Mathematics has its limits. Extraneous roots and some interesting fraction properties reveal this.
Solve the following for x:
Now use the Quadratic Formula to solve for x:
Quadratic Equations are written as such:
Determine a, b, & c:
Now plugging into the Quadratic Formula, we get the following:
Now, let's check this in our original equation. Start with x = 7:
It checks out!
Now, let's check this in our original equation. Start with x = -3:
It DOES NOT check out!
This is our extraneous root.
This is true, right?
If we convert to decimals, we get the following:
Regardless of which one you solve, you get the following:
However, we know that they are not exactly the same. This seems to be the case for certain others as well (e.g. 11*1/11=1). However, they are essentially the same. For some more fun, visit http://mathforum.org/library/drmath/view/52365.html.
The reason why this is not true and does not work is that a=b=1, then a-b=0. Yet, in the "solution," dividing by (a-b) is used, which is the same as dividing by 0, which is not allowed.
The problem with this one is that if a=b, then a-b=0. Yet, the "solution" breaks the rules by dividing by a-b. You cannot divide by 0!
What's the fallacy? In the last step to get to 2 = 1, you have to divide both sides by (a2 - ab).
Since a = b, a2
Here is the problem: miscalculation of square root. When you take the square root of the sides, you should calculate to the absolute value as follows: