Solveeit Logo
Login

Question

Question: How do you solve \[\dfrac{{b + 6}}{{4{b^2}}} + \dfrac{3}{{2{b^2}}} = \dfrac{{b + 4}}{{2{b^2}}}\] a...

How do you solve b+64b2+32b2=b+42b2\dfrac{{b + 6}}{{4{b^2}}} + \dfrac{3}{{2{b^2}}} = \dfrac{{b + 4}}{{2{b^2}}} and check for extraneous solutions?

Explanation

Solution

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

Complete step by step answer:
Let us write the given equation as
b+64b2+32b2=b+42b2\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 bbfor this, multiply both the sides of the equation by 4b24{b^2}.
Hence the given equation becomes:
4b2b+64b2+4b232b2=4b2b+42b24{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+8b + 6 + 6 = 2b + 8
b+12=2b+8b + 12 = 2b + 8
As the both the terms of bb are common let us shift to LHS and others values to RHS as shown
b−2b=8−12b - 2b = 8 - 12
−b=−4- b = - 4
Therefore, the value of bb we got is:
b=4b = 4
Now let us check for an extraneous solution, for this just substitute the value of bb in the given equation.
As the given equation is
b+64b2+32b2=b+42b2\dfrac{{b + 6}}{{4{b^2}}} + \dfrac{3}{{2{b^2}}} = \dfrac{{b + 4}}{{2{b^2}}}
Substitute the value of bbi.e., b=4b = 4 in the given equation

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.