Question

What are the possible values of x in x2 + 3x + 3 = 0?

A.
`(-2 stackrel(+)(-) isqrt(3))/(3)`

B.
`(-3 stackrel(+)(-) isqrt(12))/(2)`

C.
`(-3 stackrel(+)(-) isqrt(3))/(2)`

D.
`(3 stackrel(+)(-) isqrt(3))/(2)`

Answer 1

To solve a quadratic equation like
`ax^2+bx+c=0`, you can use the quadratic formula

`x_(1,2) = \cfrac{-b\pm√(b^2-4ac)}{2a}`

In your case,
`a = 1, b = c = 3`, so the formula becomes

`x_(1,2) = \cfrac{-3\pm√((-3)^2-4\cdot 1\cdot 3)}{2\cdot 1}`

We can simplify the expression:

`x_(1,2) = \cfrac{-3\pm√(9-12)}{2} = \cfrac{-3\pm√(-3)}{2}`

Since -3 is negative, its square root is computed as

`√(-3) = √(-1\cdot 3) = √(-1)√(3) = √(3)i`

So, the solutions are

`x = \cfrac{-3+i√(3)}{2} \text{ or } x = \cfrac{-3-i√(3)}{2}`

Answer 2

Explanation:

The answer is C. `(-3 stackrel(+)(-) isqrt(3))/(2)`