Problem 44 - Algebra and Trigonometry with Analytic Geometry

Question

Find the solutions of the equation $$ x^{2}+3 x+6=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{-3 \pm i\sqrt{15}}{2} \).

Step 1: Identify the Equation Type

The given equation is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 3 \), and \( c = 6 \). We need to use the quadratic formula to find the roots.

Step 2: Recall the Quadratic Formula

The quadratic formula for solving \( ax^2 + bx + c = 0 \) is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). We will apply this formula using our values of \( a \), \( b \), and \( c \).

Step 3: Calculate the Discriminant

The discriminant of a quadratic equation is given by \( b^2 - 4ac \). For our equation, substitute \( b = 3 \), \( a = 1 \), and \( c = 6 \) into the expression: \( (3)^2 - 4 \times 1 \times 6 = 9 - 24 = -15 \).

Step 4: Determine the Nature of the Roots

Since the discriminant is \(-15\), which is less than zero, the equation does not have real roots. Instead, it has two complex roots.

Step 5: Solve for the Imaginary Roots

Using the quadratic formula, substitute \( a = 1 \), \( b = 3 \), and the discriminant of \(-15\). The solutions are \( x = \frac{-3 \pm \sqrt{-15}}{2} = \frac{-3 \pm i\sqrt{15}}{2} \).

Step 6: Final Roots

The final solutions for the equation are \( x = \frac{-3 + i\sqrt{15}}{2} \) and \( x = \frac{-3 - i\sqrt{15}}{2} \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find solutions to quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). To find the roots of such equations, you can use the quadratic formula:

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

This formula works with any quadratic equation and provides both real and complex solutions, dependent on the discriminant. By substituting the values of \( a \), \( b \), and \( c \) from the equation into this formula, you can solve for \( x \). Remember the "\( \pm \)" in the formula, which indicates that there are generally two solutions for \( x \). These solutions represent the x-values where the quadratic graph crosses the x-axis, or if the roots are complex, where the graph would conceptually intersect the plane of real and imaginary numbers.

Complex Roots

Complex roots occur when the solutions to a quadratic equation are not real numbers. This happens when the discriminant (the value inside the square root in the quadratic formula) is negative.
In our problem, the discriminant was \(-15\), which indicates the presence of complex roots. Complex numbers are numbers that have both a real and an imaginary component. They are expressed in the form \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. The letter 'i' stands for the imaginary unit, which is the square root of \(-1\).
For the exercise, our solution turned out to be:

• \( x = \frac{-3 \pm i\sqrt{15}}{2} \)

This means the equation has two complex roots: \( \frac{-3 + i\sqrt{15}}{2} \) and \( \frac{-3 - i\sqrt{15}}{2} \). Complex roots always come in conjugate pairs, as seen here.

Discriminant Calculation

Calculating the discriminant is a crucial step to understanding the nature of the roots of a quadratic equation. The discriminant is determined using the formula:

• \( b^2 - 4ac \)

This value is taken from the quadratic equation's coefficients \( a \), \( b \), and \( c \). In our case, \( a = 1 \), \( b = 3 \), and \( c = 6 \).
When these values are plugged into the formula, we calculate:

• \( (3)^2 - 4 \times 1 \times 6 = 9 - 24 = -15 \)

This negative discriminant signifies that the quadratic equation does not intersect the x-axis and thus has no real roots, hinting at complex solutions instead.

Nature of Roots

Understanding the nature of roots is essential for interpreting the solutions of a quadratic equation. The nature is determined primarily by the discriminant:

• If the discriminant is positive, there are two distinct real roots.
• If the discriminant is zero, there is exactly one real root, or a repeated root.
• If the discriminant is negative, as in our exercise, the roots are complex.

In the given problem, we found the discriminant to be \(-15\), a negative value, which conclusively leads to the roots being complex. Hence, the roots were not real numbers but complex numbers, and they appear as conjugate pairs. This concept helps us understand the deeper nature of the graph related to the quadratic equation, particularly its roots' behavior within the real and imaginary coordinate plane.