Problem 30 - 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. $$3.2 x^{2}=2.5 x+7.6$$

Step-by-Step Solution

Answer

The solutions are \(x \approx 1.98\) and \(x \approx -1.20\).

Step 1: Write the equation in standard form

Start by rewriting the given quadratic equation in the standard form, which is \( ax^2 + bx + c = 0 \). For the given equation, subtract \(2.5x\) and \(7.6\) from both sides to obtain:\[ 3.2x^2 - 2.5x - 7.6 = 0 \]

Step 2: Identify the coefficients

Identify the values of the coefficients from the standard form equation:\[ a = 3.2, \quad b = -2.5, \quad c = -7.6 \]

Step 3: Apply the quadratic formula

The quadratic formula for solving \( ax^2 + bx + c = 0 \) is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute the coefficients \(a\), \(b\), and \(c\) into this formula.

Step 4: Calculate the discriminant

The discriminant \( D \) is given by \( b^2 - 4ac \). Calculate this value:\[ D = (-2.5)^2 - 4 \times 3.2 \times (-7.6) \]\[ D = 6.25 + 97.28 = 103.53 \]

Step 5: Compute the roots using the quadratic formula

Use the quadratic formula with the discriminant to find the roots:\[ x = \frac{-(-2.5) \pm \sqrt{103.53}}{2 \times 3.2} \]\[ x = \frac{2.5 \pm 10.17}{6.4} \]

Step 6: Find the two possible solutions

Calculate the values of \(x\) using the plus and minus options in the formula:For \(x_1\):\[ x_1 = \frac{2.5 + 10.17}{6.4} \]\[ x_1 = \frac{12.67}{6.4} \approx 1.98 \]For \(x_2\):\[ x_2 = \frac{2.5 - 10.17}{6.4} \]\[ x_2 = \frac{-7.67}{6.4} \approx -1.20 \]

Key Concepts

Discriminant in Quadratic Equations

The discriminant is a very vital part of the quadratic formula. It tells us about the nature of the roots of a quadratic equation. The discriminant is the expression inside the square root of the quadratic formula: \( b^2 - 4ac \). This value helps determine:

• The Type of Roots: If the discriminant is positive, the quadratic equation has two distinct real roots. A zero discriminant indicates exactly one real root, also called a repeated or double root. If it's negative, the equation results in two complex roots.

In our context, we calculated the discriminant to be 103.53. Since this is a positive number, we expected two distinct real roots for our quadratic equation.
It's crucial to calculate and interpret the discriminant as it guides us on what to expect when solving the equation.

Standard Form of Quadratic Equations

Quadratic equations can be recognized by their specific format, which is called the standard form: \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). The variable \( x \) is what we're solving for.
To employ the quadratic formula effectively, our first task is to ensure that the equation we start with is in this standard form. In our exercise, the equation \( 3.2x^2 = 2.5x + 7.6 \) was rearranged to become \( 3.2x^2 - 2.5x - 7.6 = 0 \).
This step is crucial because it allows us to easily identify the coefficients \( a = 3.2 \), \( b = -2.5 \), and \( c = -7.6 \), which we need for the quadratic formula. Thus, putting the equation in its standard form is a foundational step before any other solution method.

Solving Quadratic Equations

Solving quadratic equations involves finding the values of \( x \) that satisfy the equation. One of the most reliable methods for this is the quadratic formula:

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

By substituting the coefficients \( a \), \( b \), and \( c \) into the formula, you can solve for \( x \).
Following the step-by-step method, we used \( a = 3.2 \), \( b = -2.5 \), and \( c = -7.6 \) to compute the two possible solutions for \( x \). This involves calculating the discriminant and using it to derive the roots through the quadratic formula.
Remember, solving these equations using the quadratic formula is especially helpful when other methods, like factoring, are not straightforward or possible.

Roots of Quadratic Equations

The roots of a quadratic equation are the values of \( x \) that make the equation true, essentially the solutions. Depending on the discriminant, these roots can be real or complex.
In our example, we computed the roots as:

• Root 1 (\( x_1 \)): \( x_1 = \frac{12.67}{6.4} \approx 1.98 \)
• Root 2 (\( x_2 \)): \( x_2 = \frac{-7.67}{6.4} \approx -1.20 \)

These roots are real and distinct, which means that the graph of the quadratic equation crosses the x-axis at these points.
Identifying the roots is an essential part of solving any quadratic equation because it reveals the x-values where the quadratic equals zero, providing insight into its graph and behavior.