Problem 95 - 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. $$ 3 x^{2}=1-x $$

Step-by-Step Solution

Answer

The equation has two real solutions: \(x = \frac{-1 \pm \sqrt{13}}{6}\).

Step 1: Rearrange the Equation

The given equation is \(3x^2 = 1 - x\). To rewrite it in the standard quadratic form \(ax^2 + bx + c = 0\), bring all terms to one side. Add \(x\) to both sides and subtract \(1\):\[ 3x^2 + x - 1 = 0 \] The equation is now in standard form with \(a = 3\), \(b = 1\), \(c = -1\).

Step 2: Calculate the Discriminant

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\). Substitute \(a = 3\), \(b = 1\), and \(c = -1\):\[ \Delta = 1^2 - 4 \times 3 \times (-1) \]\[ \Delta = 1 + 12 = 13 \] Since \(\Delta = 13 > 0\), the equation has two distinct real solutions.

Step 3: Solve the Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the solutions. Substitute \(a = 3\), \(b = 1\), and \(c = -1\):\[ x = \frac{-1 \pm \sqrt{13}}{6} \] Therefore, the solutions are \(x_1 = \frac{-1 + \sqrt{13}}{6}\) and \(x_2 = \frac{-1 - \sqrt{13}}{6}\).

Key Concepts

Understanding the Discriminant

The discriminant is a crucial component of quadratic equations. It's represented by the formula \(b^2 - 4ac\). This value helps us determine the nature and number of solutions for a quadratic equation, which is in the form \(ax^2 + bx + c = 0\).
For our specific example, where \(a = 3\), \(b = 1\), and \(c = -1\), the discriminant calculation is: \ \(\Delta = 1^2 - 4 \times 3 \times (-1)\).
This simplifies to \(1 + 12 = 13\).

• If the discriminant (\(\Delta\)) is greater than zero, there are two distinct real solutions.
• If \(\Delta\) is exactly zero, there is one real solution, also known as a repeated root.
• If \(\Delta\) is less than zero, there are no real solutions – the roots are complex or imaginary.

In our problem, since \(\Delta = 13\) which is greater than 0, there are two real solutions.

Exploring the Quadratic Formula

The quadratic formula is an amazing tool for solving quadratic equations. It is expressed as: \ \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula can be applied to any quadratic equation, making it very versatile. Let's break it down step by step.

• \(\pm\) indicates there will be two solutions: one for the positive root and another for the negative root.
• \(b^2 - 4ac\) is simply the discriminant we discussed earlier.
• The whole expression is divided by \(2a\), which is the double of the coefficient of \(x^2\).

For our equation \(3x^2 + x - 1 = 0\), substituting \(a = 3\), \(b = 1\), and \(c = -1\) into the formula gives us: \ \(x = \frac{-1 \pm \sqrt{13}}{6}\).
This yields two solutions, \(x_1 = \frac{-1 + \sqrt{13}}{6}\) and \(x_2 = \frac{-1 - \sqrt{13}}{6}\). These are the distinct real solutions for the equation.

Number of Real Solutions

In mathematics, discovering the number of real solutions for a quadratic equation is vital. We determine this using the discriminant \(b^2 - 4ac\).
As mentioned before, the value of the discriminant tells us:

• If the discriminant is positive, the equation has two distinct real solutions.
• If it is zero, the quadratic has exactly one real solution.
• If it's negative, there are no real solutions.

In our given equation, \(3x^2 + x - 1 = 0\), we found that \(\Delta = 13\).
Since 13 is positive, it confirms that the equation has two distinct real solutions. These types of problems harness the uniqueness of quadratic behavior, showcasing how foundational mathematics utilizes consistent rules to reveal patterns and solutions.