Problem 90 - 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. $$ 3 x^{2}+3=5 x $$

Step-by-Step Solution

Answer

Rewriting gives \(3x^2 - 5x + 3 = 0\), discriminant is \(-11\), no real solutions but complex solutions are \(\frac{5 \pm i\sqrt{11}}{6}\).

Step 1: Rearrange the equation

Start with the given equation: \(3x^2 + 3 = 5x\). We need to rewrite this equation in the standard quadratic form \(ax^2 + bx + c = 0\). To do this, subtract \(5x\) from both sides to get \(3x^2 - 5x + 3 = 0\). Here, \(a = 3\), \(b = -5\), and \(c = 3\).

Step 2: Calculate the discriminant

The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula \(D = b^2 - 4ac\). Substitute the values \(a = 3\), \(b = -5\), and \(c = 3\) into this formula: \(D = (-5)^2 - 4 \cdot 3 \cdot 3\). Simplifying gives \(D = 25 - 36 = -11\).

Step 3: Determine the number of real solutions

The number of real solutions of a quadratic equation is determined by the discriminant \(D\). If \(D > 0\), there are 2 distinct real solutions. If \(D = 0\), there is exactly 1 real solution. If \(D < 0\), there are no real solutions. Since \(D = -11 < 0\), the equation has no real solutions.

Step 4: Solve the equation (complex solutions)

Even though there are no real solutions, we can find the complex solutions using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). With \(a = 3\), \(b = -5\), and \(c = 3\), substitute into the formula: \(x = \frac{-(-5) \pm \sqrt{-11}}{2 \times 3}\). This simplifies to \(x = \frac{5 \pm \sqrt{-11}}{6}\). The solutions are \(x = \frac{5 + i\sqrt{11}}{6}\) and \(x = \frac{5 - i\sqrt{11}}{6}\).

Key Concepts

Discriminant

When solving quadratic equations, the discriminant is an important component to consider. In a quadratic equation with the form \( ax^2 + bx + c = 0 \), the discriminant \( D \) is calculated as \( D = b^2 - 4ac \). The value of \( D \) helps determine the type and number of solutions the equation has. Here’s why:

• If \( D > 0 \), the equation has two distinct real solutions.
• If \( D = 0 \), there is exactly one real solution, meaning the solutions are repeated or identical.
• If \( D < 0 \), the equation has no real solutions but instead, two complex solutions.

Understanding the discriminant allows you to anticipate the nature of a quadratic equation's solutions without fully solving the equation itself. For example, if you found that \( D = -11 \), this negative value indicates that the solutions are complex.

Complex Solutions

Quadratic equations can sometimes have complex solutions, especially when the discriminant \( D \) is negative. Complex numbers are numbers that have both a real and an imaginary part. The imaginary unit \( i \) is defined as \( \sqrt{-1} \).

In cases where \( D < 0 \), like with the discriminant \( D = -11 \), the square root of the discriminant becomes an imaginary number. You will then deal with complex solutions of the form \( a \pm bi \), where \( a \) is the real part, and \( b \) is the imaginary part multiplied by \( i \).

Using the quadratic formula when \( D < 0 \) allows you to reach these solutions. For our example, the solutions were found to be \( \frac{5 + i\sqrt{11}}{6} \) and \( \frac{5 - i\sqrt{11}}{6} \). Here, \( 5/6 \) is the real part, and \( \pm\frac{\sqrt{11}}{6} \) is the imaginary part.

Quadratic Formula

The quadratic formula is a powerful tool for finding the solutions of any quadratic equation. Given a quadratic equation in the form \( ax^2 + bx + c = 0 \), the formula for the solutions is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula can handle quadratic equations with real or complex solutions depending on the value of the discriminant \( D = b^2 - 4ac \).

To use the quadratic formula, simply substitute the coefficients \( a \), \( b \), and \( c \) from your equation into the formula. It’s essential to correctly handle the square root, especially if \( D < 0 \), which involves imaginary numbers. The quadratic formula guarantees you will find the roots of the equation whether they are real or complex.

In our exercise, substituting \( a = 3 \), \( b = -5 \), and \( c = 3 \) into the formula yields complex roots, illustrating the versatility of the quadratic formula in solving a variety of quadratic equations.