Question

The approximate value of $\sqrt{6560}$ is

• A. 80.9939
• B. 80.9838
• C. 78.9939
• D. 78.9838

Hint:

When using linear approximation, choosing the closest known point (like the nearest perfect square for a square root) minimizes the error in the approximation. The smaller the $|\Delta x|$, the more accurate the result.

Answer 1

The Correct Option is A

Step 1: Identify perfect square closest to 6560 \( 80^2 = 6400 \). \( 81^2 = (80+1)^2 = 6400 + 160 + 1 = 6561 \). So, we can write \( 6560 = 6561 - 1 \).
Step 2: Use Linear Approximation Let \( f(x) = \sqrt{x} \). We want \( f(x + \Delta x) \approx f(x) + f'(x)\Delta x \). Take \( x = 6561 \) and \( \Delta x = -1 \). \( f(x) = \sqrt{6561} = 81 \). \( f'(x) = \frac{1}{2\sqrt{x}} = \frac{1}{2(81)} = \frac{1}{162} \).
Step 3: Calculate the value \[ \sqrt{6560} \approx 81 + \frac{1}{162}(-1) = 81 - \frac{1}{162} \] Calculate \( \frac{1}{162} \): \( \frac{1}{162} \approx 0.0061728 \). \( \sqrt{6560} \approx 81 - 0.0061728 = 80.993827... \) Rounding to four decimal places gives 80.9938. The closest option is 80.9939 (likely due to rounding differences in the question creation).