The approximate value of square root of 25.2 is
Let f (x) = √xNow, f(x+δ x)−f(x)=f′(x).δ x=δx2√xWe may write, 25.2 = 25 + 0.2Taking x = 25 and δx=0.2 We havef(25.2)−f(25)=0.22√25=0.02∴f(25.2)=f(25)+0.02=√25+0.02=5.02⇒√(25.2)=5.02
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