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KEAM Approximations — practice questions

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Find the approximate value of $\sqrt{25.5}$ using differentials (given $\sqrt{25} = 5$):Read the answer and worked solution →

Worked practice questions

Find the approximate value of $\sqrt{25.5}$ using differentials (given $\sqrt{25} = 5$):

  1. 5.05
  2. 5.5
  3. 5.0498 (more precise)
  4. 5.1
Show answer and solution

Correct answer: A. 5.05

Let $f(x) = \sqrt x$. $f'(x) = 1/(2\sqrt x)$. $\Delta f \approx f'(25)\cdot 0.5 = (1/10)(0.5) = 0.05$. **So $\sqrt{25.5} \approx 5 + 0.05 = 5.05$.**

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