Let \sqrt{6}+\sqrt{6}+\sqrt{6}...∞ = x

So, x = \sqrt{6} + x

x^{2} = 6 + x

So, we have got a quadratic equation and we can solve it to get x = 3 and -2. We write x = -2 is not possible. So, x = 3. But, why is this so?Why is x = -2 is not possible?

The square root of any number can be both positive and negative, and x is in the form of a square root so the root can also be -2.So, why do we neglect x = -2 ?

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TopNewestYeah, we solve this by squaring, but remember that squaring an equation can yield other, unwanted solutions that have to be checked against the original problem to ensure that they make sense.

In this case, it's what Nishant said: the radical symbol denotes only the positive square root of an expression, by definition. So the negative root of the squared equation does not make sense in the original context.

Think of it as multiplying a given equation on both sides to eliminate a denominator of (x -1) and getting x = 1 as a root. Will that be a valid solution? – Raj Magesh · 3 years, 4 months ago

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Actually, -2 is an extraneous solution. – Satvik Golechha · 3 years ago

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– Prakkash Manohar · 3 years ago

But how do we confirm that -2 is an extraneous solution. By putting it in the original equation, we again get -2 and this continues.Log in to reply

I think the square root we usually talk about refers to the principal square root which is always positive. – Nishant Sharma · 3 years, 4 months ago

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I want to ask that why do we take x = √ 6 + x ?? – Ayush Porwal · 3 years ago

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I might be wrong.. but i think that you have been misinformed .. cause whenever I do these kind of questions there is always both the solutions in the options of MCQ..!! I have this exact same ques in a book and the given options are: A) 3,-2 B)3,2 C)-3,-2 D) None of the above and the correct option in the answers was 'A' itself! – Arpit Shukla · 3 years ago

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we can check it by putting,when we put -2,we get again 2 0r -2 as possible values of x,but 2 is not the solution of x. – Aditya Kumar · 3 years ago

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