Question

How do you solve \[\dfrac{{b + 6}}{{4{b^2}}} + \dfrac{3}{{2{b^2}}} = \dfrac{{b +
4}}{{2{b^2}}}\] and check for extraneous solutions?

Answer 1

Hint:In this equation we need to get the value of \[b\] and check for extraneous solution in which, substitute the value of \[b\]in the given equation and see that the solutions both LHS and RHS are the same. This implies the value of \[b\] is an extraneous solution.

Complete step by step answer:
Let us write the given equation as
\[\dfrac{{b + 6}}{{4{b^2}}} + \dfrac{3}{{2{b^2}}} = \dfrac{{b + 4}}{{2{b^2}}}\]
Here we need to find the value of \[b\]for this, multiply both the sides of the equation by \[4{b^2}\].
Hence the given equation becomes:
\[4{b^2}\dfrac{{b + 6}}{{4{b^2}}} + 4{b^2}\dfrac{3}{{2{b^2}}} = 4{b^2}\dfrac{{b + 4}}{{2{b^2}}}\]
Simplify all the terms with respect to its numerator and denominator terms we get
\[b + 6 + 6 = 2b + 8\]
\[b + 12 = 2b + 8\]
As the both the terms of \[b\] are common let us shift to LHS and others values to RHS as shown
\[b - 2b = 8 - 12\]
\[ - b = - 4\]
Therefore, the value of \[b\] we got is:
\[b = 4\]
Now let us check for an extraneous solution, for this just substitute the value of \[b\] in the given equation.
As the given equation is
\[\dfrac{{b + 6}}{{4{b^2}}} + \dfrac{3}{{2{b^2}}} = \dfrac{{b + 4}}{{2{b^2}}}\]
Substitute the value of \[b\]i.e., \[b = 4\] in the given equation
\[\dfrac{{4 + 6}}{{4\left( {{4^2}} \right)}} + \dfrac{3}{{2\left( {{4^2}} \right)}} = \dfrac{{4 +
4}}{{2\left( {{4^2}} \right)}}\]
Simplifying the terms, we get
\[\dfrac{{10}}{{64}} + \dfrac{3}{{32}} = \dfrac{8}{{32}}\]
\[\dfrac{3}{{32}} + \dfrac{3}{{32}} = \dfrac{8}{{32}}\]
\[\dfrac{8}{{32}} = \dfrac{8}{{32}}\]
Hence after simplification of the given equation, we got the values as LHS=RHS, which implies that 4 is an extraneous solution for the given equation.

Additional information:
An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation.

Note: The key point to solve the given equation is we need to find the value of b and check for extraneous solutions by substituting the value of b in the given equation. And to find whether your solutions are extraneous or not, you need to plug each of them back into your given equation and see if they work.