Question

How can we found value of root 1

Answer 1

Let,

$x = \sqrt{1} \\ {x}^{2} = 1 \\ {x}^{2} - 1 = 0 \\ (x + 1)(x - 1) = 0 \\ x + 1 = 0 \: or \: x - 1 = 0 \\ x = - 1 \: or \: x = 1$

Thus, value of root 1 is 1,-1.

Answer 2

Answer:

±1

Step-by-step Explanation :

Let 1 be 'x'.

$x = \sqrt{1} \\ {x}^{2} = 1 \\ {x}^{2} - 1 = 0 \\ (x + 1)(x - 1) = 0 \\ \\ so \: (x + 1) = 0 \: or \: \: (x - 1) = 0 \\ \\ if \: x + 1 = 0 \\ \: \: \: \: \: x = - 1 \\ \\ if \: \: x - 1 = 0 \\ \: \: \: \: \: \: x = 1 \\ \\the \: value \: of \: x \: is \: 1 \: and \: - 1$

So, it can be said that the answer is : ‌‍±1

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