FIND THE SQUARE ROOT OF 11.666667 or 35/3 CORRECT UPTO TWO PLACES OF DECIMAL ..
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We know that 1^2 = 1 and 2^2 =4. This implies that the square root x of 3 should be such that 1 < x < 2. So, let 3^0.5 = ( 2 - h) where 0 < h < 1. => (2 - h)^2 = 3 => 4 -4h +h^2 = 3 => -4h < 3 - 4 ignoring h^2.
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