Question

How do you solve and check for extraneous solution in \[\left| x+6 \right|=2x\] ?

Answer 1

Hint: Problems of this type can be solved by using the property of absolute value. This problem will be done in two parts, the first part will be solving the equation by taking $\left( x+6 \right)>0$ and in the second part we must take $\left( x+6 \right)<0$ and solve the equation. After getting the value of $x$ we check whether the value of $x$ satisfies the given equation to check for extraneous solutions.

Complete step by step answer:
The problem can be solved by using the property of absolute value. So, we need to take two cases into account and solve the equation separately in two parts.
If $\left( x+6 \right)>0$, we will have \[\left| x+6 \right|=x+6\]
Hence, the equation becomes:
 $x+6=2x$
$\Rightarrow 2x-x=6$
$\Rightarrow x=6$

Now, we take into account the second case, which is: \[\left( x+6 \right)<0\]
We will have:\[~~\left| x+6 \right|=-\left( x+6 \right)\]
The equation becomes:
$-\left( x+6 \right)=2x$
$\Rightarrow x+6=-2x$
$\Rightarrow 3x=-6$
$\Rightarrow x=-2$

So, now we have two solutions of $x$ which are $6$ and $-2$
To find the extraneous solution we will put both the solved values of $x$ in the given equation and check whether they satisfy the equation.
Let’s, take $x=6$,
$\left| 6+6 \right|=2\cdot 6$
Hence, the solution $x=6$ is not an extraneous solution.
Now, taking $x=-2$ ,
$\left| -2+6 \right|\ne 2\cdot \left( -2 \right)$
Hence, the solution $x=-2$ is an extraneous solution for the given equation.

Therefore, we can conclude that the given equation has an extraneous solution $x=-2$ and the other non-extraneous solution is $x=6$ .

Note: While solving problems of absolute solutions we must keep in mind that both the cases $\left( x+6 \right)>0$ and \[\left( x+6 \right)<0\] must be taken into account, otherwise we will not be getting the extraneous solutions. Also, while solving the equations we have to be extra careful about putting the positive $\left( + \right)$ and negative $\left( - \right)$ signs appropriately and at the end, verify that the solutions that we get satisfy the equation or not.