Problem 88 - 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. $$ 2 x^{2}+x=2 $$

Step-by-Step Solution

Answer

Two distinct real solutions: \(x_1 = \frac{-1 + \sqrt{17}}{4}\), \(x_2 = \frac{-1 - \sqrt{17}}{4}\).

Step 1: Rearrange the Equation

Start by writing the given equation in the standard quadratic form \( ax^2 + bx + c = 0 \). The given equation is \( 2x^2 + x = 2 \). Subtract 2 from both sides to get:\[ 2x^2 + x - 2 = 0 \]Here, \( a = 2 \), \( b = 1 \), and \( c = -2 \).

Step 2: Calculate the Discriminant

The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( \Delta = b^2 - 4ac \). Substitute the values of \( a \), \( b \), and \( c \) into this formula:\[ \Delta = 1^2 - 4 \cdot 2 \cdot (-2) \]\[ \Delta = 1 + 16 = 17 \]

Step 3: Determine the Number of Real Solutions

The discriminant \( \Delta = 17 \) is positive, which indicates that the quadratic equation has two distinct real solutions.

Step 4: Solve the Equation

Since the discriminant is positive, use the quadratic formula to solve for \( x \):\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \]Substitute the known values:\[ x = \frac{-1 \pm \sqrt{17}}{2 \cdot 2} \]\[ x = \frac{-1 \pm \sqrt{17}}{4} \]This results in two solutions:\[ x_1 = \frac{-1 + \sqrt{17}}{4} \]\[ x_2 = \frac{-1 - \sqrt{17}}{4} \]

Key Concepts

Discriminant

The discriminant is vital in understanding the nature of solutions for a quadratic equation. It's found using the formula \( \Delta = b^2 - 4ac \). Here’s why it matters:

• A positive discriminant indicates two distinct real solutions.
• A zero discriminant implies exactly one real solution, meaning the equation has a repeated root.
• A negative discriminant means there are no real solutions; instead, the equation has two complex solutions.

In our example, the equation was first rearranged into the standard form \( 2x^2 + x - 2 = 0 \). From this, we identified \( a = 2 \), \( b = 1 \), and \( c = -2 \). By substituting these into the discriminant formula: \[ \Delta = 1^2 - 4 \cdot 2 \cdot (-2) = 17 \] Since 17 is positive, it confirms that our quadratic equation has two distinct real solutions.

Real Solutions

Understanding real solutions in a quadratic equation helps us predict how its graph will behave on an x-y plane. These solutions correspond to the x-values where the graph of the equation crosses the x-axis. The number of real solutions can be found using the discriminant:

• A positive discriminant means the graph crosses the x-axis twice, thus having two real solutions.
• A zero discriminant signifies the graph just touches the x-axis at one point, giving one real solution.
• A negative discriminant reveals the graph does not touch the x-axis, having no real solutions.

For our problem, since the discriminant is 17, which is positive, it tells us there are two distinct points where the graph of the quadratic equation intersects the x-axis. Thus, there are two distinct real solutions to the equation.

Quadratic Formula

The quadratic formula is a powerful method for finding the roots of any quadratic equation, especially when factoring is complex or impossible. It is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula allows you to calculate the precise values of \( x \) by substituting \( a \), \( b \), and \( c \) into it. The "\( \pm \)" symbol indicates that there will be two possible solutions. In our exercise, the equation was \( 2x^2 + x - 2 = 0 \). By substituting \( a = 2 \), \( b = 1 \), and \( c = -2 \), and since we already know that the discriminant \( \Delta = 17 \), we apply these into the quadratic formula: \[ x = \frac{-1 \pm \sqrt{17}}{4} \] Hence, the two roots are computed as:

• \( x_1 = \frac{-1 + \sqrt{17}}{4} \)
• \( x_2 = \frac{-1 - \sqrt{17}}{4} \)

This method provides a definitive solution whether or not the factors can be easily identified.