Approximation Math Example 1

Solution

1. 1
   Identify perfect squares on either side: $7^2 = 49$ and $8^2 = 64$. So $7 < \sqrt{50} < 8$.
2. 2
   $\sqrt{50}$ is very close to $\sqrt{49} = 7$. The gap between $49$ and $64$ is $15$; $50$ is $1$ above $49$, so $\sqrt{50} \approx 7 + \dfrac{1}{15} \approx 7.07$.
3. 3
   To one decimal place: $\sqrt{50} \approx 7.1$.