Problem 94 - 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. $$ 4 x=6+x^{2} $$

Step-by-Step Solution

Answer

The equation has no real solutions; it has two complex solutions: \(x = 2 \pm i\sqrt{2}\).

Step 1: Rearrange the Equation

We start with the equation \(4x = 6 + x^2\). To express it in the standard quadratic form \(ax^2 + bx + c = 0\) where \(a > 0\), we need to rearrange the terms. Move everything to one side of the equation.\[x^2 - 4x + 6 = 0\] Here, \(a = 1\), \(b = -4\), and \(c = 6\).

Step 2: Calculate the Discriminant

The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\). Substitute \(a = 1\), \(b = -4\), and \(c = 6\) into the formula:\[(-4)^2 - 4 \cdot 1 \cdot 6\] Calculate the value:\[16 - 24 = -8\] The result is \(-8\), indicating that there are no real solutions, as the discriminant is negative.

Step 3: Solve the Equation

Due to the negative discriminant, the quadratic equation does not have any real solutions. However, it will have two complex solutions. We use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find these solutions. Substitute the known values:\[x = \frac{-(-4) \pm \sqrt{-8}}{2 \cdot 1}\] Simplify:\[x = \frac{4 \pm \sqrt{-8}}{2}\]\[x = \frac{4 \pm 2i\sqrt{2}}{2}\]\[x = 2 \pm i\sqrt{2}\] Therefore, the solutions are \(x = 2 + i\sqrt{2}\) and \(x = 2 - i\sqrt{2}\).

Key Concepts

Discriminant

The discriminant is a key part of the quadratic equation, often represented by the formula \(b^2 - 4ac\). It helps determine the nature of the roots of the quadratic equation, \(ax^2 + bx + c = 0\). Think of it as a tool that gives you insight about the solutions to a quadratic equation without actually solving the equation. Here's what the discriminant tells us:

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If it is zero, there is exactly one real solution (also known as a repeated or double root).
• If the discriminant is negative, as in our example where it was \(-8\), the quadratic equation has no real solutions but instead has two complex solutions.

To calculate the discriminant for the equation \(x^2 - 4x + 6 = 0\), we substitute \(a = 1\), \(b = -4\), and \(c = 6\) into the discriminant formula: \((-4)^2 - 4 \, \times \, 1 \, \times \, 6 = 16 - 24 = -8\). This negative value informs us that the solutions will be complex.

Complex Solutions

When a quadratic equation yields a negative discriminant, this indicates that the equation has complex solutions. Complex numbers feature a real part and an imaginary part, where the imaginary part involves the square root of a negative number.In the context of our quadratic equation \(x^2 - 4x + 6 = 0\), the discriminant was \(-8\). When calculating the square root of negative numbers, we use the imaginary unit \(i\), defined as \(i = \sqrt{-1}\). Thus, \(\sqrt{-8} = 2i \sqrt{2}\), factoring out \(\sqrt{4}\) as \(2\) and leaving \(i\sqrt{2}\).For this particular quadratic equation, the solutions turn out to be those involving this imaginary component: \(x = 2 \pm i\sqrt{2}\). This tells us that:

• \(x = 2 + i\sqrt{2}\) is one of the complex solutions.
• \(x = 2 - i\sqrt{2}\) is the other complex solution.

Complex solutions often occur in pairs called conjugates, representing both the addition and subtraction of the imaginary part.

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of any quadratic equation, \(ax^2 + bx + c = 0\). The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula not only provides solutions to the quadratic equation but also neatly incorporates the discriminant \(b^2 - 4ac\) as a part of the calculation for determining the nature and form of the solutions.In our case, for the equation \(x^2 - 4x + 6 = 0\), we used:

• \(a = 1\)
• \(b = -4\)
• \(c = 6\)

Substituting these into the formula: \(x = \frac{-(-4) \pm \sqrt{-8}}{2 \times 1}\), simplifies to \(x = \frac{4 \pm 2i\sqrt{2}}{2}\) leading to the solutions \(x = 2 + i\sqrt{2}\) and \(x = 2 - i\sqrt{2}\).Remember, the quadratic formula can solve any quadratic equation, and analyzing the discriminant can tell you if the solutions are real or complex before you even complete the formula.