Problem 16
Question
In Exercises \(15-22,\) use the given root to find the solution set of the polynomial equation. $$x^{4}+13 x^{2}+36=0 ; 3 i$$
Step-by-Step Solution
Verified Answer
The solution set of the polynomial equation \(x^{4}+13 x^{2}+36=0\) given the root \(3i\) is \(x=2i, -2i, 3i, -3i\).
1Step 1: Recognize the Roots
Note that the given root of the equation \(x^{4}+13 x^{2}+36=0\) is \(3i\). Complex roots of polynomial equations always come in conjugate pairs if the coefficients of the polynomial are all real numbers. In this case all of the coefficients are real numbers thus the conjugate of \(3i\) which is \(-3i\) is also a root.
2Step 2: Factorization
Knowing that \(3i\) and \(-3i\) are roots of the equation, these roots correspond to factors of \(x-3i\) and \(x+3i\) respectively. Squaring both equations gives us \((x^{2}+9)\) factors. Hence, equation can be rewritten as \((x^{2}+9)(x^{2}+a)=0\) where \(a\) is a real number. To find the value of \(a\), we expand the expression to \(x^{4}+ax^{2}+9x^{2}+9a=0\). Matching coefficients with the original equation \(x^{4}+13x^{2}+36=0\) gives \(a+9=13\) and \(9a=36\). From \(a+9=13\), we find \(a=4\).
3Step 3: Find all the Roots
Substitute \(a=4\) into equation \((x^{2}+9)(x^{2}+4)=0\) resulting in two equations \(x^{2}+9=0\) and \(x^{2}+4=0\). Solving \(x^{2}+9=0\) gives \(x=3i, -3i\) which are the given roots. Solving \(x^{2}+4=0\) gives \(x=2i, -2i\) which makes the solution of the polynomial equation \(x=2i, -2i, 3i, -3i\).
Key Concepts
Complex RootsFactorizationConjugate PairsReal Coefficients
Complex Roots
Complex roots are solutions to polynomial equations that are not real numbers. Instead, they consist of a real part and an imaginary part. In this exercise, the complex root given is \(3i\). Here, \(i\) represents the imaginary unit, which is defined as the square root of \(-1\). Complex numbers are often expressed in the form \(a + bi\), where \(a\) is the real part, and \(b\) is the imaginary part.
When solving polynomial equations with real coefficients, it's important to remember that complex roots appear in pairs known as conjugate pairs. This means if a polynomial with real coefficients has a root \(3i\), it must also have the root \(-3i\). This is because the imaginary parts need to cancel out for the polynomial to have only real coefficients.
When solving polynomial equations with real coefficients, it's important to remember that complex roots appear in pairs known as conjugate pairs. This means if a polynomial with real coefficients has a root \(3i\), it must also have the root \(-3i\). This is because the imaginary parts need to cancel out for the polynomial to have only real coefficients.
Factorization
Factorization is the process of breaking down a polynomial equation into simpler terms or factors, which when multiplied together, give the original polynomial. When you know the roots of a polynomial, you can use them to find its factors.
In this exercise, the roots \(3i\) and \(-3i\) correspond to the factors \((x - 3i)(x + 3i)\). By analyzing these, we can convert them into a product of polynomials with real coefficients. Notice that multiplying \((x - 3i)\) and \((x + 3i)\) yields \(x^2 + 9\).
Further factorizing the given polynomial, the equation is restructured into terms that are simpler to work with, allowing one to solve for other complex roots and ultimately the whole equation.
In this exercise, the roots \(3i\) and \(-3i\) correspond to the factors \((x - 3i)(x + 3i)\). By analyzing these, we can convert them into a product of polynomials with real coefficients. Notice that multiplying \((x - 3i)\) and \((x + 3i)\) yields \(x^2 + 9\).
Further factorizing the given polynomial, the equation is restructured into terms that are simpler to work with, allowing one to solve for other complex roots and ultimately the whole equation.
Conjugate Pairs
Conjugate pairs are a key concept when dealing with polynomial equations with real coefficients. This is because, when you have one complex root, you must also have its conjugate to ensure that the polynomial has real coefficients only.
In our problem, the given root is \(3i\). Its conjugate is \(-3i\). Together, they form a conjugate pair. The idea is that the imaginary parts cancel each other out in the multiplication process, leaving only real values, as seen in the factor \((x - 3i)(x + 3i) = x^2 + 9\).
Conjugate pairs protect the polynomial's integrity in terms of having real coefficients, even while accommodating complex roots.
In our problem, the given root is \(3i\). Its conjugate is \(-3i\). Together, they form a conjugate pair. The idea is that the imaginary parts cancel each other out in the multiplication process, leaving only real values, as seen in the factor \((x - 3i)(x + 3i) = x^2 + 9\).
Conjugate pairs protect the polynomial's integrity in terms of having real coefficients, even while accommodating complex roots.
Real Coefficients
A polynomial with real coefficients means that each constant multiplying the variables in the polynomial is a real number. This is crucial because the nature of the equation's roots is directly connected to this property. Whenever a polynomial equation, like the one in the exercise, displays real coefficients, any complex roots must occur in conjugate pairs.
This was why when \(3i\) was discovered as a root of \(x^4 + 13x^2 + 36 = 0\), its conjugate pair \(-3i\) was also a root. When expanded and simplified, they yield a factor with real coefficients: \(x^2 + 9\). This ensures that, despite having complex roots, the polynomial retains its real nature.
Therefore, real coefficients are a fundamental aspect ensuring that complex roots are managed and balanced properly within the polynomial structure.
This was why when \(3i\) was discovered as a root of \(x^4 + 13x^2 + 36 = 0\), its conjugate pair \(-3i\) was also a root. When expanded and simplified, they yield a factor with real coefficients: \(x^2 + 9\). This ensures that, despite having complex roots, the polynomial retains its real nature.
Therefore, real coefficients are a fundamental aspect ensuring that complex roots are managed and balanced properly within the polynomial structure.
Other exercises in this chapter
Problem 15
Find the coordinates of the vertex for the parabola defined by the given quadratic function. \(f(x)=-x^{2}-2 x+8\)
View solution Problem 16
Determine the constant of variation for each stated condition. \(y\) varies directly as \(x^{2},\) and \(y=72\) when \(x=6\)
View solution Problem 16
a. List all possible rational roots. b. Use synthetic division to test the possible rational roots and find an actual root. c. Use the root from part (b) and so
View solution Problem 16
Divide using long division. State the quotient, \(q(x),\) and the remainder, \(r(x)\). $$\frac{2 x^{5}-8 x^{4}+2 x^{3}+x^{2}}{2 x^{3}+1}$$
View solution