Problem 46 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. \(39-56:\) Find the solutions of the equation. $$ -3 x^{2}+x-5=0 $$

Step-by-Step Solution

Answer

The solutions are complex: \(x = \frac{1}{6} \pm \frac{i\sqrt{59}}{6}\).

Step 1: Identify the Components of the Quadratic Equation

The given equation is \(-3x^2 + x - 5 = 0\). This is a quadratic equation of the form \(ax^2 + bx + c = 0\), where \(a = -3\), \(b = 1\), and \(c = -5\).

Step 2: Use the Quadratic Formula

The quadratic formula is \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\). We'll use this formula to find the solutions of the equation.

Step 3: Calculate the Discriminant

Compute the discriminant \(b^2 - 4ac\). Substitute the values of \(a\), \(b\), and \(c\): \[b^2 - 4ac = 1^2 - 4(-3)(-5) = 1 - 60 = -59\]Since the discriminant is negative, the solutions will be complex numbers.

Step 4: Compute the Solutions Using the Quadratic Formula

Substitute the values of \(a\), \(b\), and the calculated discriminant into the quadratic formula:\[x = \frac{{-1 \pm \sqrt{-59}}}{-6}\]Since \(\sqrt{-59} = i\sqrt{59}\), where \(i\) is the imaginary unit, the solutions become:\[x = \frac{{-1 \pm i\sqrt{59}}}{-6}\]This can be simplified to:\[x = \frac{-1}{-6} \pm \frac{i\sqrt{59}}{-6}\]\[x = \frac{1}{6} \pm \frac{i\sqrt{59}}{6}\].

Step 5: Write the Final Solutions

The solutions to the equation \(-3x^2 + x - 5 = 0\) are:\[x = \frac{1}{6} + \frac{i\sqrt{59}}{6}\] and \[x = \frac{1}{6} - \frac{i\sqrt{59}}{6}\].

Key Concepts

Complex Solutions

Quadratic equations can sometimes have solutions that are not real numbers. When equations have solutions involving imaginary numbers, these solutions are called complex solutions. Complex solutions arise when the discriminant, which is part of the quadratic formula, is negative. This means that when we try to take the square root of a negative number, we end up with an imaginary number.
The imaginary unit is represented by the symbol \(i\), where \(i = \sqrt{-1}\). In the given equation \(-3x^2 + x - 5 = 0\), we calculated the discriminant as \(-59\). Since this number is negative, our equation will have two complex solutions.
These complex solutions are expressed in the form \(a + bi\) and \(a - bi\), where \(a\) and \(b\) are real numbers. Here, our solutions are written as \[x = \frac{1}{6} + \frac{i\sqrt{59}}{6}\] and \[x = \frac{1}{6} - \frac{i\sqrt{59}}{6}\]. With this understanding, whenever you encounter a negative discriminant, be prepared to deal with complex numbers.

Discriminant

The discriminant in a quadratic equation plays a vital role in determining the nature of the solutions. The discriminant is part of the quadratic formula and is given by \(b^2 - 4ac\). It tells us not only how many solutions there are but also what type they will be.
The value of the discriminant can provide us with *three possibilities*:

• If the discriminant is positive, there are two distinct real solutions.
• If it's zero, there is exactly one real solution (or a repeated solution).
• If the discriminant is negative, like in our problem with a value of \(-59\), there are no real solutions, but two complex solutions.

Calculating the discriminant helps us understand what to expect without actually plugging values into the equation. So, it’s always useful to start with this calculation when solving quadratic equations.

Quadratic Formula

The quadratic formula is a universal tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It is expressed as \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\). This formula provides the solutions to any quadratic equation, no matter whether they are real or complex.
The formula incorporates the discriminant \(b^2 - 4ac\), meaning that the nature of the solutions - whether they are real or complex - is determined as part of the solution process.
In our specific problem of \(-3x^2 + x - 5 = 0\), values for \(a\), \(b\), and \(c\) are \(-3\), \(1\), and \(-5\) respectively. Using the quadratic formula allows us to solve neatly and accurately for any quadratic equation you might encounter.
It is important to become comfortable with this formula because it will appear frequently in both academic and practical settings, making it essential for anyone studying math or related fields.