has two solutions x = 0 and x = 3. The solution x = 0 does not satisfy the original equation and is therefore extraneous.

Often in the algebra process there is no way to avoid introducing extraneous solutions. The only recourse is to check all our solutions by substituting them back into the original equation. Any that don’t satisfy the original equation are extraneous and must be dropped.



Lost solutions

Suppose that x is the variable to be solved for in some equation. If we decrease the degree of x in the equation (for example by dividing through by x − a or by taking the square root of both sides of the equation) then we run the risk of losing solutions. But there is an easy way to avoid this problem. Never divide an equation through by any expression containing x that could equal zero. Instead simply keep the factors.

For example, the equation
x (x − 1) (x 2 + 1) = 0
can be simplified by dividing both sides by 5 and by x 2 + 1 since neither quantity can equal zero. Doing so gives us the equation
x (x − 1) = 0.
But we can’t divide through by x or x − 1 because then we will lose the solutions x = 0 and x = 1.





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