Solving Cubic Equation X³ - 2x² - 7x + 5 With Root X = 1/2 - √3

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In this comprehensive guide, we will delve into the process of solving the cubic equation x³ - 2x² - 7x + 5 = 0, given that one of the roots is x = 1/2 - √3. This problem combines algebraic manipulation, the application of the quadratic formula, and a deep understanding of polynomial roots. Our primary focus will be on presenting a step-by-step solution that is both clear and accessible. We aim to enhance your problem-solving skills and provide a solid foundation for tackling similar mathematical challenges. By breaking down the problem into manageable segments, we ensure that each step is thoroughly explained, enabling you to grasp the underlying concepts and techniques effectively.

Understanding the Problem

When we are given a cubic equation like x³ - 2x² - 7x + 5 = 0, and one of its roots, our task is to find the remaining roots. Knowing one root is particularly helpful because it allows us to reduce the cubic equation into a simpler quadratic equation. This simplification is achieved through polynomial division or synthetic division. In this specific case, we are given that one root is x = 1/2 - √3. The presence of a radical suggests that another root might be its conjugate, which is a crucial insight that can significantly simplify our calculations. Understanding these nuances is the first step in effectively solving the problem. This initial assessment not only guides our approach but also ensures we are using the most efficient methods to arrive at the solution. The ability to recognize such patterns and relationships is a hallmark of strong mathematical proficiency.

Utilizing the Conjugate Root Theorem

The conjugate root theorem is a fundamental concept in algebra, stating that if a polynomial equation with real coefficients has a complex or irrational root of the form a + √b, then its conjugate a - √b is also a root. In our case, one root is x = 1/2 - √3. Applying the conjugate root theorem, we can deduce that x = 1/2 + √3 is also a root of the equation. This theorem dramatically simplifies the problem because instead of dealing with a cubic equation directly, we can now work with the quadratic factor formed by these two roots. The conjugate root theorem is not just a shortcut; it is a powerful tool that highlights the inherent symmetry in polynomial equations with real coefficients. Recognizing and applying this theorem is a key step in efficiently solving such problems, saving time and reducing the complexity of calculations.

Constructing the Quadratic Factor

Now that we have identified two roots, x = 1/2 - √3 and x = 1/2 + √3, we can construct a quadratic factor. To do this, we first express the roots as (x - (1/2 - √3)) and (x - (1/2 + √3)). Multiplying these two factors will give us the quadratic expression. This process involves careful algebraic manipulation to ensure accuracy. Let’s perform the multiplication:

(x - (1/2 - √3)) * (x - (1/2 + √3))

= (x - 1/2 + √3) * (x - 1/2 - √3)

= (x - 1/2)² - (√3)²

= x² - x + 1/4 - 3

= x² - x - 11/4

To eliminate the fraction, we multiply the entire quadratic by 4, resulting in 4x² - 4x - 11. This quadratic factor is a significant step forward, as it allows us to reduce the cubic equation into a more manageable form. The ability to construct factors from roots is a critical skill in polynomial algebra, enabling us to solve higher-degree equations by breaking them down into simpler components.

Dividing the Cubic Equation

With the quadratic factor 4x² - 4x - 11 in hand, we can now divide the original cubic equation x³ - 2x² - 7x + 5 by this factor. Polynomial division will give us the remaining linear factor, which will lead us to the third root. The process of polynomial division involves carefully aligning terms and subtracting them systematically. It may seem tedious, but it is a reliable method to reduce the degree of the polynomial. Let's perform the polynomial division:

 x/4  - 1/4
4x² - 4x - 11 | x³  - 2x²  - 7x  + 5
              -(x³  -  x² - 11x/4)
              ---------------------
                   -x² - 17x/4 + 5
                   -(-x² +  x  + 11/4)
                   ---------------------
                        -21x/4 + 9/4

Upon performing the division, we find that the quotient is x/4 - 1/4 and the remainder is 0, which confirms that 4x² - 4x - 11 is indeed a factor of the cubic equation. Thus, the other factor is x/4 - 1/4. To simplify, we multiply this by 4 to get x - 1. The result of this division is a critical step in solving the cubic equation, providing us with the final linear factor necessary to find all the roots.

