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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...

Note by Archiet Dev
10 months, 2 weeks ago

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The symbol \(\sqrt x\) denotes the prinicpal square root of the non-negative number \(x.\) That means that it is the non-negative solution 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 usually mean 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 \(x.\) Eli Ross Staff · 10 months, 2 weeks ago

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@Eli Ross 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...? Archiet Dev · 10 months, 2 weeks ago

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@Archiet Dev Correct. \(s^2=4\) has two solutions, \(2\) and \(-2\), while \(\sqrt{4} = 2\) and only 2. Eli Ross Staff · 10 months, 2 weeks ago

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@Eli Ross Thank you very much ,Sir Archiet Dev · 10 months, 2 weeks ago

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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-negative root 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 the non-negative value of such root ONLY, and NOT the negative one. Kay Xspre · 10 months, 2 weeks ago

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