What are the complex solutions to the following equation? -3x^2 + 2x - 3 = 0
it can also include i, for
imaginary
not only can it include i, it must include i.
Apply the quadratic fomrula.
x = (-b±√(b^2-4ac))/(2a)
= (-2±√(2^2 - 4*3*3))/(2(-3)) = (-2±√-32)/-6 = (1±2√2 i)/3
To find the complex solutions to the equation -3x^2 + 2x - 3 = 0, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this equation, a = -3, b = 2, and c = -3. Substituting these values into the quadratic formula, we have:
x = (-2 ± sqrt(2^2 - 4(-3)(-3))) / (2(-3))
x = (-2 ± sqrt(4 - 36)) / (-6)
x = (-2 ± sqrt(-32)) / (-6)
Since we have a negative value inside the square root (sqrt(-32)), this indicates that there are no real solutions to the equation. However, we can find the complex solutions by simplifying the square root of the negative number.
To simplify sqrt(-32), we can write it as sqrt(32) * sqrt(-1), where sqrt(-1) is denoted by the imaginary unit "i". Now, let's simplify sqrt(32):
sqrt(32) = sqrt(16 * 2) = sqrt(16) * sqrt(2) = 4 * sqrt(2)
Therefore, we have:
x = (-2 ± 4 * sqrt(2) * i) / (-6)
x = (1 ± 2 * sqrt(2) * i) / 3
So, the complex solutions to the equation -3x^2 + 2x - 3 = 0 are:
x = (1 + 2 * sqrt(2) * i) / 3
and
x = (1 - 2 * sqrt(2) * i) / 3