Problem 36 - College Algebra and Calculus: An Applied Approach

Question

Solve the quadratic equation by extracting square roots. List both the exact answer and a decimal answer that has been rounded to two decimal places. $$ 15 x^{2}=620 $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = ±\sqrt{41.3333} = ±6.43\) (to 2 decimal places).

Step 1: Divide by the Coefficient of x^2

Divide both sides of the equation by 15 to isolate \(x^{2}\) on one side of the equation: \[x^{2} = \frac{620}{15}\]

Step 2: Simplify the Right Side

Calculate the division on the right side of the equation: \[x^{2} = 41.3333\]

Step 3: Extract the Square Root

Take the square root of both sides: \[x = ±\sqrt{41.3333}\]

Step 4: Calculate the Exact and Approximate Values

Calculate the exact and approximate values of \(x\). Keep in mind that there are two solutions, a positive and a negative one. The positive value is \[\sqrt{41.3333}\] and the negative value is \[-\sqrt{41.3333}\] When calculating the square root of 41.3333, it results in \(x = 6.43\) or \(x = -6.43\) (rounded to two decimal places) as positive and negative solutions respectively.

Key Concepts

Extracting Square Roots

Extracting square roots is like reversing the squaring process. Suppose you have an equation where a number is squared, like \(x^2 = a\). To find the value of \(x\), you need to extract, or take, the square root of \(a\). This is because squaring a number and taking the square root are inverse operations. For example, if \(x^2 = 41.3333\), extracting the square root means finding \(x = ±\sqrt{41.3333}\). Remember that extracting the square roots gives both a positive and a negative solution. This is because both positive and negative numbers become positive when squared. Therefore, the solutions include both \(6.43\) and \(-6.43\).

Exact Solutions

An exact solution represents the most precise answer you can provide without compromising on any values. Typically, this means leaving the solution in terms of a mathematical expression rather than a rounded decimal. In the quadratic equation we have,\[x = ±\sqrt{41.3333}\]Here, \(\sqrt{41.3333}\) is considered an exact solution. It is untouched by approximation or rounding which retains accuracy for further mathematical operations or proofs. Keeping solutions in this form can be particularly useful in more advanced mathematics where precision is key.

Approximate Solutions

An approximate solution is derived when exact values are converted into a more manageable form, often involving rounding. This allows users to work with numbers that are easier to understand and use in everyday contexts.In our scenario \[x = ±6.43\]These are approximate solutions to the original square root value of \(41.3333\). Approximation is beneficial when exact values are unnecessarily complex for practical applications. However, it is important to recognize the decreased precision that comes with approximation. This is especially relevant in fields requiring stringent accuracy.

Decimal Rounding

Decimal rounding makes numbers simpler by reducing the number of decimal places. This process is particularly useful when a high degree of precision isn't required. For instance, when you find the square root of a number like \(41.3333\), you might initially get a long decimal.To communicate it easily, you round it to two decimal places, resulting in\[x \approx 6.43\].The process of rounding involves looking at the number in the third decimal place. If it's 5 or higher, round up. If it's less than 5, leave the second decimal place as is. By rounding, you get a number that's easier to work with, though some precision will be lost.