Problem 86 - College Algebra with Modeling and Visualization

Question

Complete the following. (a) Write the equation as \(a x^{2}+b x+c=0\) with \(a>0\) (b) Calculate the discriminant \(b^{2}-4 a c\) and determine the number of real solutions. (c) Solve the equation. $$ 16 x^{2}+9=24 x $$

Step-by-Step Solution

Answer

The equation is already in the form with \( a=16, b=-24, c=9 \). The discriminant is 0, indicating one real solution: \( x = \frac{3}{4} \).

Step 1: Rewrite the Equation

Start by rewriting the equation in the form \( ax^2 + bx + c = 0 \) with \( a > 0 \). The initial equation is \( 16x^2 + 9 = 24x \). Subtract \( 24x \) from both sides to obtain \( 16x^2 - 24x + 9 = 0 \). Here, \( a = 16 \), \( b = -24 \), and \( c = 9 \).

Step 2: Calculate the Discriminant

The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( b^2 - 4ac \). Substitute the values \( a = 16 \), \( b = -24 \), and \( c = 9 \) into the formula: \[ \Delta = (-24)^2 - 4 \cdot 16 \cdot 9 \]. Evaluate it to find \( \Delta = 576 - 576 = 0 \).

Step 3: Determine Number of Solutions

Based on the discriminant, \( \Delta = 0 \), there is exactly one real solution, which means the quadratic equation has a double root.

Step 4: Solve the Equation Using the Quadratic Formula

For \( ax^2 + bx + c = 0 \), the roots are given by the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Using \( b = -24 \), \( a = 16 \), and \( \Delta = 0 \), the formula reduces to: \[ x = \frac{-(-24) \pm \sqrt{0}}{2 \cdot 16} \]. Simplifying gives \( x = \frac{24}{32} = \frac{3}{4} \). The sole solution is \( x = \frac{3}{4} \).

Key Concepts

Understanding the Discriminant

The discriminant is a special component when dealing with quadratic equations. It helps us to determine the nature and number of solutions for a quadratic equation. In a typical quadratic equation \[ ax^2 + bx + c = 0, \] the discriminant \( \Delta \) is calculated using the formula:\[ \Delta = b^2 - 4ac. \]The computed discriminant value indicates:

• \( \Delta > 0 \): Two distinct real solutions.
• \( \Delta = 0 \): Exactly one real solution, indicating a repeated or double root.
• \( \Delta < 0 \): No real solutions, only complex ones.

For the equation from the problem, substituting the values \( a = 16 \), \( b = -24 \), and \( c = 9 \) resulted in a discriminant of 0. Therefore, this equation has one real solution. Understanding the discriminant helps in foreseeing the nature of solutions without actually solving the equation.

The Quadratic Formula

The quadratic formula offers a way to find the roots of any quadratic equation. It is given as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]This formula is versatile and can handle any quadratic, whether or not the solutions are real. By inserting the values of \( a \), \( b \), and \( c \), you can solve for the variable \( x \). The portion under the square root, \( b^2 - 4ac \), is known as the discriminant and determines the nature of the roots, as discussed earlier.In the given exercise, the formula was simplified due to the discriminant being 0:\[ x = \frac{-(-24) \pm \sqrt{0}}{2 \times 16}, \]which reduced further to \[ x = \frac{24}{32} = \frac{3}{4}. \] Thus, the single real solution \( x = \frac{3}{4} \) was found efficiently. Mastery of the quadratic formula is powerful for tackling various quadratic problems, regardless of complexity.

What Are Real Solutions?

Real solutions refer to the roots of the equation that are real numbers, not involving imaginary parts. A real solution means there’s a tangible point on the graph \( y = ax^2 + bx + c \) that intersects the x-axis. This occurs when the discriminant is zero or positive:

• If \( \Delta = 0 \): There is one real solution, suggesting the vertex of the parabola touches the x-axis, forming a "touch" rather than a proper intersection.
• If \( \Delta > 0 \): There are two distinct points where the graph crosses the x-axis, corresponding to two different real solutions.

Real solutions are what many algebra problems aim to find since they represent actual values for which the original equation holds true. In our problem, with the discriminant equaling zero, we encountered one real solution, demonstrating a special case where the quadratic graph just brushes the axis, at the point \( x = \frac{3}{4} \). Understanding real solutions helps visualize the problem and assures us of the equation's behavior intuitively.