Write a quadratic equation with roots 2 - 3i and 2 + 3i.
I just need confirmation I did it correctly. I got y = x^2-4x+13
Your calculations are correct
(sum of roots = 2+3i + 2-3i = 4
product of roots = (2+3i)(2-3i) = 4-(-9) = 13 )
It said to write the quadratic equation, which is
x^2 - 4x + 13 = 0
you had y = x^2 - 4x + 13 , which is the quadratic function.
To write a quadratic equation with the given roots, we can use the fact that complex roots always come in conjugate pairs. In this case, the roots are 2 - 3i and 2 + 3i.
To find the equation, we need to multiply the factors corresponding to each root together. So the factor corresponding to 2 - 3i is (x - (2 - 3i)) = (x - 2 + 3i), and the factor corresponding to 2 + 3i is (x - (2 + 3i)) = (x - 2 - 3i).
Now, multiply the two factors together:
(x - 2 + 3i)(x - 2 - 3i)
Using the difference of squares formula, (a - b)(a + b) = a^2 - b^2, we can simplify this expression:
(x - 2)^2 - (3i)^2
(x - 2)^2 - 9i^2
(x - 2)^2 - 9(-1)
(x - 2)^2 + 9
Therefore, the quadratic equation with roots 2 - 3i and 2 + 3i is:
y = (x - 2)^2 + 9
So, the equation y = x^2 - 4x + 13 is incorrect. The correct equation is y = (x - 2)^2 + 9.
To confirm if you have correctly written the quadratic equation with roots 2 - 3i and 2 + 3i, you can use the fact that complex roots occur in conjugate pairs for quadratic equations with real coefficients.
Let's start by assuming the quadratic equation is of the form:
y = ax^2 + bx + c
Next, we substitute the given roots into the equation:
For the root 2 - 3i:
2 - 3i = a(2 - 3i)^2 + b(2 - 3i) + c
To simplify, we use the fact that (a + bi)^2 = a^2 + 2abi - b^2:
2 - 3i = a(4 - 12i + 9i^2) + b(2 - 3i) + c
2 - 3i = a(4 - 12i - 9) + b(2 - 3i) + c
2 - 3i = -5a - 12ai + 2b - 3bi + c
Separating the real and imaginary parts, we get:
(1) Re: 2 = -5a + 2b + c
(2) Im: -3 = -12a - 3b
Similarly, for the root 2 + 3i:
2 + 3i = -5a + 2b + c
-3 = -12a + 3b
Now, we can solve these two sets of equations simultaneously to find the values of a, b, and c.
By adding equation (1) and equation (2) together, we eliminate a:
2 - 3 = -5a + 2b + c -12a - 3b
-1 = -17a - b + c + 2b
-1 = -17a + b + c
By subtracting equation (1) from equation (2), we eliminate c:
-3 - 2 = -12a + 3b - (-5a + 2b + c)
-5 = -7a + b + c
Now, we have a system of two linear equations:
-1 = -17a + b + c [Equation 3]
-5 = -7a + b + c [Equation 4]
We can solve equations (3) and (4) to find a, b, and c.
By subtracting equation (4) from equation (3), we eliminate c:
(-1) - (-5) = (-17a + b + c) - (-7a + b + c)
4 = -17a + 7a
4 = -10a
a = -4/10
a = -2/5
Substituting the value of a back into equation (3):
-1 = -17 (-2/5) + b + c
-1 = 34/5 + b + c
-5/5 = 34/5 + b + c
-5/5 - 34/5 = b + c
-39/5 = b + c
Now, let's pick any value for either b or c in terms of the other variable. Let's say we set b = 1.
Substituting b = 1 into equation (4):
-5 = -7a + 1 + c
-5 = -7 (-2/5) + 1 + c
-5 = 14/5 + 1 + c
-25/5 = 14/5 + 1 + c
-25/5 - 14/5 - 1 = c
-40/5 - 14/5 - 1 = c
-55/5 = c
c = -11
Therefore, we have found the values of a, b, and c as follows:
a = -2/5
b = 1
c = -11
Finally, we can now write the quadratic equation using the found values of a, b, and c:
y = (-2/5)x^2 + x - 11
Therefore, the quadratic equation with roots 2 - 3i and 2 + 3i is indeed:
y = x^2 - 4x + 13.
You have correctly written the quadratic equation!