Finding The Real Root Of A Cubic Equation With Complex Roots

In the fascinating realm of algebra, polynomial equations hold a central position, their solutions, or roots, revealing profound insights into the nature of mathematical relationships. Among these equations, cubic polynomials, characterized by their highest power of three, present a captivating blend of complexity and elegance. Today, we embark on a journey to unravel the mysteries of a specific cubic equation, employing our algebraic prowess to pinpoint its elusive roots. Let's dive in!

Delving into the Cubic Equation and its Complex Roots

Our quest begins with the cubic equation x³ - 4x² + 2x + 10 = x² - 5x - 3. This equation, at first glance, appears as a jumble of terms, but beneath the surface lies a hidden structure waiting to be unveiled. To begin our exploration, we must first consolidate the terms and express the equation in its standard form. By transposing the terms from the right-hand side to the left, we arrive at the simplified form:

x³ - 5x² + 7x + 13 = 0

Now, our equation takes on a more familiar guise, a cubic polynomial set equal to zero. This form lends itself more readily to analysis and root-finding techniques. We are given a crucial piece of information: the equation possesses complex roots of the form 3 ± 2i. These complex roots, with their real and imaginary components, hint at the intricate nature of the equation's solutions. But what exactly are complex roots, and how do they influence our search for the remaining root?

Complex roots arise when the solutions to a polynomial equation involve the imaginary unit i, defined as the square root of -1. These roots always occur in conjugate pairs, meaning that if a + bi is a root, then a - bi is also a root. This fundamental property stems from the nature of polynomial equations with real coefficients. In our case, the given complex roots, 3 + 2i and 3 - 2i, beautifully exemplify this conjugate relationship. This knowledge provides a significant stepping stone in our root-finding endeavor. Knowing two of the roots allows us to deduce the third, as the total number of roots in a polynomial equation corresponds to its degree – three in the case of a cubic equation.

Employing Vieta's Formulas: A Powerful Tool

To unearth the hidden root, we turn to a powerful set of relationships known as Vieta's formulas. These formulas, named after the renowned French mathematician François Viète, establish connections between the coefficients of a polynomial and its roots. For a cubic equation of the form ax³ + bx² + cx + d = 0, Vieta's formulas state the following:

• Sum of roots: -b/a
• Sum of pairwise products of roots: c/a
• Product of roots: -d/a

These formulas provide a direct link between the coefficients of our equation and the algebraic combinations of its roots. In our specific equation, x³ - 5x² + 7x + 13 = 0, we can identify the coefficients as follows: a = 1, b = -5, c = 7, and d = 13. We already know two of the roots: 3 + 2i and 3 - 2i. Let's denote the unknown third root as r. Now, we can invoke Vieta's formulas to establish relationships involving r.

The most relevant formula for our purpose is the sum of the roots: r + (3 + 2i) + (3 - 2i) = -(-5)/1. This equation elegantly connects the unknown root r with the known complex roots and the coefficient of the x² term. By simplifying this equation, we can isolate r and reveal its value. The beauty of Vieta's formulas lies in their ability to transform a complex root-finding problem into a manageable algebraic exercise. We are on the verge of discovering the missing piece of our puzzle.

Unveiling the Real Root: A Step-by-Step Solution

Let's put Vieta's formulas into action and solve for the unknown root r. From the sum of roots formula, we have:

r + (3 + 2i) + (3 - 2i) = 5

Notice that the imaginary terms, +2i and -2i, gracefully cancel each other out, leaving us with:

r + 3 + 3 = 5

Simplifying further, we get:

r + 6 = 5

Now, a simple subtraction reveals the value of r:

r = 5 - 6 = -1

Thus, the missing real root of the cubic equation is r = -1. We have successfully navigated the complexities of the equation and unearthed its hidden root. This solution not only completes our understanding of the equation's roots but also highlights the power of Vieta's formulas in solving polynomial equations. The real root, r = -1, stands in stark contrast to the complex roots, 3 + 2i and 3 - 2i, showcasing the diverse nature of polynomial solutions.

Verification and Conclusion: Completing the Puzzle

To ensure the accuracy of our solution, let's verify that x = -1 indeed satisfies the original cubic equation x³ - 5x² + 7x + 13 = 0. Substituting x = -1 into the equation, we get:

(-1)³ - 5(-1)² + 7(-1) + 13 = -1 - 5 - 7 + 13 = 0

The equation holds true, confirming that x = -1 is indeed a root. We have successfully identified all three roots of the cubic equation: the complex roots 3 + 2i and 3 - 2i, and the real root -1. This comprehensive solution underscores the interplay between complex and real numbers in polynomial equations. The initial cubic equation, which seemed daunting at first glance, has now yielded its secrets, thanks to our understanding of complex roots and the application of Vieta's formulas.

In conclusion, the polynomial equation x³ - 4x² + 2x + 10 = x² - 5x - 3 possesses a rich mathematical structure, with roots that span both the complex and real number systems. By leveraging the information about the complex roots and employing Vieta's formulas, we successfully determined the remaining real root, x = -1. This journey through the realm of cubic equations exemplifies the power of algebraic techniques in unraveling mathematical mysteries. The solutions to polynomial equations offer a glimpse into the intricate relationships that govern mathematical expressions, and our exploration today has illuminated one such relationship in a compelling manner.

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