Problem 63 - Algebra and Trigonometry

Question

Find all solutions of the equation and express them in the form \(a+b i .\) $$ 6 x^{2}+12 x+7=0 $$

Step-by-Step Solution

Answer

The solutions are \(-1 + \frac{\sqrt{6}}{6}i\) and \(-1 - \frac{\sqrt{6}}{6}i\).

Step 1: Identify the Type of Equation

The given equation is a quadratic equation in the form of \(ax^2 + bx + c = 0\) where \(a = 6\), \(b = 12\), and \(c = 7\). We will use the quadratic formula to find the solutions.

Step 2: Apply the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute \(a = 6\), \(b = 12\), and \(c = 7\) into the formula.

Step 3: Calculate the Discriminant

The discriminant \(D\) is calculated as \(b^2 - 4ac\). Substitute the values to find \(D = 12^2 - 4 \times 6 \times 7 = 144 - 168 = -24\). Since \(D < 0\), the solutions will involve complex numbers.

Step 4: Compute the Square Root of the Discriminant

Since the discriminant is negative, \(\sqrt{D} = \sqrt{-24} = \sqrt{24}i\). Further simplify \(\sqrt{24} = 2\sqrt{6}\), thus \(\sqrt{-24} = 2\sqrt{6}i\).

Step 5: Find the Roots Using the Quadratic Formula

Substitute \(b = 12\), \(\sqrt{-24} = 2\sqrt{6}i\), and \(a = 6\) back into the quadratic formula: \(x = \frac{-12 \pm 2\sqrt{6}i}{12}\). Simplify to get \(x = \frac{-12}{12} \pm \frac{2\sqrt{6}i}{12}\).

Step 6: Simplify the Expression

Simplify the expression: \(x = -1 \pm \frac{\sqrt{6}}{6}i\). Thus, the solutions are \(-1 + \frac{\sqrt{6}}{6}i\) and \(-1 - \frac{\sqrt{6}}{6}i\).

Key Concepts

Complex Numbers

Complex numbers are an extension of the real numbers and include a real part and an imaginary part. A complex number is usually written in the form \(a + bi\), where \(a\) is the real part, and \(bi\) is the imaginary part. The imaginary unit \(i\) is defined by the property \(i^2 = -1\). This means that \(i\) is the square root of \(-1\). Complex numbers are extremely useful in mathematics, especially when dealing with equations that do not have real solutions.
In the context of quadratic equations, when the discriminant is negative, the solutions will be complex numbers because taking a square root of a negative number involves the imaginary unit \(i\). For example, if the quadratic equation has a discriminant of \(-24\), then \(\sqrt{-24} = 2\sqrt{6}i\). This reflects the fact that the solutions of the quadratic equation cannot be found within the set of real numbers alone but instead, they require an extension into the complex plane.

Quadratic Formula

The quadratic formula is a powerful tool used for finding the roots of quadratic equations. Any equation of the form \(ax^2 + bx + c = 0\) can be solved using:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

This formula provides the values of \(x\) that make the quadratic equation true. It works universally for any quadratic equation, no matter whether the coefficients lead to real or complex solutions.
The "\(\pm\)" symbol in the formula indicates that there are generally two solutions, which correspond to the two directions (positive and negative) of the square root term. It's important to properly substitute all values for \(a, b,\) and \(c\) and to carefully compute each step, especially the square root and division calculations. For equations with a negative discriminant, as in the exercise given, this will result in solutions having a complex part from the term involving \(i\).

Discriminant

The discriminant is a special part of the quadratic formula. It is the portion under the square root: \(D = b^2 - 4ac\). Calculating the discriminant helps identify the nature of the roots of a quadratic equation. There are three different cases to consider:

• If \(D > 0\), the quadratic equation has two distinct real roots.
• If \(D = 0\), the quadratic equation has exactly one real root, also known as a repeated or double root.
• If \(D < 0\), the equation has two complex roots, which are conjugates of each other.

In the given exercise, the discriminant was calculated as \(-24\), which is less than zero, indicating that the solutions will be complex numbers. This is consistent with the expressions \(-1 + \frac{\sqrt{6}}{6}i\) and \(-1 - \frac{\sqrt{6}}{6}i\). The imaginary unit \(i\) emphasizes that these solutions lie in the complex plane rather than being purely real numbers.