Question

Prove that square root of 101 lies between 10 and 10.05

Answer 1

√(101) lies between 10 & 10.05

Step-by-step explanation:

√101 = √(100 + 1)

√(100 + 1) > √100

=> √(100 + 1) > √10²

=> √(101) > 10

√101 = √(100 + 1)

=> √(100 + 1) = √(10² + 2*10*0.5  + (0.05)² - (0.05)²)

=> √(100 + 1) = √((10 + 0.05)² - (0.05)²)

=> √(100 + 1) = √((10.05)² - (0.05)²)

=> √(100 + 1) < √(10.05)²

=> √(101) <  10.05

Hence

10 < √(101)  < 10.05

√(101) lies between 10 & 10.05

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