Problem 99 - College Algebra Essentials

Question

Solve each equation in Exercises \(83-108\) by the method of your choice. $$ x^{2}-6 x+13=0 $$

Step-by-Step Solution

Answer

The solution to the equation \(x^{2}-6 x+13=0\) are two complex roots \(x = 3+ 2i\) and \(x = 3 - 2i\).

Step 1: Identifying Coefficients

The coefficients of the quadratic equation are identified as \(a=1\), \(b=-6\), and \(c=13\).

Step 2: Using Quadratic Formula

We put the coefficient values into the quadratic formula like this: \(x = \frac{-(-6) \pm \sqrt{(-6)^{2} - 4(1)(13)}}{2(1)}\). This simplifies to \(x = \frac{6 \pm \sqrt{36 - 52}}{2}\).

Step 3: Determining Roots

The discriminant is less than zero (\(\sqrt{-16}\)), thus the roots of the equation are complex numbers: \(x = \frac{6 \pm \sqrt{-16}}{2}\) = \(3 \pm \sqrt{-4}\) = \(3 \pm 2i\)

Key Concepts

Quadratic Formula

When faced with a quadratic equation, such as \(x^{2}-6x+13=0\), one tool that is universally useful is the quadratic formula. The formula is \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation in the form \(ax^{2} + bx + c = 0\).

To apply the quadratic formula, you first identify the coefficients from the equation as the solution provided: \(a = 1\), \(b = -6\), and \(c = 13\). Plugging these values into the quadratic formula, you will deal with operations including squaring, multiplication, and finding the square root, ultimately leading you to the solutions of the equation.

Discriminant in Quadratics

A key element of the quadratic formula is the discriminant, which is the component under the square root sign, expressed as \(D = b^{2} - 4ac\). The discriminant tells us about the nature of the roots of the quadratic equation without actually having to calculate them. If \(D > 0\), the equation has two distinct real roots. If \(D = 0\), there's exactly one real root, and if \(D < 0\), the equation has complex roots.

In your example, after placing the coefficients into the discriminant formula, \(D = (-6)^{2} - 4(1)(13)\), you find that \(D = 36 - 52 = -16\), which is less than zero. This immediately indicates that the roots of the equation will not be real numbers but rather complex.

Complex Numbers

When the discriminant of a quadratic equation is negative, as seen with \(D = -16\), this leads us to the realm of complex numbers. A complex number is of the form \(a + bi\), where \(a\) is the real part, \(b\) is the imaginary part, and \(i\) is the imaginary unit, satisfying \(i^{2} = -1\).

For the given equation, once you've determined that the discriminant is negative, you know that the roots take the form of complex numbers. By simplifying \(3 \pm \sqrt{-4}\), we use the fact that \(\sqrt{-1} = i\) to write the roots as \(3 \pm 2i\). This pair of complex numbers are the solutions to the quadratic equation when no real solutions exist.