Problem 52 - College Algebra and Calculus: An Applied Approach

Question

Solve the quadratic equation using any convenient method. \(80+6 x=9 x^{2}\)

Step-by-Step Solution

Answer

The roots of the quadratic equation are \(x = \frac{10}{3}\) and \(x = -8\).

Step 1: Rearrange the Equation

Rearrange the quadratic equation to be in the standard form \(ax^{2}+bx+c=0\). This is done by subtracting \(80+6x\) from both sides of the equation. The result will be: \(9x^{2}-6x-80=0\).

Step 2: Identify Coefficients

The quadratic equation is now in the standard form. Identify the coefficients \(a\), \(b\), and \(c\) where \(a\) is the coefficient of \(x^{2}\), \(b\) is the coefficient of \(x\) and \(c\) is the constant term. So, in the given equation, \(a=9\), \(b=-6\), and \(c=-80\).

Step 3: Use the Quadratic Formula

Substitute the values of \(a\), \(b\), and \(c\) into the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). This gives \(x = \frac{6 \pm \sqrt{(-6)^{2}-4*9*(-80)}}{2*9}\). Simplify that to get the solutions for \(x\).

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations. Whenever you encounter an equation of the form \(ax^2 + bx + c = 0\), the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) can help you find the solutions for \(x\). This formula works for any quadratic equation, regardless of whether the roots are real or complex.

To apply the quadratic formula effectively, follow these steps:

• Identify the coefficients in your equation. These will be the values of \(a\), \(b\), and \(c\).
• Create the expression \(b^2 - 4ac\). This is known as the discriminant. It will determine the nature of the roots.
• Calculate the terms in the formula: the square root \(\sqrt{b^2-4ac}\), which influences the solutions, and the division by \(2a\), which scales the solutions.

This formula elegantly combines these components to produce two possible values for \(x\), representing the roots of the quadratic equation.

Standard Form of Quadratic Equation

A quadratic equation can usually be expressed in what is called the "standard form." This form looks like \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The standard form is crucial for recognizing the type of equation you are dealing with and for using various methods to solve it efficiently.

To convert any given quadratic equation into its standard form, you would typically rearrange the terms. In our original problem, we started with \(80 + 6x = 9x^2\). By moving all terms to one side of the equation – specifically, the left – we get \(9x^2 - 6x - 80 = 0\). Placing zero on one side makes it easier to apply methods like factoring, completing the square, or using the quadratic formula.

Once in standard form, you can better visualize the quadratic nature of the polynomial and plan the most effective solving strategy.

Identifying Coefficients

Determining which values belong to \(a\), \(b\), and \(c\) in a quadratic equation is a foundational step for solving it. This process involves identifying these constants directly from the equation when it's in standard form \(ax^2 + bx + c = 0\).

In the context of our example \(9x^2 - 6x - 80 = 0\), to pinpoint the coefficients:

• The coefficient \(a\) is linked to \(x^2\), thus \(a = 9\).
• The coefficient \(b\) is linked to \(x\), thus \(b = -6\).
• The constant term \(c\) does not involve \(x\), thus \(c = -80\).

Properly identifying these coefficients is necessary before substituting them into the quadratic formula or any other solving method. This step sets the stage for finding the equation's roots accurately.