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

Answer:

Step-by-step explanation:

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Middle SchoolMathematics 5 points

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)`

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Answer

Answered by

sarahtony826pdkcef

+1

Kaneppeleqw and 1 more users found this answer helpful

Answer:

Step-by-step explanation:

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

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Answer

Answered by

FalconRob

FalconRob

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

x_{1,2} = \cfrac{-b\pm\sqrt{b^2-4ac}}{2a}

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

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

We can simplify the expression:

x_{1,2} = \cfrac{-3\pm\sqrt{9-12}}{2} = \cfrac{-3\pm\sqrt{-3}}{2}

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

\sqrt{-3} = \sqrt{-1\cdot 3} = \sqrt{-1}\sqrt{3} = \sqrt{3}i

So, the solutions are

x = \cfrac{-3+i\sqrt{3}}{2} \text{ or } x = \cfrac{-3-i\sqrt{3}}{2}