Problem 66 - Elementary Algebra

Question

For the following problems, round each result to the nearest hundredth. \(0.048 x^{2}=2.01\)

Step-by-Step Solution

Answer

Answer: The two values for \(x\) are approximately 6.47 and -6.47.

Step 1: Solve for \(x^2\)#

Divide both the left and right sides of the equation by \(0.048\) to isolate the \(x^2\) term: \(x^2 = \dfrac{2.01}{0.048}\) Now, calculate the value of the fraction: \(x^2 ≈ \dfrac{2.01}{0.048} ≈ 41.875\)

Step 2: Solve for \(x\)#

To find the value of \(x\), we need to take the square root of \(41.875\): \(x = \pm\sqrt{41.875}\) Now, calculate the square root of \(41.875\): \(x ≈ \pm\sqrt{41.875} ≈ \pm 6.47\)

Step 3: Round to the nearest hundredth#

We have found two values for \(x\), \(6.47\) and \(-6.47\). Since the question asks to round the result to the nearest hundredth, neither value needs any further adjustment. Therefore, \(x \approx 6.47\) or \(x \approx -6.47\).

Key Concepts

Quadratic Formula

The quadratic formula is a universal tool used to find the solutions of quadratic equations, which are equations shaped like \(ax^2 + bx + c = 0\) where \(a\), \(b\), and \(c\) are constants. This explicit formula is \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\), derived from completing the square in algebra. When facing any quadratic equation, you apply the formula by plugging the equation's coefficients into \(a\), \(b\), and \(c\) correspondingly.

The beauty of the quadratic formula is its ability to find solutions regardless of whether they are real or imaginary numbers. If the part under the square root, known as the discriminant \(b^2-4ac\), is positive, you end up with two different real solutions. If it's zero, there is exactly one real solution. In the case of it being negative, this indicates the existence of two complex solutions. Understanding the quadratic formula is crucial because it not only offers a way to solve for \(x\) but also provides insight into the nature of the solutions of a quadratic equation.

Square Roots

The square root of a number is a value that, when multiplied by itself, gives the original number. Symbolically, if \(x^2 = a\), then \(x\) is the square root of \(a\), written as \(x = \sqrt{a}\). Square roots apply in a variety of mathematical contexts, but they play a central role in solving quadratic equations as seen in the step-by-step solution above.

When solving a quadratic equation, after isolating the \(x^2\) term, you often take the square root of both sides to find \(x\). A critical point to remember is that every positive number has two square roots: a positive one and a negative one. That's why you see the \(\pm\) symbol in the expression \(x = \pm\sqrt{41.875}\); it indicates that there are two solutions: \(x = \sqrt{41.875}\) and \(x = -\sqrt{41.875}\). For negative numbers, their square roots are not real numbers. They are imaginary, expressed as multiples of the imaginary unit \(i\), such as \(\sqrt{-1} = i\).

Rounding Decimals

Rounding decimals is a numerical process used to simplify a decimal number to a nearby value, making it easier to work with while preserving an acceptable accuracy level. When rounding to the nearest hundredth, you look at the third decimal place. If the third decimal place is 5 or higher, you increase the second place by one. If it's less than 5, you leave the second place unchanged and drop all the decimal places that follow.

In the given solution, \(6.475\) was rounded to \(6.47\) because the third decimal place is a 5, leaving the second decimal place unchanged. This practice of rounding is commonly used in everyday life, such as when dealing with currency, where it is necessary to round to two decimal places, because that's the smallest denomination in many currencies. While rounding is sufficient for practical purposes, in mathematical settings, especially when high precision is required, it's important to maintain the number in its most accurate form as long as possible.