Question

How do I check for extraneous solutions?

Answer 1

Identify the extraneous solution:

1. An extraneous solution is a solution that emerges while solving problems but is an invalid solution for that problem.
2. This usually gets introduced while solving the problem we mistakenly change the original equation which is not equivalent to the original equation like multiplying an equation by x or squaring both sides etc.
3. The only way to check for an extraneous solution is to put back the solution in the original problem and check whether it satisfies or not.

As an example,

$\sqrt{2x+35}=x$ $...\left[1\right]$

$\Rightarrow$ $2x+35=x^2$ [Squaring both sides]

$\Rightarrow$ $x^2-2x-35=0$

$\Rightarrow$ $x^2-7x+5x-35=0$

$\Rightarrow$$x(x-7)+5(x-7)=0$

$\Rightarrow$ $(x-7)(x+5)=0$

$\Rightarrow$ $x-7=0$ or $x+5=0$

$\Rightarrow$ $x=7,x=-5$

Put $x=7$ in $\left[1\right]$

$\sqrt{2\left(7\right)+35}=7$

$7=7$

$\Rightarrow$$x=7$ is the solution of the original equation $\left[1\right]$

Put $x=-5$ in $\left[1\right]$

$\sqrt{2(-5)+35}=-5$

$5=-5$

$\Rightarrow$$x=-5$ is not a solution of the original equation $\left[1\right]$

$x=-5$ is an extraneous solution.

We actually solved the negated equation $-\sqrt{2x+35}=x$ of which $x=-5$ is the solution.

The radical symbol represents only the positive square root, it is this restriction on the range that caused our $x=-5$ not to be a solution.

Hence, this is how we check for extraneous solutions.