Solving for b
Date: 05/10/2000 at 18:14:43 From: Rebecca Armstrong Subject: Solving for "b" in square root problem Help (please)! My son is hung up on a problem and no matter how we try to solve it, we're stuck. Here's the problem: 10 + 2 * sqrt(b) = 0 First we subtracted 10 from both sides -10 + 10 +2 * sqrt(b) = 0 -10 2 * sqrt(b) = -10 then we squared both sides of the equation 2^2 * sqrt(b)^2 = -10^2 4b = 100 divided by 4 on both sides b = 25 but when we insert 25 into the formula, we end up with 10 + 2*sqrt(25) = 0 10 + 2*5 = 0 20 = 0 FALSE Thanks for your help. Signed, "Stumped and too old for algebra..."
Date: 05/11/2000 at 07:39:57 From: Doctor Floor Subject: Re: Solving for "b" in square root problem Hi, Rebecca, thanks for writing. You have done everything right. When you take squares in solving an equation, you have to check whether your results really fit. You concluded that your solution didn't fit. That means there are no solutions to the given equation. Indeed the equation 10 + 2*sqrt(b) = 0 should not have solutions. A square root is always zero or positive, so 10 + 2*sqrt(b) must be greater than zero. You did it right. Why do you have to check after squaring? Let's give a simple example. Suppose you have the "equation" t = 10. Of course you know that this is in fact the solution. But if we square both sides, we get t^2 = 100, and this equation has two solutions, t = 10 and t = -10. So by squaring we seem to have added a "solution." We can only rule this solution out by checking. If you have more questions, just write back. Best regards, -Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum