7. Solve for x and check
First clear denominators. The smallest common denominator for all of the denominators is x(x + a). Multiply both sides by x(x + a).
Recall that the way you multiply a side by something like x(x + a) is by putting parentheses around the side and an x(x + a) next to the parentheses. The reason for the parentheses is that we had the sum of the two fraction on the left together before we thought about multiplying by x(x + a). However, the parentheses tell us how to proceed. Multiply each term on the left by x(x + a).
This cleans out all of our denominators. A very common mistake with these types of problems is that students forget to multiply n the right as well. Remove the parentheses around the surviving factors.
Combine like terms
After we simplify, we see that we have a quadratic equation. The first step in solving a quadratic equation is to transpose all terms to the same side leaving a 0 on the other. Since the square term is positive on the right, transpose all terms to the right.
This will not factor except for very special values of a, so we must use the quadratic formula. First, we must combine the x terms. That is accomplished by factoring an x out of the second and third terms.
Now we can use the quadratic formula
This can be simplified. First, square the binomial under the radical
When we simplify the -4(1)(-a), two of the terms under the radical cancel. We can also remove the parentheses outside the radical.
This gives us a final answer of
To check this result, we copy down the original equation
except that wherever you see the unknown, x, copy down the solution in parentheses. There are two solutions here, one where the radical is positive and one where it is negative. It can be shown that if the solution where the radical is positive will check, then the one with the negative radical will also check. So we check the solution with the positive radical.
this gives us two compound fractions. The one on the left is ready to invert and multiply, but the one on the right isn't. We find common denominators for the two terms in the bottom.
Now that we have common denominators, we can add the tops in the left compound fraction. There are two like terms that will combine on the top of the bottom.
Now we are ready to invert and multiply.
We need to rationalize both denominators.
Multiplying a sum times a difference gives us a difference of squares.
When we remove the parentheses on the bottom, most of the terms cancel.
We need only multiply the top and bottom of the first fraction by -1 to get common denominators.
Now that we have common denominators, we can add the tops. This means remove parentheses and combine like terms.
When we combine the surviving terms on the tops, we get
so long as a is not 0.
If a = 0, then the story is a little different. The original equation would have been
in which case
is the only solution.
It checks quite easily
If we use the formula for the solution
when a = 0, we get
which gives us two answers
The solution x = 2 is the only valid solution. The x = 0 will not check. We would get a zero denominator if we tried to substitute it into the equation for x.
The reason we get an extraneouos solution in this case is that if a = 0, then when we cleared denominatros in our solution procedure, we multiplied by two factors of x when in this case only one factor would have sufficed. This illustrates that if you multiply by a common denominator which is not the one of smallest degree, you always introduce extraneous roots. This shows up when we try to check our result.