To solve the partial fraction of -x + 1/(x^2 + 2x + 4), we need to decompose the expression into simpler fractions. First, we factor the denominator of the fraction x^2 + 2x + 4.
The quadratic equation x^2 + 2x + 4 does not factorize neatly. So, we use the quadratic formula which states that the roots of the quadratic equation ax^2 + bx + c = 0 are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation x^2 + 2x + 4 = 0, a = 1, b = 2, and c = 4. Plugging these values into the quadratic formula, we have:
x = (-2 ± √(2^2 - 4(1)(4))) / (2(1))
x = (-2 ± √(4 - 16)) / 2
x = (-2 ± √(-12)) / 2
x = (-2 ± √(12)i) / 2
x = (-2 ± 2√(3)i) / 2
Simplifying further gives:
x = -1 ± √(3)i
Since the roots are complex, we can write the quadratic equation as:
x^2 + 2x + 4 = (x - (-1 + √(3)i))(x - (-1 - √(3)i))
So, the partial fraction decomposition will be in the form:
-1x + 1/(x^2 + 2x + 4) = A/(x - (-1 + √(3)i)) + B/(x - (-1 - √(3)i))
We need to find the values of A and B. To do this, we can add the fractions on the right side of the equation by finding a common denominator:
-1x + 1/(x^2 + 2x + 4) = [A(x - (-1 - √(3)i)) + B(x - (-1 + √(3)i))] / [(x - (-1 + √(3)i))(x - (-1 - √(3)i))]
Simplifying further:
-1x + 1/(x^2 + 2x + 4) = (Ax + A(1 + √(3)i) + Bx + B(1 - √(3)i)) / [x^2 - x(-1 + √(3)i) - x(-1 - √(3)i) + (-1 + √(3)i)(-1 - √(3)i)]
-1x + 1/(x^2 + 2x + 4) = (Ax + A + A√(3)i + Bx + B - B√(3)i) / [x^2 + x - √(3)xi + x + √(3)xi - 3]
-1x + 1/(x^2 + 2x + 4) = (Ax + A + Bx + B) / (x^2 + 2x - 2)
From here, we can equate the numerators on both sides to find the values of A and B:
-1x + 1 = Ax + Bx + A + B
Comparing the coefficients of x and the constants on both sides:
A + B = -1
A + B = 1
Since the coefficients of x and the constants are equal, we can conclude that A + B = -1 = 1. This is not possible, so we made an error in the decomposition. Let's correct it.
We need to verify the factorization of the denominator x^2 + 2x + 4 again using the quadratic formula:
x = (-2 ± √(2^2 - 4(1)(4))) / (2(1))
x = (-2 ± √(4 - 16)) / 2
x = (-2 ± √(-12)) / 2
x = (-2 ± 2√3i) / 2
x = -1 ± √3i
The factorization of the denominator is indeed (x - (-1 + √3)i)(x - (-1 - √3)i). Let's proceed with the correct decomposition:
-1x + 1/(x^2 + 2x + 4) = A/(x - (-1 + √3)i) + B/(x - (-1 - √3)i)
Multiplying the fractions on the right side by their respective denominators:
-1x + 1/(x^2 + 2x + 4) = A * [(x - (-1 - √3)i) / [(x - (-1 + √3)i)(x - (-1 - √3)i)]] + B * [(x - (-1 + √3)i) / [(x - (-1 + √3)i)(x - (-1 - √3)i)]]
Simplifying further:
-1x + 1/(x^2 + 2x + 4) = A(x - (-1 - √3)i) / (x^2 + 2x + 4) + B(x - (-1 + √3)i) / (x^2 + 2x + 4)
Now, we can equate the numerators on both sides:
1 = A(x - (-1 - √3)i) + B(x - (-1 + √3)i)
Expanding the terms:
1 = Ax - A(-1 - √3)i + Bx - B(-1 + √3)i
Rearranging the terms:
1 = (A + B)x + A + B + A√3i - B√3i
Comparing the coefficients of x and the constants on both sides:
A + B = 0 (coefficient of x)
A + B = 1 (constant term)
A√3i - B√3i = 0 (coefficient of √3i)
From the first equation, A = -B. Substituting this into the second equation:
-B + B = 1
0 = 1
This is not possible. The error lies in the original decomposition. Let's correct it one more time.
We factorize the denominator again:
x^2 + 2x + 4 = (x + 1 + √3i)(x + 1 - √3i)
So, the correct partial fraction decomposition is:
-1x + 1/(x^2 + 2x + 4) = A/(x + 1 + √3i) + B/(x + 1 - √3i)
We need to find the values of A and B. To do this, we can add the fractions on the right side:
-1x + 1/(x^2 + 2x + 4) = [A(x + 1 - √3i) + B(x + 1 + √3i)] / [(x + 1 + √3i)(x + 1 - √3i)]
Simplifying further:
-1x + 1/(x^2 + 2x + 4) = (Ax + A + A√3i + Bx + B + B√3i) / [(x + 1 + √3i)(x + 1 - √3i)]
-1x + 1/(x^2 + 2x + 4) = (Ax + Bx + A + B + A√3i + B√3i) / (x^2 + x + x + 1 - √3ix + 1 + √3ix - √3ix - √3i + √3ix - √3i - 3)
-1x + 1/(x^2 + 2x + 4) = (Ax + Bx + A + B + A√3i + B√3i) / (x^2 + 2x - 2)
From here, we can equate the numerators on both sides to find the values of A and B:
-1x + 1 = Ax + Bx + A + B + A√3i + B√3i
Comparing the coefficients of x and the constants on both sides:
A + B = -1
A√3i + B√3i = 1
From the equation A + B = -1, we have A = -B. Substituting this into the equation A√3i + B√3i = 1:
-A√3i + B√3i = 1
Simplifying further, we have:
√3i(B - A) = 1
√3i(B + A) = 1
From these equations, we can conclude that B - A = 1/√3i and B + A = √3i. Simplifying further:
B - A = 1/√3i ---> Multiplying both sides by √3i gives B√3i - A√3i = 1
B + A = √3i
Adding these equations together, we have:
2B√3i = 1 + √3i
Dividing both sides by 2√3i, we get:
B = (1 + √3i)/(2√3i)
To simplify this expression, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of 2√3i, which is -2√3i:
B = (1 + √3i)/(2√3i) * (-2√3i)/(-2√3i)
B = (-2√3i - 2√3)/(12i)
B = (-√3 - 1)/(6i)
Similarly, subtracting the two equations B - A = 1/√3i and B + A = √3i gives:
2A = √3i - 1/√3i
Multiplying both sides by √3i gives:
2A√3i = 3i - 1
Dividing both sides by 2√3i, we get:
A = (3i - 1)/(2√3i)
To simplify this expression, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of 2√3i, which is -2√3i:
A = (3i - 1)/(2√3i) * (-2√3i)/(-2√3i)
A = (6i√3 - 2√3)/(6i)
A = (√3 - 1)/(3i)
So, the partial fraction decomposition of -x + 1/(x^2 + 2x + 4) is:
-x + 1/(x^2 + 2x + 4) = [(-√3 - 1)/(6i)]/(x + 1 + √3i) + [(√3 - 1)/(3i)]/(x + 1 - √3i)
Solve the partial fraction -1X+1/(X^2+2X+4)
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