Problem 12 - Analysis 1: Ein Lehr- und Arbeitsbuch für Studienanfänger

Question

Berechnen Sie \(\sqrt{38}\) auf drei Stellen genau.

Step-by-Step Solution

Answer

\(\sqrt{38} \approx 6.164\) (to three decimal places).

Step 1: Identify the Range

First, let's determine the two whole numbers between which the square root of 38 lies. We know that \(6^2 = 36\) and \(7^2 = 49\). Since 38 is between 36 and 49, \(\sqrt{38}\) must be between 6 and 7.

Step 2: Use Estimation

To find a more precise value, we'll start with an estimate. Testing with 6.1 and 6.2:- \(6.1^2 = 37.21\), which is less than 38.- \(6.2^2 = 38.44\), which is more than 38.So, \(\sqrt{38}\) is between 6.1 and 6.2.

Step 3: Narrow the Estimation

Continue by checking between 6.15 and 6.16:- \(6.15^2 = 37.8225\), which is less than 38.- \(6.16^2 = 37.9856\), which is less than 38.Thus, \(\sqrt{38}\) is between 6.16 and 6.2.

Step 4: Refine the Estimate

Now, try between 6.164 and 6.165:- \(6.164^2 = 37.976896\), which is less than 38.- \(6.165^2 = 37.989225\), which is still less than 38.So, test a slightly higher bound.

Step 5: Final Estimation

Test 6.17:- \(6.17^2 = 38.0489\), which is more than 38.Thus, \(\sqrt{38}\) is closer to 6.164 or 6.165.

Step 6: Selecting Final Value

Since 6.165 was closer to producing a result under 38, let's conclude that to three decimals, \(\sqrt{38} \approx 6.164\).

Key Concepts

Numerical Estimation

Numerical estimation is a critical skill in mathematics that helps you find approximate values for solutions that cannot be easily calculated mentally. It's like becoming a detective, making educated guesses and refining them until you're as close as possible to the true value. In the context of square roots, numerical estimation becomes very handy. When faced with a number like 38, which doesn't have a perfect square root, estimation allows us to find an approximate value.

• Step-by-step refining: Start with broad approximations and incrementally narrow down the range.
• Precision: Decide how many decimal points you need to approximate.

When estimating, we typically look at the squares of whole numbers nearest to the target number. For 38:

• We know that 6 squared is 36 and 7 squared is 49.
• This tells us that the square root of 38 is somewhere between 6 and 7.

Mathematical Problem Solving

Mathematical problem solving involves a methodical approach to finding solutions, often using estimation, calculation, and reasoning. The process involves dividing problems into manageable steps and addressing each thoughtfully. When calculating square roots without a calculator, problem-solving skills are crucial. Here's a glance at how these skills come into play:

• Range identification: First, identify the broad range using known squares, like determining that \(6^2 = 36\) and \(7^2 = 49\) to figure the range for \(\sqrt{38}\).
• Iterative testing: Select numbers between the identified range to narrow down the estimate.
• Precision checks: With each step, calculations become more refined, like testing numbers such as 6.1 and 6.2.

These steps ensure that you continuously hone in on the more precise value, enhancing both understanding and skill in mathematical analysis.

Step by Step Solution

Utilizing a step-by-step solution is an excellent way to grasp complex calculations. It makes the process of finding square roots manageable, even for those who find math challenging. Here's how this works in the context of calculating \(\sqrt{38}\):

• Identify the starting range: Here, we noticed that \(38\) is between \(6^2\) and \(7^2\).
• Make initial estimations: Start with values between 6 and 7, such as 6.1 and 6.2. Testing each, you find which one's square is closest to 38.
• Refine iteratively: With each test, narrow the range further. If \(6.164^2\) is less than 38, and \(6.17^2\) is more, you know that \(\sqrt{38}\) is between these values.

Finally, by refining repeatedly, you can confidently approximate \(\sqrt{38} \approx 6.164\). A step-by-step solution allows anyone to follow along, comprehend, and become adept at similar calculations in the future.