Question

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

Answer 1

Hint: In this question, we are given an equation and we have been asked to solve it. But, if we observe, we will see that one side of the equation has a square root in it. We cannot solve the question with a square root in it. So, square both the sides. You will get a quadratic. Solve the equation and keep the values in the given equation to find the values.

Complete step-by-step solution:
We are given an equation in this question and we have to solve that question.
$ \Rightarrow x - 3 = \sqrt {x - 1} $ …. (given)
Since the right-hand side of the question has a square root in it, we will square both the sides to remove the square root.
Squaring both the sides, we get,
$ \Rightarrow {\left( {x - 3} \right)^2} = {\left( {\sqrt {x - 1} } \right)^2}$
Simplifying the equation,
$ \Rightarrow {x^2} + 9 - 6x = x - 1$
Rearranging the terms,
$ \Rightarrow {x^2} + 9 - 6x - x + 1 = 0$
On add the term and we get,
$ \Rightarrow {x^2} - 7x + 10 = 0$
Now, we will use splitting the middle term to factorise the given equation.
Think of two such factors of$10$, which when added or subtracted will give us $ - 7$. Now, such factors are $2$ and $5$. Now, if we add them, we will get $7$. But we want $ - 7$. So, we will take both the factors as negative.
$ \Rightarrow {x^2} - 2x - 5x + 10 = 0$
Now, we will take x common out of the first two terms and 5 common out of last two terms.
$ \Rightarrow x\left( {x - 2} \right) - 5\left( {x - 2} \right) = 0$
Making the factors,
$ \Rightarrow \left( {x - 5} \right)\left( {x - 2} \right) = 0$
Keep each factor equal to 0 to find the values of x,
$ \Rightarrow x - 5 = 0,x - 2 = 0$
$ \Rightarrow x = 5,2$
Now, in order to check these values, we will put them in the given equation,
$ \Rightarrow $Putting $x = 5$ in $x - 3 = \sqrt {x - 1} $ ,
$5 - 3 = \sqrt {5 - 1} = \sqrt 4 = 2$
$2 = 2$
Hence, the value is satisfied.
$ \Rightarrow $Putting $x = 2$ in $x - 3 = \sqrt {x - 1} $ ,
$2 - 3 = \sqrt {2 - 1} = \sqrt 1 = 1$
$ - 1 \ne 1$
Hence, this value is extraneous.

Note: In mathematics, extraneous solution is a solution which came out as a solution from the entire operation but it is not a valid answer as it does not satisfy the equation. For example: in the above question, when we put $x = 2$, our left-hand side was different from the right-hand side. This tells us that $x = 2$ is an extraneous solution.