Problem 29 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations, using the quadratic formula. Exercises \(5-8\) are the same as Exercises \(11-14\) of Section 7.2. $$x^{2}-0.20 x-0.40=0$$

Step-by-Step Solution

Answer

The solutions are \(x = 0.74\) and \(x = -0.54\).

Step 1: Identify Coefficients

The given quadratic equation is \(x^{2} - 0.20x - 0.40 = 0\). Identify the coefficients \(a\), \(b\), and \(c\) for the quadratic formula \(ax^2 + bx + c = 0\). In this equation, \(a = 1\), \(b = -0.20\), and \(c = -0.40\).

Step 2: Write the Quadratic Formula

The quadratic formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula will be used to find the solutions for \(x\).

Step 3: Calculate the Discriminant

The discriminant is \(b^2 - 4ac\). Substitute the values \(b = -0.20\), \(a = 1\), and \(c = -0.40\) into the discriminant: \((-0.20)^2 - 4(1)(-0.40)\). Calculate it to get the value of the discriminant.

Step 4: Compute the Discriminant

Calculate the expression from Step 3. \((-0.20)^2 = 0.04\) and \(4 \times 1 \times -0.40 = -1.6\). The discriminant is: \(0.04 + 1.6 = 1.64\).

Step 5: Apply the Quadratic Formula

Substitute the values into the quadratic formula: \(x = \frac{-(-0.20) \pm \sqrt{1.64}}{2(1)}\). Simplify to \(x = \frac{0.20 \pm \sqrt{1.64}}{2}\).

Step 6: Simplify the Square Root

Calculate \(\sqrt{1.64}\), which is approximately 1.28. Update the expression: \(x = \frac{0.20 \pm 1.28}{2}\).

Step 7: Calculate the Solutions

Solve for the two possible solutions. First solution: \(x_1 = \frac{0.20 + 1.28}{2} = \frac{1.48}{2} = 0.74\). Second solution: \(x_2 = \frac{0.20 - 1.28}{2} = \frac{-1.08}{2} = -0.54\).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental mathematical expressions that take the form of \( ax^2 + bx + c = 0 \). These equations are key in both algebra and calculus because they represent parabolas when graphed. The coefficients, \(a\), \(b\), and \(c\), play specific roles:

• \(a\) is the coefficient of the quadratic term \(x^2\).
• \(b\) is the coefficient of the linear term \(x\).
• \(c\) is the constant term.

Identifying these coefficients helps set the stage for applying various methods to find the solutions of the equation. It's crucial to understand that the solutions, or "roots," are the \(x\)-values where the quadratic equals zero.

In our original exercise, we had the equation \(x^2 - 0.20x - 0.40 = 0\). Here, \(a = 1\), \(b = -0.20\), and \(c = -0.40\). These coefficients will be particularly important as we move further to solve using the quadratic formula.

Discriminant

The discriminant is a specific part of a quadratic equation that can give us insight into the nature of the equation's solutions without actually solving it. Subset of the quadratic formula, the discriminant is expressed as \(b^2 - 4ac\).

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If it is zero, the equation has exactly one real solution, meaning the parabola touches the x-axis at a single point.
• If the discriminant is negative, there are no real solutions, but rather two complex solutions.

Calculating the discriminant was one of the key steps in the provided solution. We calculated it as \((-0.20)^2 - 4 \times 1 \times (-0.40)\), which simplifies to \(1.64\). Since the discriminant here is positive, we find that the equation has two distinct real roots.

Solving Quadratic Equations

Solving quadratic equations can be approached in different ways: graphing, factoring, completing the square, and the quadratic formula. The most universal method is the quadratic formula, \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. The formula is derived from rewriting the standard form as a perfect square.

In practice, it offers a formulaic approach, allowing us to plug in the coefficients directly. After identifying the coefficients \(a = 1\), \(b = -0.20\), and \(c = -0.40\), we calculated the discriminant, as discussed earlier, and inserted these into the formula. This method yielded two solutions: \(x_1 = 0.74\) and \(x_2 = -0.54\). These values represent the points where our quadratic equation crosses the x-axis.

Choosing the quadratic formula can mitigate errors and is particularly useful when equations do not factor easily. Thus, understanding how to apply it is crucial in algebraic problem-solving.