Square Roots in Equations
Date: 06/12/97 at 23:32:10 From: Carrie Carlson Subject: Algebra I'm supposed to solve this: 2(x-4)^2 - 3 = 13 Math is my worst subject. Here's what I think: 2(x-4) - 3 = sqrt(13) or 2(x-4) - 3 = sqrt-(13) 2x - 8 - 3 = sqrt(13) or 2x - 8 - 3 = sqrt-(13) 2x - 5 = sqrt(13) or 2x - 5 = sqrt-(13) 2x = 5 + sqrt(13) or 2x = 5 + sqrt-(13) x = 5 + sqrt(13) or x = 5 + sqrt-(13) ----------- ----------- 2 2 Did I do this right, or did I mess up at the beginning with the exponent? The exponent kind of throws me off because I'm not sure if I'm allowed to just take it out of there like that before I multiply the parentheses. Also, could you help me with this problem: 3x^2 - 16 = 0 I'm supposed to solve it like the last one, but I'm lost. Thanks. This is cool that you people would help me. Of course I try to do them first. Carrie
Date: 06/13/97 at 10:33:39 From: Doctor Wilkinson Subject: Re: Algebra I think you recognize that what you have done isn't right. This is good; knowing when you are wrong is the first step towards learning to get it right. So let's take another look at that equation: 2(x-4)^2 - 3 = 13 Every equation has certain difficulties that it puts in the way of finding the answer. The biggest difficulty that this equation gives us is that exponent. It's natural that you should want to get rid of the exponent as quickly as possible, but to do that you have to take the square root, and now there's a barrier in the way of just taking the square root, because whatever you do in an equation you have to do to ALL of both sides. If we take the square root as our first step, the result is: sqrt[2(x-4)^2 - 3] = sqrt(13) As you can see, this isn't very helpful because all that stuff that you're supposed to take the square root of on the lefthand side is pretty messy. So that means that we need to do a little more rearranging of the equation before we can take any square roots. Now you can easily take the square root of (x-4)^2 or even of 2(x-4)^2, but that's not what you have on the left side of the equation: you have that -3 term also. So before you can go taking square roots, you have to get rid of that -3. You can do that by adding 3 to both sides of the equation. This gives you: 2(x-4)^2 = 16 Next you probably want to get rid of the factor of 2 on the left side. That's easy since you can just divide both sides by 2. This gives: (x-4)^2 = 8 NOW you have something squared all by itself on the left side, and you can go ahead and take square roots. So you get: (x-4) = sqrt(8) OR (x-4) = -sqrt(8) This is because there are TWO numbers whose square is 8, one of them positive and the other negative (which you seem to already know about). This gives the two solutions for x: x = 4 + sqrt(8) and x = 4 - sqrt(8) With this, you should be able to solve the other problem that you gave. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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