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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 2, Problem 99

Solve each equation in Exercises 83–108 by the method of your choice. x26x+13=0x^2 - 6x + 13 = 0

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1
Identify the type of equation: This is a quadratic equation in the form \(x^2 - 6x + 13 = 0\), where \(a = 1\), \(b = -6\), and \(c = 13\).
Calculate the discriminant using the formula \(\Delta = b^2 - 4ac\). Substitute the values to get \(\Delta = (-6)^2 - 4(1)(13)\).
Evaluate the discriminant to determine the nature of the roots: if \(\Delta > 0\), there are two real roots; if \(\Delta = 0\), one real root; if \(\Delta < 0\), two complex roots.
Since the discriminant is less than zero, use the quadratic formula to find the complex roots: \(x = \frac{-b \pm \sqrt{\Delta}}{2a}\).
Substitute \(a\), \(b\), and \(\Delta\) into the quadratic formula and simplify to express the solutions in terms of real and imaginary parts.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Quadratic Equations

A quadratic equation is a second-degree polynomial equation in the form ax² + bx + c = 0, where a ≠ 0. It represents a parabola when graphed and can have zero, one, or two real solutions depending on the discriminant.
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Discriminant and Nature of Roots

The discriminant, given by b² - 4ac, determines the nature of the roots of a quadratic equation. If it is positive, there are two distinct real roots; if zero, one real root; and if negative, two complex conjugate roots.
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Solving Quadratic Equations Using the Quadratic Formula

The quadratic formula x = (-b ± √(b² - 4ac)) / (2a) provides the solutions to any quadratic equation. It is especially useful when factoring is difficult or impossible, and it accounts for all types of roots based on the discriminant.
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Solving Quadratic Equations Using The Quadratic Formula