Problem 19 - 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. $$30 y^{2}+23 y-40=0$$

Step-by-Step Solution

Answer

The roots are \(y = \frac{5}{6}\) and \(y = -\frac{8}{5}\).

Step 1: Identify coefficients

The given quadratic equation is \(30y^2 + 23y - 40 = 0\), which is in the standard form \(ay^2 + by + c = 0\). Here, the coefficients are \(a = 30\), \(b = 23\), and \(c = -40\).

Step 2: Write the quadratic formula

The quadratic formula to find the roots of a quadratic equation \(ay^2 + by + c = 0\) is given by: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Step 3: Calculate the discriminant

The discriminant \(D\) is calculated using the formula \(D = b^2 - 4ac\). Plug in the values of \(a\), \(b\), and \(c\): \[ D = 23^2 - 4 \cdot 30 \cdot (-40) = 529 + 4800 = 5329 \]

Step 4: Compute the square root of the discriminant

Find the square root of the discriminant \(D = 5329\): \[ \sqrt{5329} = 73 \]

Step 5: Substitute values into the quadratic formula

Substitute \(b = 23\), \(a = 30\), and \(\sqrt{D} = 73\) into the quadratic formula: \[ y = \frac{-23 \pm 73}{60} \]

Step 6: Solve for the roots

Calculate the two possible values for \(y\):1. \( y_1 = \frac{-23 + 73}{60} = \frac{50}{60} = \frac{5}{6} \)2. \( y_2 = \frac{-23 - 73}{60} = \frac{-96}{60} = -\frac{8}{5} \)

Key Concepts

Discriminant

The discriminant is a significant component of the quadratic formula that helps determine the nature of the roots of a quadratic equation. It is denoted by the letter \(D\) and calculated using the formula:

• \(D = b^2 - 4ac\)

Where \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation \(ay^2 + by + c = 0\).
The discriminant reveals important information about the roots:

• If \(D > 0\), the equation has two distinct real roots.
• If \(D = 0\), the equation has exactly one real root, also known as a repeated or double root.
• If \(D < 0\), the equation has two complex roots, which means the roots are not real numbers but include imaginary components.

In the given example, we calculated the discriminant as 5329, which is greater than zero. This means the quadratic equation has two distinct real roots.

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree. It can typically be expressed in the standard form:

• \(ay^2 + by + c = 0\)

Here, \(a\), \(b\), and \(c\) are constants or coefficients, with \(a eq 0\). The presence of \(a\) ensures that the equation is indeed quadratic, as opposed to linear or another polynomial form.
The standard form is essential because it sets the groundwork for many methods used to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. Different quadratic equations can have varying characteristics:

• Depending on the coefficients, the graph of the equation forms a parabola that opens upwards if \(a > 0\), and downwards if \(a < 0\).
• The vertex of this parabola represents the maximum or minimum point of the function.
• The roots of the equation correspond to the x-intercepts of the parabola if they exist.

In the example exercise, the quadratic equation is \(30y^2 + 23y - 40 = 0\), where \(a = 30\), \(b = 23\), and \(c = -40\).

Roots of a Quadratic Equation

The roots of a quadratic equation are the solutions that satisfy the condition \(ay^2 + by + c = 0\). They are calculated using the quadratic formula:

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

This formula derives from completing the square technique and provides a straightforward approach to find the roots, regardless of whether they are rational, irrational, or complex numbers.
Let's consider how the roots were calculated for the given exercise. The quadratic formula yields two possible values for \(y\) because of the "\(\pm\)" sign, which represents the two potential solutions:

• \( y_1 = \frac{-23 + 73}{60} = \frac{50}{60} = \frac{5}{6} \)
• \( y_2 = \frac{-23 - 73}{60} = \frac{-96}{60} = -\frac{8}{5} \)

These calculations produced two distinct real roots for our example, consistent with the positive discriminant earlier derived. Understanding the results of the roots provides insights into the graph of the equation and its intercepts with the x-axis.