Question

How do you solve \[\dfrac{1}{6a}+\dfrac{1}{3{{a}^{2}}}=\dfrac{1}{3a}\] and find any extraneous solutions?

Answer 1

Hint: For the question \[\dfrac{1}{6a}+\dfrac{1}{3{{a}^{2}}}=\dfrac{1}{3a}\] it is asked to solve for the value of x and find if there are any extraneous solutions. For this question by multiplying the both sides with the variable a and do some arithmetical calculations using operations like multiplication and addition and thus find the solution. To find whether it has extraneous solutions or not we will substitute the answer we got in the same given question and check whether it is extraneous or not which is discussed and done below according to the given question.

Complete step by step solution:
So, firstly for the equation we will multiply with a on both sides of the equation. The equation will be reduced as follows.
\[\Rightarrow \dfrac{1a}{6a}+\dfrac{1a}{3{{a}^{2}}}=\dfrac{1a}{3a}\]
For this equation we will cancel the numerator and denominator which are the same then the equation will be reduced as follows.
\[\Rightarrow \dfrac{1}{6}+\dfrac{1}{3a}=\dfrac{1}{3}\]
Here we will do the basic arithmetic property of maths, that is we will shift the left hand side term to the right hand side of the equation then after doing it the equation will be reduced as follows.
\[\Rightarrow \dfrac{1}{3a}=\dfrac{1}{3}-\dfrac{1}{6}\]
Here for this equation we will add the numerator on the left hand side of the equation then it will be reduced as follows.
\[\Rightarrow \dfrac{1}{3a}=\dfrac{1}{6}\]
Here we will multiply the whole equation with 3 that is on both sides of the equation and get the required solution that is the value of a as follows.
\[\Rightarrow a=2\]
This is the only solution. All that is left to do is check whether or not it is extraneous (that is, it doesn’t actually work when we plug it back in due causing division by 0 or something of that nature). To do this, we have to just plug it back that is 2 for a into the original equation and check to make sure we get the right answer.

Here in this we will substitute 5 which we got as a solution in the question then the equation will be reduced as follows.
\[\Rightarrow \dfrac{1}{6a}+\dfrac{1}{3{{a}^{2}}}=\dfrac{1}{3a}\]
\[\Rightarrow \dfrac{1}{12}+\dfrac{1}{3\times 4}=\dfrac{1}{3\times 2}\]
\[\Rightarrow \dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{6}\]
\[\Rightarrow \dfrac{1}{6}=\dfrac{1}{6}\]
Here the solution works, so it is not extraneous.

Note: Students must be very careful in doing the calculations. To find whether the given question has an extraneous solution or not we have to substitute the value which we got in the same equation if it satisfies the given question is not extraneous if it doesn’t satisfy then it will be extraneous solutions. We should check if the function has an extraneous solution or not as an extraneous solution is the root of the equation which is transformed one which is not a root of the original given question because it is not there in the domain of the function.