Problem 53 - Contemporary Precalculus

Question

Use a calculator to find approximate solutions of the equation. $$4.42 x^{2}-10.14 x+3.79=0$$

Step-by-Step Solution

Answer

Answer: The approximate solutions to the given quadratic equation are $$x_1 \approx 0.4912$$ and $$x_2 \approx 1.7274$$.

Step 1: Identify the coefficients in the quadratic equation

First, identify the coefficients a, b, and c in the given quadratic equation $$4.42x^2 - 10.14x + 3.79 = 0$$. The coefficients are as follows: $$a = 4.42$$ $$b = -10.14$$ $$c = 3.79$$

Step 2: Use quadratic formula for the approximate solutions

The quadratic formula is given by: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ Substitute the values of a, b, and c into the formula to find the approximate solutions: $$x = \frac{-(-10.14) \pm \sqrt{(-10.14)^2-4(4.42)(3.79)}}{2(4.42)}$$

Step 3: Calculate the solutions using the calculator

To find the approximate solutions, calculate the value of x as per the quadratic formula on your calculator. Ensure that you calculate both the positive and negative root. After plugging in the values and calculating, you should get the following approximate solutions: $$x_1 \approx 0.4912$$ $$x_2 \approx 1.7274$$

Step 4: Write down the approximate solutions

The two approximate solutions to the quadratic equation $$4.42x^2 - 10.14x + 3.79$$ are: $$x_1 \approx 0.4912$$ $$x_2 \approx 1.7274$$

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used in algebra to find the solutions of quadratic equations, which are equations in the form \(ax^2 + bx + c = 0\). This formula provides the roots, or solutions, of the equation, even if they are complex or involve irrational numbers. The standard quadratic formula is expressed as:

• \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Using this formula, you substitute the coefficients of the equation for \(a\), \(b\), and \(c\), making it a universal solution for any quadratic equation. The \(\pm\) symbol indicates there are generally two solutions, which correspond to the two possible values of \(x\). These solutions are the points at which the graph of the quadratic equation intersects the x-axis.
When using the formula, it’s important to handle the calculations carefully, especially when working with decimals. In our example, with the equation \(4.42x^2 - 10.14x + 3.79 = 0\), using the quadratic formula allows us to solve for \(x\) without needing to factor the equation or use complex graphing techniques.

Coefficients in Quadratic Equations

In a quadratic equation of the form \(ax^2 + bx + c = 0\), the values of \(a\), \(b\), and \(c\) are called coefficients. These coefficients play a crucial role in determining the nature and the solutions of the equation.

• \(a\) is the coefficient of \(x^2\)
• \(b\) is the coefficient of \(x\)
• \(c\) is the constant term, not associated with \(x\)

Identifying these coefficients is a crucial first step in using the quadratic formula. For the given quadratic equation, \(4.42x^2 - 10.14x + 3.79 = 0\), we have:

• \(a = 4.42\)
• \(b = -10.14\)
• \(c = 3.79\)

These coefficients affect the discriminant \(b^2 - 4ac\), which helps determine whether the solutions are real or complex. In simpler terms, they define the shape and position of the parabola represented by the quadratic equation.

Approximate Solutions

When solving quadratic equations, finding approximate solutions is often necessary, especially when dealing with irrational numbers or inexact coefficients. Approximate solutions can be found using the quadratic formula and a calculator to evaluate the expression.
When you input the values of \(a\), \(b\), and \(c\) into the quadratic formula, you perform the calculations up to a certain number of decimal places for accuracy. For example, solving the quadratic equation \(4.42x^2 - 10.14x + 3.79 = 0\) gives us:

• \(x_1 \approx 0.4912\)
• \(x_2 \approx 1.7274\)

Using a calculator helps ensure that we get these values with precision, which is especially important in fields like engineering or physics where accuracy is key. The precision of solutions is determined by how many decimal places you decide to use, and in many cases, you can increase precision if needed to improve accuracy for practical applications.