Problem 32 - A Graphical Approach to College Algebra

Question

Solve each equation. For equations with real solutions, support your answers graphically. $$-5+16 x-3 x^{2}=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{3}\) and \(x = 5\).

Step 1: Identify the Equation

The equation given is a quadratic equation in the standard form, represented as \(-5 + 16x - 3x^2 = 0\). This can also be written as \(-3x^2 + 16x - 5 = 0\).

Step 2: Rearrange the Equation

To solve the quadratic equation \(-3x^2 + 16x - 5 = 0\), it may help to rearrange it to the standard form \(ax^2 + bx + c = 0\) for solving: \(-3x^2 + 16x - 5 = 0\). Here we have \(a = -3\), \(b = 16\), \(c = -5\).

Step 3: Use the Quadratic Formula

Apply the quadratic formula to find the values of \(x\): \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Plugging in the values of \(a\), \(b\), and \(c\):\[x = \frac{-16 \pm \sqrt{(16)^2 - 4(-3)(-5)}}{2(-3)}\]

Step 4: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\): \[(16)^2 - 4(-3)(-5) = 256 - 60 = 196\]Since 196 is positive, there are two distinct real solutions.

Step 5: Solve for x

Substitute the value of the discriminant back into the quadratic formula:\[x = \frac{-16 \pm \sqrt{196}}{-6}\] \[x = \frac{-16 \pm 14}{-6}\]

Step 6: Find the Solutions

Calculate the two possible values for \(x\): 1. \[x = \frac{-16 + 14}{-6} = \frac{-2}{-6} = \frac{1}{3}\] 2. \[x = \frac{-16 - 14}{-6} = \frac{-30}{-6} = 5\] Thus, the solutions are \(x = \frac{1}{3}\) and \(x = 5\).

Step 7: Graphically Support the Solutions

To verify graphically, we sketch the parabola given by the equation \(y = -3x^2 + 16x - 5\). The vertex should lie between the two roots, and the parabola opens downwards. The graph will intersect the x-axis at \(x = \frac{1}{3}\) and \(x = 5\), confirming the solutions.

Key Concepts

Quadratic Formula

The quadratic formula is a tool used to find the solutions of a quadratic equation, which is an equation of the form \( ax^2 + bx + c = 0 \). The formula itself is:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula will provide us with the solutions for \( x \), which are the points where the graph of the quadratic function intersects the x-axis. The \( a \), \( b \), and \( c \) are the coefficients from the standard form of the quadratic equation. Using this formula, you can quickly solve any quadratic equation without needing to factor or complete the square, making it a very efficient method.

Discriminant

The discriminant is a part of the quadratic formula which is under the square root, \( b^2 - 4ac \). It tells us how many real solutions a quadratic equation has:

• If the discriminant is positive, there are two distinct real solutions.
• If the discriminant is zero, there is exactly one real solution, which is also called a repeated or double root.
• If the discriminant is negative, there are no real solutions, and the solutions are complex or imaginary numbers.

In our exercise, the discriminant was calculated to be 196, which is positive, indicating that there are two real solutions: \( x = \frac{1}{3} \) and \( x = 5 \). This means the parabola will intersect the x-axis at these two points. Understanding the discriminant helps in predicting the nature of the roots without solving the equation fully.

Graphing Parabolas

When graphing quadratic equations, the graph takes the shape of a parabola. For the equation \( y = -3x^2 + 16x - 5 \), the parabola "opens" downwards because the coefficient of \( x^2 \) (which is \(-3\)) is negative.
The vertex of this parabola is the highest point on the graph, and it lies between the solutions \( x = \frac{1}{3} \) and \( x = 5 \). To graph this, you can plot these solutions on the x-axis and draw a smooth curve through these points, reaching up to the vertex.
A detailed graph can support the solutions by showing that the parabola truly intersects the x-axis at the computed solutions and how it behaves between them.

Real Solutions

Real solutions of a quadratic equation are the values of \( x \) for which the quadratic formula results in normal, non-imaginary numbers. These solutions are the \( x \)-intercepts of the parabola, and in the context of real-world problems, they can represent actual measurable quantities.

• In our exercise, the calculated real solutions were \( x = \frac{1}{3} \) and \( x = 5 \).
• These are where the graph intersects the x-axis, providing a visual confirmation of the solutions.

Real solutions are of critical importance as they often represent possible solutions to physical problems, such as distances, times, and speeds, where imaginary numbers would not be applicable.