I guess , this question has one of the most debated solutions on Brilliant...

But still my concept on the very answer is not clear,

Will somebody please help me out...

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## Comments

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TopNewestThe symbol \(\sqrt x\) denotes the

prinicpal square rootof the non-negative number \(x.\) That means that it is thenon-negativesolution to \(s^2 = x.\)In this case, the non-negative solution to \(s^2 = 4\) is \(2.\)

In general, when people refer to "the square root," they

usuallymean the principal square root.On the other hand, some texts refer to "a square root" as both the positive and negative solutions to \(s^2=x.\) But there is no lack of clarity with \(\sqrt{x}\) or \(x^{\frac{1}{2}}\) -- these precisely denote the single, non-negative, principal square root of \(x.\)

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So, when it will be in the form of equation ,we will consider it's both roots( i.e. negative & positive one)

But when it will simply asks \(\sqrt{x}\) ,then it is just -- the non - negative one...?

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Correct. \(s^2=4\) has two solutions, \(2\) and \(-2\), while \(\sqrt{4} = 2\) and only 2.

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Consider the equation of \(x^2-y = 0\) where \(y\) is a real number. If you use perfect square differences, you will find \(x = \pm\sqrt{y}\). In this sense, it is absolutely correct to say that there are two roots to the equation, but if you notice it carefully, you will see the \(-1\) and \(1\) times in front of the value (it's similar to signum function in this regard). The -1 or 1 times in front of \(\sqrt{y}\) denotes whether that root is a positive or non-positive root.

Generally when writing out the \(\sqrt{y}\), it would refer to

non-negativeroot as it is the same way as \(1\times\sqrt{y}\). If the question explicitly ask for negative root, it should write as such or use \(-\sqrt{y}\) to denote, the same way as \(-1\times\sqrt{y}\). The question asking for \(\sqrt{y}\) then ask for thenon-negativevalue of such root ONLY, and NOT the negative one.Log in to reply