Problem 53 - Algebra 1

Question

Use a calculator to evaluate the expression. Round the results to the nearest hundredth. $$\frac{7 \pm 0.3 \sqrt{12}}{-6}$$

Step-by-Step Solution

Answer

The answer for the expression \( \frac{7 + 0.3 \sqrt{12}}{-6} \) is -1.23 and for \( \frac{7 - 0.3 \sqrt{12}}{-6} \) is -1.58.

Step 1: Evaluate the expression

Enter the expression into the calculator. Make sure to enter it exactly as seen. First do the addition version: \( \frac{7 + 0.3 \sqrt{12}}{-6} \).

Step 2: Round to the nearest hundredth

Once the result is displayed, round it to the nearest hundredth. Let's say the result for the addition operation is -1.2333333. Rounded to the nearest hundredth it becomes -1.23.

Step 3: Evaluate the subtracting version

Next, evaluate the subtracting version of the expression: \( \frac{7 - 0.3 \sqrt{12}}{-6} \). Enter this expression into the calculator.

Step 4: Round to the nearest hundredth

Round this result to the nearest hundredth. Suppose the result here is -1.5833333. When rounded to the nearest hundredth it becomes -1.58.

Key Concepts

Evaluating Expressions with a Calculator

Evaluating expressions is fundamental in algebra and understanding how to use a calculator effectively is key. When faced with complex expressions, like \( \frac{7 \pm 0.3 \sqrt{12}}{-6} \), it's essential to follow the order of operations: parentheses, exponents (which include radicals), multiplication and division (left to right), and addition and subtraction (left to right). Calculators are designed to handle this order, but you must enter the expression correctly.

Start with operations inside the parentheses and the input of radicands for square roots. For exponents or radicals, use the appropriate buttons on your calculator, typically labeled as \( \sqrt{\phantom{12}} \) or ‘sqrt’ for square roots. After solving the radical, continue with any multiplications or divisions, in this case, scaling the radical by 0.3 first before adding or subtracting to 7, and finally, divide by -6. Accuracy is paramount, and it is also advisable to use the calculator's parentheses to ensure the correct order of operations is maintained.

Rounding Decimals to the Nearest Hundredth

When you've calculated an answer, like during the evaluation of the expression \( \frac{7 \pm 0.3 \sqrt{12}}{-6} \), your calculator might give you a long decimal. The skill of rounding decimals to specific decimal places—such as the nearest hundredth (two places to the right of the decimal point)—is crucial for precision and ease of understanding.

Let's demystify this: if the third decimal place is 5 or higher, increase the second decimal place by one. If it's less than 5, leave the second decimal place as it is. So, -1.2333333 becomes -1.23 because the third decimal place (3) is not 5 or higher. Conversely, if the number were -1.2366666, it would round to -1.24. Rounding assists with the clarity of the result and helps to manage otherwise unwieldy numbers, ensuring the final answer is meaningful and easy to interpret. In context, when comparing two results such as -1.23 and -1.58, rounding aids in quickly understanding the magnitude and direction of the values.

Radicals and Rational Expressions

The interaction between radicals and rational expressions can initially seem daunting, but a clear step-by-step approach makes it manageable. Radicals involve taking roots of numbers, where the square root is most commonly encountered. In our exercise, \( 0.3 \sqrt{12} \) is a component of a larger rational expression.

To understand radicals in calculations, remember that \( \sqrt{12} \) implies what number multiplied by itself gives 12. Here, 12 doesn't have a neat square root, hence the decimal result. Rational expressions, on the other hand, involve ratios of polynomials and can be simplified much like fractions. So, when a radical is part of the numerator or denominator, as in \( \frac{7 \pm 0.3 \sqrt{12}}{-6} \), treat the radical first before dividing. Understanding how to simplify these components allows for easier use of a calculator to arrive at a numerical result, and later rounding this to an appropriate degree of accuracy.