Form a quadratic equation ax^2+bx+c=0 with a=1 that has the given roots:
( −3±i 2)/ 2
sum of roots = (-3 + 2i)/2 + (-3 - 2i)/2
= -3
product of roots = (-3 + 2i)/2 * (-3 - 2i)/2
= (9 - 4i^2)/4 = 13/4
equation: x^2 + 3x + 13/4 = 0
You wanted a = 1, so we leave it like that
To form a quadratic equation with given roots, we can use the fact that the roots of a quadratic equation ax^2 + bx + c = 0 are given by the solutions to x = (-b ± √(b^2 - 4ac))/(2a).
In this case, we are given the roots as (-3 ± i2)/2. Let's simplify this expression first:
(-3 ± i2)/2 = -3/2 ± i/2
Now, we need to equate this expression to x:
x = -3/2 ± i/2
To find the quadratic equation with these roots, we can rewrite x by replacing the ± symbol with a minus sign (-) and a plus sign (+) respectively:
For x = -3/2 - i/2, the corresponding factor will be (x - (-3/2 - i/2))
For x = -3/2 + i/2, the corresponding factor will be (x - (-3/2 + i/2))
Now, let's multiply these factors:
(x - (-3/2 - i/2))(x - (-3/2 + i/2))
= (x + 3/2 + i/2)(x + 3/2 - i/2)
Expanding this expression:
= (x + 3/2 + i/2)(x + 3/2 - i/2)
= x(x + 3/2 - i/2) + (3/2 + i/2)(x + 3/2 - i/2)
= x^2 + 3/2x - i/2x + 3/2x + 9/4 - 3/4 - 3i/4 + 3i/4 - i^2/4
Since a = 1, the quadratic equation takes the form:
x^2 + (3/2 + 3/2)x + (9/4 - 3/4 - i^2/4)
= x^2 + 3x + 6/4 - 1/4
= x^2 + 3x + 5/2
Therefore, the quadratic equation with a = 1 and the given roots (-3 ± i2)/2 is:
x^2 + 3x + 5/2 = 0