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help in numper prop. important yet confusing square root

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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help in numper prop. important yet confusing square root

by jaymin » Sun Jan 25, 2009 10:04 pm
Hi, I need help in understanding the following two:

1.Does the square root of a number have 2 values or 1 value. If a DS question is asked as follows:
What is the value of 3x+2?
a. square root 4 = x.

Then, is the first statement sufficient or is x= +,- 2.? I interpret it this way- root 4=x => 4=x square => x= +,-2. I read the og11 math review definition but could not get a clear understanding, since it mentions that root 4=x has 2 values viz. + and - 2, but the positive root of 4 has only one value +2. So, with this definition, will statement (1) in above Question be suffcient?

2. Also, Is 0 considered a perfect square? OR does the perfect square series begin with 1,2,4,9....

Thanx
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by DanaJ » Mon Jan 26, 2009 6:26 am
1. The square root of a number can only be positive. This means that sqrt(9) = 3, sqrt(25) = 5, sqrt(196) = 14 and so on. But watch out, there are certain equations that can only be correctly answered if you consider both the positive and the negative number. For instance, let's say you have the following equation:

If sqrt(x^2) = 3, find x.

This equation is equivalent to x^2 = 3^2 (I just raised the first equation to the second power) or x^2 = 9. Now, there are two possible solutions, since x^2 = 9 is the same as x^2 - 9 = 0 or (x - 3)(x +3) = 0 [don't forget about the formula (x - y)(x + y) = x^2 - y^2].

Conclusion: sqrt(x) will always be a positive value, but some equations will provide both the positive and the negative values as solutions.

2. Yes! 0 = 0^2. The perfect square series for 0 - 20:
0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400.
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Confusing

by Vemuri » Mon Jan 26, 2009 9:09 am
DanaJ wrote:1. The square root of a number can only be positive. This means that sqrt(9) = 3, sqrt(25) = 5, sqrt(196) = 14 and so on. But watch out, there are certain equations that can only be correctly answered if you consider both the positive and the negative number. For instance, let's say you have the following equation:

If sqrt(x^2) = 3, find x.

This equation is equivalent to x^2 = 3^2 (I just raised the first equation to the second power) or x^2 = 9. Now, there are two possible solutions, since x^2 = 9 is the same as x^2 - 9 = 0 or (x - 3)(x +3) = 0 [don't forget about the formula (x - y)(x + y) = x^2 - y^2].

Conclusion: sqrt(x) will always be a positive value, but some equations will provide both the positive and the negative values as solutions.

2. Yes! 0 = 0^2. The perfect square series for 0 - 20:
0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400.
Hi DanaJ,

Your response is very confusing, esp. in the context of DS. As per the original question, statement 1 very clearly says sqrt of 4=x. Based on your response (1st para), the statement should be sufficient as the sqrt of 4 is a positive 2 only. Why do we have to move the sqrt to the right side of the equality (which contradicts the statement). I believe statements are to be taken as facts & should not be modified.

Please correct my understanding, if wrong.

Cheers
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by DanaJ » Mon Jan 26, 2009 11:16 pm
My example was just an illustration of an equation that requires both the negative and the positive values of the square root. Indeed, the original equation is sufficient to answer the DS question.
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by r_walid » Wed Jan 28, 2009 2:27 pm
the point is the the sqrt root of 4 = 2 or -2.
because 2*2 = 4 and -2*-2 = 4
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by DanaJ » Wed Jan 28, 2009 10:22 pm
I'm sorry, r_walid, but I have to disagree. We must make a distinction between the square root of a number and the numbers that, raised to the second power, will equal that number. In other words:

1. The square root of a number will always be positive (well, or 0). ALWAYS!

2. If we are looking for the numbers that raised to the second power equal a given number, then indeed we also have a negative value. For instance:

If x^2 = 4, what is x? You have to ways of solving this:

a. you notice that 4 = 2^2, which means that sqrt(4) = 2, so you figure out that x is either 2 OR the negative of the sqrt(4), which is -2.
b. you use our trusty formula (x - y)(x + y) = x^2 - y^2. Since x^2 = 4, this means that x^2 - 4 = 0 or that (x - 2)(x + 2) = 0. This equation has two solutions, which will again be 2 or -2.


Hope this helps...
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by r_walid » Thu Jan 29, 2009 10:22 am
Sorry, i was tying fast.
I agree Dana. sqrt of x is always positive. But x^2 =4 means that x is either 2 or -2.
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