Question

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

Answer 1

Hint: We will square both the terms in the given equation and then we will try to find the value of \[x\]. On doing some simplification we get the required answer.

Formula Used:
Let say, two inequalities are there as following:
\[ax + b \geqslant 0\]
And, \[mx + n \geqslant 0\].
Then we need to find the domain of the inequality so that we can confirm the range of the function.
If any function is given in the non-negative format, it shows that the domain of the variable should lie in a particular range.
If we take the square root of any expression then it can have either positive value or a negative valued sign in front of it.

Complete step by step answer:
In the above question following three equations are given:
The following expression is given in the question: \[\sqrt {(3x + 1)} = x - 1\].
As the term in the L.H.S is a non-negative term, we must need to find the domain for the L.H.S .
So, we can derive the following inequality from the above equation, as \[\sqrt {(3x + 1)} \] is a non-negative term.
Therefore,
\[ \Rightarrow 3x + 1 \geqslant 0\].
By solving the inequality, we can draw the following inequality:
\[ \Rightarrow x \geqslant - \dfrac{1}{3}.\]
Now, for the given equation the term in R.H.S should always be greater than or equal to zero also.
So,
\[ \Rightarrow x - 1 \geqslant 0\].
So, by comparing both the inequality, we can derive that following inequality: \[x\]shall be greater than or equal to zero.
So, \[x \geqslant 1.\]
So, \[x \geqslant 1\] is the required domain for the above equation.
Now, taking the square on the both sides of the given equation, we get:
\[ \Rightarrow {\left( {\sqrt {(3x + 1)} } \right)^2} = {(x - 1)^2}\].
Now, by doing the further calculations, we get:
\[ \Rightarrow \left( {3x + 1} \right) = ({x^2} - 2x + 1)\].
Now, taking all the terms into one side of the equation, we get:
\[ \Rightarrow {x^2} - 2x + 1 - 3x - 1 = 0.\]
Now, doing the simplification, we get:
\[ \Rightarrow {x^2} - 5x = 0.\]
Now, by doing the further simplification, we get:
\[ \Rightarrow x(x - 5) = 0\].
So, we can say that either \[x\] is equal to \[0\] or\[(x - 5) = 0\].
So, \[x = 0\] or\[x = 5\].
But we already found that\[x \geqslant 1\], it must be true.

Therefore, the final solution of the equation is \[x = 5\]. And, \[x = 0\] is the extraneous solution of the equation.

Note: Points to remember:
If any equation is showing that it has multiple solutions but it does not define the equation’s general solution, then we can use that solution in the real equation whether the solution is feasible or not.