4. Solve for x and check.
This is a rational equation so we run through the steps for solving rational equations. First, clear denominators. Multiply both sides by a common denominator. In this case the smallest common denominator is x(x - 2).
Since the left side has two terms, we need to multiply them both.
Assemble the surviving factors
3x +3x2 - 6x = 2x - 4
3x2 - 3x = 2x - 4
When we combine like terms we see that we have a quadratic. Transpose all the terms to the left leaving a 0 on the right.
3x2 - 3x - 2x + 4 = 0
3x2 - 5x + 4 = 0
Now that we have a zero on one side we see if the left side factors. It doesn't, so let's use the quadratic formula.
and we find a negative number under the rtadical. This gives us an imaginary answer. A negative inside the radical is the same as an i outside the radical, where i2 = -1.
Check: Even though we have a complex solution, we can still check it. Indeed, these checks provide us with a great deal of fun.
One can prove that if the soution wth the positive radical will check, then so will the solution with the negative radical, so we need only check the solution with the positive radical.
Copy down the original equation
except where we see an x, copy down the solution in parentheses.
Notice that this is just copying, and copying is easy. It gives us a very interesting simiplification problem. The first term on the left is a compound fraction, so we follow the procedures for compound fractions. Find common denminators for the fractions on the botom, and reduce the fraction in the bottom on the right.
We are now ready to invert and multiply.
It is now time to rationalize the deniminator. To clear radicals, mult iply the binomial in the bottom by its conjugate, which is what we call itwhen we change the sign in the middle of the binomial.
(a +ib)(a - ib) = a2 - i2b2 = a2 + b2
since i2 = -1. So our check becomes
Now we need to find common denominators on the left.
and, amazingly, since -7 + 12 = 5, it checks. The check for the solution which has the negative sign on the radical, is exactly the same except that at each step, the signs on the radicals will be different.