Conjugate Root Theorem is a fundamental concept in algebra that applies to polynomials with real coefficients. It states that if a polynomial has a complex number as a root, its conjugate must also be a root. This theorem ensures that complex roots of polynomials always occur in conjugate pairs when the coefficients of the polynomial are real.
In this article, we will discuss Conjugate Root Theorem in detail.
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What is the Conjugate Root Theorem?
Conjugate Root Theorem states that if a polynomial has real coefficients and a non-real complex number a+bi (where i is the imaginary unit and b ≠ 0) as a root, then its complex conjugate a − bi is also a root of the polynomial.
In simpler terms, if a polynomial with real coefficients has a complex root, the complex conjugate of that root must also be a root of the polynomial.
For example:
- If 2+3i is a root of a polynomial with real coefficients, then 2−3i must also be a root.
Statement of Conjugate Root Theorem
In other words, if P(x) is a polynomial with real coefficients, and a + bi (where a and b are real numbers, and i is the imaginary unit) is a root of the polynomial, then its conjugate a − bi is also a root of P(x).
Example of Conjugate Roots
Consider the polynomial P(x) = x2 + 1.
The roots of this polynomial are x = +-i.
According to the Conjugate Root Theorem, since i is a root, its conjugate -i must also be a root, which is indeed confirmed.
Examine the polynomial P(x) = x2 - 4x + 13.
The roots can be determined using the quadratic formula:
x = 4 + ((-4)2 - 4(1)(13))1/2 / 2(1)
= 4 + (16 - 52)1/2 / 2
= 4 + (-36)1/2 /2 = 2 + 3i or 2 - 3i
In this case, the roots are 2 + 3i and 2 - 3i, which are complex conjugates, consistent with the Conjugate Root Theorem..
Proof of Conjugate Root Theorem
Let's walk through the proof to understand why this theorem holds.
- Start with a Polynomial: Consider a polynomial P(x) with real coefficients: where all ai are real numbers.
P(x) = an xn + a(n-1)x(n-1) + ... + a1 x + a0
- Assume a Complex Root: Now let z = a + bi where ‘a’ and ‘b’ are real and ‘i’ is the imaginary unit be a root of P(x).
- Conjugate the Polynomial: By using the complex conjugate of the whole polynomial we take into account of the fact that the coefficients a_i are real numbers that’s why they don’t change when the conjugate is taken:
P(z') = an z'n + an-1 z'n-1 + ... + a1z' + a0 = 0
Therefore, z' is also a root of the polynomial.
How to Apply Conjugate Root Theorem?
Understanding how to apply the Conjugate Root Theorem in problem-solving is crucial. Here’s a step-by-step guide.
Problem: Determine the roots of the polynomial P(x) = x² - 6x + 25.
Solution:
Step 1: Identify the coefficients of the polynomial.
- The coefficients are a = 1, b = -6, and c = 25.
Step 2: Apply the quadratic formula:
- x = (-b ± √(b² - 4ac)) / (2a).
Step 3: Substitute the identified values:
x = -(-6) ± √((-6)² - 4(1)(25)) / (2(1)) = 6 ± √(36 - 100) / 2 = 6 ± √(-64) / 2 = 3 ± 4i.
Result: The roots are 3 + 4i and 3 - 4i.
Solved Examples
Q 1. Consider the polynomial P(x) = x² - 6x + 10, and determine its roots.
Solution:
Applying the quadratic formula:
x = -(-6) ± √((-6)² - 4(1)(10)) / (2(1))
x = 6 ± √(36 - 40) / 2
x = 6 ± √(-4) / 2
x = 6 ± 2i / 2
x = 3 ± i
The roots are 3 + i and 3 - i. According to the Conjugate Root Theorem, since 3 + i is a root, it follows that 3 - i must also be a root.
Q 2. Examine whether the polynomial P(x) = x³ - 3x² + 4x - 2 possesses any complex roots, and if so, identify them.
Solution:
To begin, we will check for a real root by evaluating potential values. x = 1 satisfies the polynomial:
1³ - 3(1)² + 4(1) - 2 = 0
Thus, x - 1 is a factor of P(x). By dividing the polynomial by x - 1, we obtain:
P(x) = (x - 1)(x² - 2x + 2)
Next, we solve x² - 2x + 2 = 0 using the quadratic formula:
x = 2 ± √(4 - 8) / 2
x = 1 ± i
The roots are 1 ± i, confirming that the polynomial indeed has complex roots, which are conjugates.
Q 3. If a polynomial P(x) = x4 + 2x3 + 5x2 + 2x + 6 has one root as 2 + i, find all other roots.
Solution:
By the Conjugate Root Theorem, 2 − i must also be a root. To find the other two roots, factor out the quadratic formed by these roots from P(x):
(x − (2 + i))(x − (2 − i)) = (x − 2 − i)(x − 2 + i) = (x − 2) 2 + 1 = x2 − 4x + 5
Now, divide P(x) by −4x + 5 to get the remaining quadratic factor. After the division, we can solve for the remaining roots, which will be real since the quotient is another quadratic polynomial with real coefficients.
