help in numper prop. important yet confusing square root

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.

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

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.

the point is the the sqrt root of 4 = 2 or -2.
because 2*2 = 4 and -2*-2 = 4

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...

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.