
How do I check for extraneous solutions?
Answer
547.8k+ views
Hint: In the given question, we have been asked how can we check for an extraneous solution. To attempt this question, we must first know what an extraneous solution is. An extraneous solution to an equation is a solution which is obtained by solving the equation using the appropriate method, but the solution is not fit for the equation. It is not that the method is wrong, but it just is not acceptable in the given range.
Complete step-by-step answer:
In some questions, when we solve them, we end up with an extraneous solution – the solutions which when calculated through the proper methodology gives an answer which is not fit for the range of the given question, for example, the answer yields a negative value inside a square root, or we get an indeterminate form.
We check for these by simply putting in the calculated values and check if they break any fundamental rule.
For instance, consider the linear equation
\[\sqrt {x + 4} = x - 2\]
Let us solve this,
\[\sqrt {x + 4} = x - 2\]
Squaring both sides and cancelling out the common terms,
$\Rightarrow$ \[x + {4} = {x^2} + {4} - 4x\]
Taking all the variables to one side,
$\Rightarrow$ \[{x^2} - 5x = 0\]
Taking \[x\] common,
$\Rightarrow$ \[x\left( {x - 5} \right) = 0\]
Hence, \[x = 0,5\]
Now, putting the calculated value of \[x\] into the original equation,
$\Rightarrow$ \[\sqrt {x + 4} = x - 2\]
Checking for the extraneous solutions by putting the calculated values,
$\Rightarrow$ \[\sqrt {5 + 4} = 5 - 2\]
$\Rightarrow$ \[\sqrt 9 = 3\]
or \[3 = 3\]
Hence, \[5\] is a verified solution.
Now \[\sqrt {0 + 4} = 0 - 2\]
\[\sqrt 4 = - 2\]
or \[2 = - 2\]
But, \[2 \ne - 2\]
Hence, \[0\] is an extraneous solution.
Note: In the given question we had to check how to solve for the extraneous solutions. These are the solutions which (even though) we get from correctly solving the equation, are still wrong, because they are not fit for the equation in the given range. To check for that, we just put in the calculated answer into the equation and see if it is violating any fundamental rule, for example, the answer yields a negative value inside a square root, or we get an indeterminate form.
Complete step-by-step answer:
In some questions, when we solve them, we end up with an extraneous solution – the solutions which when calculated through the proper methodology gives an answer which is not fit for the range of the given question, for example, the answer yields a negative value inside a square root, or we get an indeterminate form.
We check for these by simply putting in the calculated values and check if they break any fundamental rule.
For instance, consider the linear equation
\[\sqrt {x + 4} = x - 2\]
Let us solve this,
\[\sqrt {x + 4} = x - 2\]
Squaring both sides and cancelling out the common terms,
$\Rightarrow$ \[x + {4} = {x^2} + {4} - 4x\]
Taking all the variables to one side,
$\Rightarrow$ \[{x^2} - 5x = 0\]
Taking \[x\] common,
$\Rightarrow$ \[x\left( {x - 5} \right) = 0\]
Hence, \[x = 0,5\]
Now, putting the calculated value of \[x\] into the original equation,
$\Rightarrow$ \[\sqrt {x + 4} = x - 2\]
Checking for the extraneous solutions by putting the calculated values,
$\Rightarrow$ \[\sqrt {5 + 4} = 5 - 2\]
$\Rightarrow$ \[\sqrt 9 = 3\]
or \[3 = 3\]
Hence, \[5\] is a verified solution.
Now \[\sqrt {0 + 4} = 0 - 2\]
\[\sqrt 4 = - 2\]
or \[2 = - 2\]
But, \[2 \ne - 2\]
Hence, \[0\] is an extraneous solution.
Note: In the given question we had to check how to solve for the extraneous solutions. These are the solutions which (even though) we get from correctly solving the equation, are still wrong, because they are not fit for the equation in the given range. To check for that, we just put in the calculated answer into the equation and see if it is violating any fundamental rule, for example, the answer yields a negative value inside a square root, or we get an indeterminate form.
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