Problem 69 - 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 \(-1 + \frac{i\sqrt{6}}{6}\) and \(-1 - \frac{i\sqrt{6}}{6}\).

Step 1: Identify Coefficients

The quadratic equation is given by \[6x^2 + 12x + 7 = 0\] where the coefficients are \(a = 6\), \(b = 12\), and \(c = 7\).

Step 2: Calculate the Discriminant

The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(D = b^2 - 4ac\). Substitute the values: \[D = 12^2 - 4(6)(7) = 144 - 168 = -24\].

Step 3: Determine the Nature of Roots

Since the discriminant \(D = -24\) is negative, the equation has two complex conjugate roots. These can be expressed in the form \(a + bi\).

Step 4: Use the Quadratic Formula

The roots of a quadratic equation are given by the formula \[x = \frac{-b \pm \sqrt{D}}{2a}\]. Substituting the known values: \[x = \frac{-12 \pm \sqrt{-24}}{12}\].

Step 5: Simplify the Expression

First, simplify \(\sqrt{-24}\): \[\sqrt{-24} = \sqrt{-1 \times 24} = i\sqrt{24} = i\sqrt{4 \times 6} = 2i\sqrt{6}\]. Now substitute: \[x = \frac{-12 \pm 2i\sqrt{6}}{12}\].

Step 6: Divide by 2a

Divide each term by 12 to simplify: \[x = \frac{-12}{12} \pm \frac{2i\sqrt{6}}{12} = -1 \pm \frac{i\sqrt{6}}{6}\]. Thus, the solutions are \(-1 + \frac{i\sqrt{6}}{6}\) and \(-1 - \frac{i\sqrt{6}}{6}\).

Key Concepts

Quadratic Equations

A quadratic equation is a polynomial equation of degree two, which means it has the form \( ax^2 + bx + c = 0 \). In this formula, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \) being essential to maintaining the quadratic nature. The solution to a quadratic equation is generally two values of \( x \), as the equation represents a parabola in algebraic geometry. By solving the equation, you effectively find the points where the parabola intersects the x-axis, known as the roots or solutions.

There are several methods to solve quadratic equations, such as factoring, completing the square, or utilizing the quadratic formula. The quadratic formula is particularly useful when an equation cannot be easily factored. This formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
It efficiently calculates the roots by accounting for the equation's discriminant, \( b^2 - 4ac \), under the square root symbol.

Discriminant

The discriminant is a key concept in understanding the nature of the roots of a quadratic equation. For a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( D \) is calculated as \( b^2 - 4ac \). This value provides crucial information:

• If \( D > 0 \), the equation has two distinct real roots.
• If \( D = 0 \), the equation has exactly one real root, which is a repeated root.
• If \( D < 0 \), the equation has two complex conjugate roots.

In our exercise, the discriminant was calculated to be \( -24 \), a negative number. This indicates the presence of complex roots, which surfaces when you cannot take the square root of a negative number in the set of real numbers. Instead, complex numbers come into play, involving the imaginary unit \( i \), where \( i^2 = -1 \).

Complex Conjugate

Complex conjugates are pairs of complex numbers that have identical real parts but opposite imaginary parts. Complex numbers take the form \( a + bi \), where \( a \) represents the real part, and \( bi \) the imaginary part. The complex conjugate of this number is \( a - bi \).

The significance of complex conjugates lies in their roles in complex number arithmetic. When a quadratic equation like the given one results in non-real roots due to a negative discriminant, those roots are complex conjugates. This means that if one root is \(-1 + \frac{i\sqrt{6}}{6}\), then the other is \(-1 - \frac{i\sqrt{6}}{6}\). These conjugate pairs have the property that they simplify multiplication and division operations, as multiplying a number by its conjugate results in a real number.