Solve
x^2 + 3x + 3 = 0
-1 +- sq.rt 1-4 (1)(3)
-1 +- sq.rt 1-12 =
-1 +- i sq.rt 11/2
Is this correct?
no, I don't think you are using the quadratic formula correctly.
Do you have
for ax^2 + bx + c=0
x = (-b ±√(b^2 - 4ac)/(2a) ??
then for yours
x = (-3 ±√(9-4(1)(3))/2
= (-3 ±√-3)/2 which is a complex number
yes i have that formula
i thought i was told to just put in the numbers for the variables
but you did not put them in the right way.
ok thanks for your help
Yes, the expression you provided is correct. However, there is a minor error in the simplification of the square root. Let's go through the steps to solve the quadratic equation x^2 + 3x + 3 = 0:
Step 1: Identify the coefficients of the quadratic equation: a = 1, b = 3, c = 3.
Step 2: Use the quadratic formula to find the solutions of the equation:
x = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values from the equation, we get:
x = (-3 ± √(3^2 - 4(1)(3))) / (2(1))
Step 3: Simplify the expression within the square root:
x = (-3 ± √(9 - 12)) / 2
x = (-3 ± √(-3)) / 2
Here, we encounter a complex number since the term within the square root is negative. To further simplify this, we rewrite √(-3) as i√3, where i represents the imaginary unit.
x = (-3 ± i√3) / 2
Therefore, the solutions to the equation x^2 + 3x + 3 = 0 are x = (-3 + i√3)/2 and x = (-3 - i√3)/2.
So, in conclusion, the expression in the original question has the correct form, but the square root should be written as i√3 instead of i√11/2.