Question

How do you solve $\sqrt {3x + 1} = x - 1$ and find any extraneous solutions?

Answer 1

Hint: We will first take the square of both the sides in the given equation and thus form a quadratic equation and find its roots. Now, we will see if the roots create any problem in the given equation and term that as an extraneous solution.

Complete step by step answer:
We are given that we are required to solve $\sqrt {3x + 1} = x - 1$ and find any extraneous solution, if it has any.
Since, we are given that $\sqrt {3x + 1} = x - 1$ ………………(1)
Taking squares on both the sides of the above equation, we will then obtain the following equation:-
$ \Rightarrow {\left( {\sqrt {3x + 1} } \right)^2} = {\left( {x - 1} \right)^2}$
Since, we know that square – root cancels the square, we will then obtain the following equation:-
$ \Rightarrow 3x + 1 = {\left( {x - 1} \right)^2}$ …………..(2)
Now, we will use the identity given by the following formula:-
$ \Rightarrow {(a - b)^2} = {a^2} + {b^2} - 2ab$
Replacing a by x and b by 1, we will then obtain the following equation:-
$ \Rightarrow {(x - 1)^2} = {x^2} + {1^2} - 2x$
Simplifying the calculations in the right hand side of the above equation, we will then obtain the following equation:-
$ \Rightarrow {(x - 1)^2} = {x^2} + 1 - 2x$
Now, if we put this in equation number 2, we will then obtain the following equation:-
$ \Rightarrow 3x + 1 = {x^2} + 1 - 2x$
Bringing all the terms from right hand side to the left hand side and multiplying it by a negative sign, we will then obtain the following equation:-
$ \Rightarrow {x^2} - 5x = 0$
Taking x common from both of these terms, we will then obtain the following equation:-
$ \Rightarrow x(x - 5) = 0$
Therefore, the possible values of x are 0 and 5.
Now, we see that if we put x = 0 in the given equation number 1, we get the LHS = 1 but the RHS = -1.
So, x = 0 is an extraneous solution.

Note:
Extraneous Solution: The solution which we obtain while solving the question but it has no relation whatsoever to the given equation. For example: x = 0 in the above example is an extraneous solution.
In other words, we have:
An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation.