Problem 52 - A Graphical Approach to College Algebra

Question

Solve each equation. For equations with real solutions, support your answers graphically. $$3 x^{2}-6 x=4$$

Step-by-Step Solution

Answer

The solutions are \(x = 1 + \frac{\sqrt{21}}{3}\) and \(x = 1 - \frac{\sqrt{21}}{3}\).

Step 1: Write the Equation in Standard Form

The original equation is \(3x^2 - 6x = 4\). To write it in standard form, we need to move all terms to one side of the equation. Subtract 4 from both sides to obtain \(3x^2 - 6x - 4 = 0\).

Step 2: Identify Coefficients for Quadratic Formula

Identify the coefficients in the standard form equation \(ax^2 + bx + c = 0\). Here, \(a = 3\), \(b = -6\), and \(c = -4\).

Step 3: Apply the Quadratic Formula

The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) can be used to find solutions. Substitute the coefficients: \(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3}\).

Step 4: Simplify the Expression Under the Square Root

First, calculate the discriminant: \((-6)^2 - 4 \cdot 3 \cdot (-4) = 36 + 48 = 84\). Now the equation becomes \(x = \frac{6 \pm \sqrt{84}}{6}\).

Step 5: Simplify the Square Root and Final Expression

The square root of 84 is \(\sqrt{84} = \sqrt{4 \cdot 21} = 2\sqrt{21}\). Thus, \(x = \frac{6 \pm 2\sqrt{21}}{6}\), which simplifies to \(x = 1 \pm \frac{\sqrt{21}}{3}\).

Step 6: Graph the Equation to Verify Solutions

Plot the function \(y = 3x^2 - 6x - 4\) on a graph. The roots of the equation are the x-values where the graph intersects the x-axis. These roots correspond to the solutions \(x = 1 + \frac{\sqrt{21}}{3}\) and \(x = 1 - \frac{\sqrt{21}}{3}\).

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations, which typically take the form \(ax^2 + bx + c = 0\). This formula allows us to find the roots of the quadratic equations, even when factoring is not straightforward or possible. The formula is given by:

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

To use this formula effectively, you just need to identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation and then substitute these values into the formula. The "\(\pm\)" symbol indicates that there are generally two possible solutions for \(x\): one involving addition and the other involving subtraction. These solutions represent the points (or "roots") at which the parabola represented by the quadratic equation intersects the x-axis.
This makes the quadratic formula incredibly useful for finding the exact solutions to quadratic problems.

Standard Form

The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). Writing the equation in this form is a crucial first step in solving it using the quadratic formula. It involves ensuring all terms are on one side of the equation with the equation set to zero.
In our original exercise, the equation was given as \(3x^2 - 6x = 4\). To convert it to the standard form, you must subtract 4 from both sides, resulting in \(3x^2 - 6x - 4 = 0\).
In this form:

• \(a\) represents the coefficient of \(x^2\) (here, \(a = 3\))
• \(b\) is the coefficient of \(x\) (here, \(b = -6\))
• \(c\) is the constant term (here, \(c = -4\))

Understanding standard form is essential because it provides a structured way to input the necessary values into the quadratic formula accurately.

Discriminant

The discriminant is a component of the quadratic formula found within the square root portion \(b^2 - 4ac\). It plays a critical role in determining the nature of the solutions for the quadratic equation.
In our exercise, the discriminant is computed as \((-6)^2 - 4 \cdot 3 \cdot (-4)\). Calculating this gives us:

• \(36 + 48 = 84\)

The value of the discriminant helps us understand:

• If the discriminant is positive, there are two distinct real solutions (as is the case here).
• If it is zero, there is exactly one real solution.
• If negative, there are no real solutions, and the solutions are complex or imaginary.

In this case, since the discriminant is 84, which is positive, we know there are two real solutions. This information helps predict the nature of the solutions even before solving them completely.

Real Solutions

Real solutions to a quadratic equation are the values of \(x\) that satisfy the equation such that the corresponding quadratic function intersects the x-axis. These solutions can be found using factors or the quadratic formula and can also be represented graphically.
In our example, the solutions were derived by applying the quadratic formula, and we found that:

• \(x = 1 + \frac{\sqrt{21}}{3}\)
• \(x = 1 - \frac{\sqrt{21}}{3}\)

These two values indicate where the graph of the equation \(y = 3x^2 - 6x - 4\) intersects the x-axis. Since both solutions are derived from a positive discriminant, they confirm the existence of two points of intersection, representing two distinct real solutions.
Understanding real solutions helps not only in solving equations algebraically but also in visualizing them graphically to interpret the behavior of quadratic functions.