Solving Equations with Square Roots

Date: 03/28/2003 at 13:01:44
From: Nancy
Subject: Solving equations with square roots

When solving x+1 = sqrt(x+3), we can square both sides then solve the 
equation; however, we get an extra solution that doesn't work. Why 
does this happen?

Date: 03/28/2003 at 16:13:09
From: Doctor Rick
Subject: Re: Solving equations with square roots

Hi, Nancy.

  x + 1 = sqrt(x + 3)

Let's start by considering another equation:

  x = 2

We can square this equation. It's a silly thing to do since it's 
already solved for x, but we can do it. We get

  x^2 = 4

However, the new equation has two solutions. It is true if  x= 2 OR if 
x = -2. Both of these numbers squared equal 4. In squaring the 
equation, we added a spurious solution.

Any equation

  (something)^2 = (something else)^2

corresponds to two different equations:

  (something) = (something else)

or

  (something) = -(something else)

You started with the first of these, and in squaring both sides, you 
ended up with an equation that is equivalent to either that or the 
second equation. The second equation gives you the spurious root.

  (x + 1)^2 = x + 3

is equivalent to

  x + 1 = sqrt(x + 3)  OR  x + 1 = -sqrt(x + 3)

Often the benefits of squaring both sides outweigh the disadvantage 
of needing to check the results to eliminate the spurious root.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/