Solving Equations with Square RootsDate: 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/ |
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