Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 3: Problem 23
Find all solutions of the equation. $$8 x^{3}+18 x^{2}+45 x+27=0$$
Short Answer
Expert verified
Solutions are \(-\frac{3}{2}, \frac{-3 + i \sqrt{141}}{8}, \frac{-3 - i \sqrt{141}}{8}\).
Step by step solution
01
Identify Possible Rational Roots
First, we find all potential rational roots using the Rational Root Theorem. This theorem suggests that any rational solution, expressed as a fraction \( \frac{p}{q} \), of the polynomial equation \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 = 0 \) is such that \( p \) is a divisor of the constant term \( a_0 \) and \( q \) is a divisor of the leading coefficient \( a_n \). Here, \( a_0 = 27 \) and \( a_n = 8 \). Thus, possible rational roots are from the divisors of 27 (\( \pm 1, \pm 3, \pm 9, \pm 27 \)) over the divisors of 8 (\( \pm 1, \pm 2, \pm 4, \pm 8 \)).
02
Test Rational Roots
Use synthetic division to test these possible rational roots: \( \pm 1, \pm \frac{1}{2}, \pm 2, \pm \frac{3}{2}, \pm 3, \pm \frac{9}{8}, \cdots \). After testing, we find that \( x = -\frac{3}{2} \) is a root. Upon dividing the original polynomial by \( x + \frac{3}{2} \), we reduce the polynomial to a quadratic: \( 8x^2 + 6x + 18 = 0 \).
03
Solve the Quadratic Equation
Next, solve the quadratic equation \( 8x^2 + 6x + 18 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a = 8 \), \( b = 6 \), and \( c = 18 \). We calculate the discriminant: \( b^2 - 4ac = 6^2 - 4 \times 8 \times 18 = -564 \), which is negative, indicating two complex solutions.
04
Find Complex Roots
With a negative discriminant \( -564 \), the square root becomes \( i \sqrt{564} = i \sqrt{4 \times 141} = 2i \sqrt{141} \). Therefore, the roots are \( x = \frac{-6 \pm 2i \sqrt{141}}{16} = \frac{-3 \pm i \sqrt{141}}{8} \).
05
Conclusion: All Solutions
The solutions to the original cubic equation are \( x = -\frac{3}{2} \), \( x = \frac{-3 + i \sqrt{141}}{8} \), and \( x = \frac{-3 - i \sqrt{141}}{8} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Solutions
In mathematics, complex solutions occur when solving equations with real coefficients that result in a negative discriminant under the square root when using the quadratic formula. This results in roots that involve the imaginary unit, represented by \(i\), where \(i^2 = -1\).
- In our exercise, the equation \(8x^2 + 6x + 18 = 0\) yields a discriminant of \(-564\). This negative value indicates that our solutions will be complex.- The square root of a negative number introduces the imaginary unit \(i\). Here, the square root of \(-564\) is computed as \(2i \sqrt{141}\).Ultimately, the complex solutions for the quadratic part of the original cubic equation are \(x = \frac{-3 + i \sqrt{141}}{8}\) and \(x = \frac{-3 - i \sqrt{141}}{8}\) after simplifying these expressions. Complex solutions often appear in conjugate pairs, such as the ones we found.
- In our exercise, the equation \(8x^2 + 6x + 18 = 0\) yields a discriminant of \(-564\). This negative value indicates that our solutions will be complex.- The square root of a negative number introduces the imaginary unit \(i\). Here, the square root of \(-564\) is computed as \(2i \sqrt{141}\).Ultimately, the complex solutions for the quadratic part of the original cubic equation are \(x = \frac{-3 + i \sqrt{141}}{8}\) and \(x = \frac{-3 - i \sqrt{141}}{8}\) after simplifying these expressions. Complex solutions often appear in conjugate pairs, such as the ones we found.
Synthetic Division
Synthetic division is a simplified process to divide a polynomial by a linear divisor of the form \(x - c\). This method allows us to find whether a candidate root, often identified through the Rational Root Theorem, is indeed a root of the polynomial. Using synthetic division, you can quickly test possible rational roots and reduce polynomial expressions without the extensive work involved in long division. It's especially useful when dealing with polynomials in higher degrees like cubic or quartic expressions.- In the exercise context, synthetic division was used to see if \(x = -\frac{3}{2}\) was a root of the polynomial \(8x^3 + 18x^2 + 45x + 27\).- Upon finding that \(x = -\frac{3}{2}\) is a root, synthetic division simplifies the cubic polynomial to the quadratic \(8x^2 + 6x + 18\). This step is crucial for further solving the equation by methods like the quadratic formula.
Quadratic Formula
The quadratic formula is a powerful tool in algebra used to solve equations of the form \(ax^2 + bx + c = 0\). It finds the roots of these equations by using:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]- The terms inside the square root \(b^2 - 4ac\) are known as the discriminant. The discriminant's value determines the nature of the roots.- A negative discriminant signals the presence of complex roots, as seen in the exercise where \(b^2 - 4ac = -564\). This is a key concept indicating that real number solutions are not possible, and the roots will be complex.Following these calculations, substituting appropriate values from the polynomial \(a = 8\), \(b = 6\), and \(c = 18\) into the quadratic formula provided the solutions to the remaining quadratic equation post-synthetic division. The roots are found to be complex, specifically \(x = \frac{-3 \pm i \sqrt{141}}{8}\). This method is effective and universally applicable to any quadratic equation.
