Expert: Bobby Soltani - 8/15/2009

Question
Hi Bobby,

I recently came acros the following type of equation:

(x - 2)/(x - 1) = (x^2 + 6x + 4)/(x - 1)  .........(1)

We have to find the roots of this equation.

Now both the L.H.S. and R.H.S. have been divided by the same factor (x - 1).
One option is to cancel this factor from both sides. Then the roots we would get would be x= -2 or x= -3.

But another way is to cross multiply and then take all the terms on one side after which we get the following equation:

(x-1)(x+3)(x+2) = 0      .........(2)

So from here x can be 1, -3 or -2

But my book says that for x=1, the equation (1) does not make sense because then x-1 would become 0 and the denominators of both the l.h.s and r.h.s. would be 0 which wouldn't make sense, and so x=1 can not be a root of the equation.

My question is that though I do agree with the book that for x=1 eq. (1) doesn't make sense but we can also express the eq(1) as eq(2) which is written above. And then eq. (2) makes perfect sense for x=1.

Is there a difference between eq(1) and eq(2) ???
Why can't we express eq(1) as eq(2) like I did above.....??

Thanks,
Shikhin

Answer
Hi Shikhin,

When you multiply both sides of an equation by an expression which
includes the variable, you run the risk of introducing extraneous
roots.  Let me give a very simple example.  Suppose you have the
equation

  x = 3

for which the solution is obviously 3.  Now multiply both sides of
this equation by x which gives

  x^2 = 3x

One of the roots of this equation is still 3, but there is also
another root, namely zero, which has been introduced.  {0,3} are roots
of the second equation, but only 3 is a root of the first. Often, in
the process of solving radical equations, we square both sides -
the equivalent of multiplying by an expression which includes the
variable. We find roots to this new equation, but sometimes not all
roots of the new equation are also roots of the original equation.
I hope that this helps.