Q 4. Identify the roots of the polynomial P(x) = x2 - 2x + 5.
Solution:
Apply the quadratic formula:
x = -(-2) + ((-2)^2 - 4(1)(5))1/2 / 2(1)
= 2 + (4 - 20)1/2 / 2
= 2 + (-16)1/2 / 2 = 1 + 2i or 1 - 2i
The roots are 1 + 2i and 1 - 2i.
Q 5. Given that 2 + 3i is a root of P(x) = x3 - 6x2 + 25x - 50, determine the remaining roots.
Solution:
According to the Conjugate Root Theorem, 2 - 3i is also a root. Conduct polynomial division of P(x) by (x - (2 + 3i))(x - (2 - 3i)) = (x2 - 4x + 13).
Dividing P(x) by x2 - 4x + 13 yields a quotient of x - 2, which represents the third root.
The roots are 2 + 3i, 2 - 3i, and 2.
Q 6. Determine the roots of P(x) = x2 + 4x + 13.
Solution:
Utilize the quadratic formula:
x = -4 + (42 - 4(1)(13))1/2 / 2(1)
= -4 + (16 - 52)1/2 / 2
= -4 (-36)1/2 / 2 = -2 + 3i or -2 - 3i
The roots are -2 + 3i and -2 - 3i.
Q 7. Determine the roots of the polynomial P(x) = x² + 6x + 10.
Solution:
Utilize the quadratic formula:
x = -6 ± √(6² - 4(1)(10)) / (2(1))
= -6 ± √(36 - 40) / 2
= -6 ± √(-4) / 2
= -3 ± i
The roots are -3 + i and -3 - i.
Q 8. Given that 4 - 5i is a root of P(x) = x³ - 12x² + 58x - 85, identify the remaining roots.
Solution:
According to the Conjugate Root Theorem, 4 + 5i is also a root. Conduct polynomial division of P(x) by (x - (4 - 5i))(x - (4 + 5i)) = (x² - 8x + 41).
Dividing P(x) by x² - 8x + 41 yields a quotient of x - 2, which represents the third root.
The roots are 4 - 5i, 4 + 5i, and 2.
Q 9. Find the roots of the polynomial P(x) = x² - 10x + 34.
Solution:
Apply the quadratic formula:
x = 10 ± √((-10)² - 4(1)(34)) / (2(1))
x = 10 ± √(100 - 136)/2
x = 10 ± √(-36)/2
x = 5 ± 3i
The roots are 5 + 3i and 5 - 3i.
Q 10. Identify the roots of the polynomial P(x) = x4 - 4x3 + 6x2 - 4x + 13.
Solution:
This polynomial can be factored through grouping. Initially, we explore possible quadratic factors:
P(x) = (x2 - 2x + 2)(x2 - 2x + 7)
Applying the quadratic formula to each quadratic factor:
For x2 - 2x + 2:
x = 2 + (4 - 8)1/2 / 2 = 1 + i
For x2 - 2x + 7:
x = 2 + (4 - 28)1/2 / 2 = 1 + (6i)1/2
The roots are 1 + i, 1 - i, 1 + 6i, and 1 - (6i)1/2
Conjugate Root Theorem: Practice Questions
Question 1: Identify the roots of the polynomial P(x) = x2 + 4x + 13.
Question 2: It is known that 3 + 2i is a root of the polynomial P(x) = x3 - 9x2 + 27x - 35. Proceed to find the other roots of this polynomial.
Question 3: Additionally, ascertain the roots of P(x) = x2 - 4x + 5.
Question 4: Next, determine the roots of P(x) = x2 + 10x + 25.
Question 5: Given that 1 - 3i is a root of the polynomial P(x) = x3 - 3x2 + 9x - 10, find the remaining roots.
Question 6: Identify the roots of P(x) = x4 - 5x3 + 11x2 - 5x + 25.
Question 7: Solve for the roots of P(x) = x2 + 6x + 10.
Question 8: Find the roots of P(x) = x2 + 2x + 2. It is also
Question 9: Given that 4 + i is a root of the polynomial P(x) = x3 - 12x2 + 48x - 65; thus, find the other roots.
Question 10: Finally, determine the roots of P(x) = x4 + 4x2 + 4.
Answer Key
- 2 + i, 2 - i
- 3 + 2i, 3 - 2i, 3
- 2 + i, 2 - i
- -5, -5
- 1 + 3i, 1 - 3i, 1
- 2 + i, 2 - i, 1, 1
- -3 + i, -3 - i
- -1 + i, -1 - i
- 4 + i, 4 - i, 4
- 1 + i, 1 - i, -1 + i, -1 - i
Conclusion
Conjugate root theorem is an essential theorem in algebra with the capability of easing the process of finding the root of the polynomial and to understand polynomials with real coefficients. Besides helping in polynomial factorization, this theorem proves useful to make sure complex roots are always in conjugate pairs; the theorem has applications in engineering, physics, and signal processing.
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