Finding the Remaining Root

The division yielded the linear factor x - 1. To find the remaining root, we set this factor equal to zero:

x - 1 = 0

Solving for x, we get:

x = 1

Thus, the third root of the cubic equation is x = 1. This simple linear equation gives us the final piece of the puzzle, completing the set of roots for the given cubic equation. Finding the roots through factors is a cornerstone of polynomial algebra, and this step underscores the importance of polynomial division in simplifying and solving complex equations.

Summarizing the Roots

We have successfully found all three roots of the cubic equation x³ - 2x² - 7x + 5 = 0. These roots are:

  1. x = 1/2 - √3 (given)
  2. x = 1/2 + √3 (conjugate root)
  3. x = 1 (from the linear factor)

These roots represent the complete solution set for the cubic equation. Summarizing the roots in this manner provides a clear and concise conclusion to the problem-solving process. It not only confirms that we have found all possible solutions but also highlights the relationships between the roots, such as the conjugate pair arising from the irrational root. This comprehensive approach ensures a thorough understanding of the solution and the methods used to obtain it.

Verification

To ensure the correctness of our solution, we can verify the roots by substituting them back into the original equation x³ - 2x² - 7x + 5 = 0. If the equation holds true for each root, our solution is correct. This step is a crucial part of problem-solving, providing a check against potential errors and reinforcing the accuracy of our methods. Let's verify each root:

  1. For x = 1/2 - √3:

    This root was given, and we used it to derive the other roots, so it should satisfy the equation.

  2. For x = 1/2 + √3:

    Substituting this into the equation will also yield 0, as it is the conjugate of the given root.

  3. For x = 1:

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

    There seems to be a slight calculation error here. Let's re-evaluate the steps.

Correcting the Error

Upon closer inspection, it seems there was a mistake in the polynomial division. The remainder should indeed be 0 for the division to be accurate. Let's correct the division process:

The correct polynomial division should yield the quotient x - 1/4 instead of x/4 - 1/4. Therefore, the correct linear factor is x - 1, which gives us the root x = 1. Let's verify again:

For x = 1:

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

It seems there is still a mistake. Let's go back to the polynomial division and check that part again. It's important to get this right to make sure the solution is correct.

Performing Polynomial Long Division

We divide x³ - 2x² - 7x + 5 by 4x² - 4x - 11:

       1/4 x - 1/4
4x²-4x-11 | x³ - 2x² - 7x + 5
          -(x³ -  x² - 11/4 x)
          ---------------------
              -x² - 17/4 x + 5
              -(-x² +  x  + 11/4)
              ---------------------
                   -21/4 x + 9/4

There seems to be a persistent error in the calculation. The mistake lies in the long division process. The correct division is: (

(x³ - 2x² - 7x + 5) / (x² - x - 11/4)

Multiply the second expression by 4, we can make it as: (4x² - 4x -11). Then,

Let's do the long divison again:

 x/4 - 1/4
4x² - 4x - 11 | x³ - 2x² - 7x + 5
 - (x³ - x² - (11/4)x)
 --------------------
 -x² - (17/4)x + 5
 - (-x² + x + 11/4)
 --------------------
 (-21/4)x + 9/4

So, there still seem some mistakes. It suggests that given root is not correct. Let's start from scratch, and try rational root theorem,

Possible rational roots: ±1, ±5. If x = 1: 1 - 2 - 7 + 5 = -3 ≠ 0. If x = -1: -1 - 2 + 7 + 5 = 9 ≠ 0. If x = 5: 125 - 50 - 35 + 5 = 45 ≠ 0. If x = -5: -125 - 50 + 35 + 5 = -135 ≠ 0.

Thus, there is no integer root. Then, the root should be wrong, Thus, the given root x = 1/2 - √3 is incorrect, or the question has a typo. With correct root, we can complete long division and verification. Due to the incorrect original question, we can't find exact answer.

Final Conclusion

Due to a likely error in the given root or the equation itself, we encountered persistent inconsistencies during the verification and polynomial division processes. The rational root theorem did not yield any integer roots, further suggesting a potential issue with the problem statement. While we have thoroughly explored the steps required to solve such a problem, the accuracy of our final solution depends on the correctness of the initial information. If the equation and root are accurate, the methods described above should lead to a valid solution once the errors in calculation are rectified. However, with the information at hand, we can only conclude that there is a discrepancy that prevents us from finding a definitive solution.