Problem 96 - 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. $$ x(5 x-3)=1 $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{3 \pm \sqrt{29}}{10}\).

Step 1: Expand the Equation

First, expand the given equation to bring it to the standard quadratic form. The given equation is \(x(5x - 3) = 1\). Expand it to: \[5x^2 - 3x = 1\]Subtract 1 from both sides to set it to zero: \[5x^2 - 3x - 1 = 0\] Now, the equation is in the form \(ax^2 + bx + c = 0\) where \(a = 5\), \(b = -3\), and \(c = -1\).

Step 2: Calculate the Discriminant

For a quadratic equation \(ax^2 + bx + c = 0\), the discriminant \(\Delta\) is calculated as: \[\Delta = b^2 - 4ac\]Substituting in the values \(a = 5\), \(b = -3\), and \(c = -1\):\[\Delta = (-3)^2 - 4 \times 5 \times (-1) = 9 + 20 = 29\]Since the discriminant is positive and not a perfect square, the equation has two distinct real solutions.

Step 3: Solve the Quadratic Equation Using the Quadratic Formula

Use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute \(a = 5\), \(b = -3\), \(c = -1\), and the discriminant \(\Delta = 29\):\[x = \frac{-(-3) \pm \sqrt{29}}{2 \times 5}\]Simplify this to find the roots:\[x = \frac{3 \pm \sqrt{29}}{10}\]Thus, the solutions for \(x\) are \(x_1 = \frac{3 + \sqrt{29}}{10}\) and \(x_2 = \frac{3 - \sqrt{29}}{10}\).

Key Concepts

Understanding the Discriminant

In quadratic equations, the discriminant plays a crucial role in determining the nature of the roots or solutions. The discriminant is denoted by the symbol \(\Delta\) and is calculated using the formula \(b^2 - 4ac\) from the quadratic equation \(ax^2 + bx + c = 0\). This helpful tool allows us to predict the number of real solutions without actually solving the equation entirely.

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

Knowing these outcomes about the discriminant can save time and give quick insights into the characteristics of your quadratic equation solutions. For instance, in the problem we worked on, \(\Delta = 29\). Since \(29\) is positive, we can confidently state that our quadratic equation has two distinct real solutions.

Identifying Real Solutions

Real solutions refer to the actual numerical values of \(x\) that satisfy the quadratic equation \(ax^2 + bx + c = 0\). These solutions are the x-intercepts of the parabola represented by the equation. The number of real solutions directly correlates with the value of the discriminant as mentioned earlier. Understanding this can help us quickly deduce whether the parabola crosses the x-axis and how many times.

• Two real solutions mean the parabola intersects the x-axis at two distinct points.
• One real solution implies the parabola touches the x-axis at one point, which is its vertex.
• No real solutions suggest that the parabola does not intersect the x-axis at all.

In practice, finding the real solutions will guide you towards the exact x-values where the equation holds true. Knowing these solutions is fundamental in many areas such as physics and engineering when determining probable outcomes or solutions to a real-world problem.

Exploring the Quadratic Formula

The quadratic formula provides a reliable method for solving quadratic equations and is expressed by the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula is universal for any quadratic equation and will yield all solutions, real or complex, based on the discriminant.

• The term \(-b\) gives the center value of the solution range along the x-axis.
• The \(\pm\) symbol indicates that there will typically be two solutions: one adding the square root and the other subtracting it.
• The square root part, \(\sqrt{b^2 - 4ac}\), shows the influence of the discriminant in determining the distance between the solutions from the center point.

Using the quadratic formula ensures that you consider all possible solutions for \(x\). In our example, substituting the known values, we get \(x = \frac{3 \pm \sqrt{29}}{10}\). This calculation confirms there are indeed two distinct solutions, matching our earlier prediction using the discriminant.