Why squaring both sides of a radical equation is an irreversible operation?

1 Answer
Feb 19, 2018

See explanation...

Explanation:

Given an equation to solve of the form:

#"left hand expression " = " right hand expression"#

we may attempt to simplify the problem by applying the same function #f(x)# to both sides to get:

#f(" left hand expression ") = f(" right hand expression ")#

Any solution of the original equation will be a solution of this new equation.

However, note that any solution of the new equation may or may not be a solution of the original one.

If #f(x)# is one to one - e.g. multiplication by a non-zero constant, cubing, adding or subtracting the same thing from both sides - then solutions of the new equation will be solutions of the original.

In the case of #f(x) = x^2#, we have a function which is not one to one. For example #f(-x) = f(x)#. So solutions of the new equation may not be solutions of the original one.

For example, given:

#sqrt(2x+1) = -sqrt(x+3)#

We can square both sides of the equation to get:

#2x+1 = x+3#

This new equation has solution #x=2#, but it is not a solution of the original equation.