Question

Can someone please help me with the problem? Please show your work.

Directions: Solve the following. Which is a solution to 3x^2 + 2 =0?

(√6/3)i

2/3

1

(2/3)i​​​

Answer 1

x = ±
`\sqrt{(2)/(3)}` i

Explanation:

I don't think any of the choices you show are correct. Here is the answer I get:

3x² + 2 = 0

3x² = -2

x² = -2/3

x = √(-2/3)

x = ±
`\sqrt{(2)/(3)}` i

Answer 2

Answer: Choice A. (√6/3)i

Work Shown

Part 1

`3\text{x}^2 + 2 = 0\\\\3\text{x}^2 = -2\\\\\text{x}^2 = -(2)/(3)\\\\\text{x} = \pm\sqrt{-(2)/(3)}\\\\\text{x} = \pm(√(-2))/(√(3))\\\\`

Part 2

`\text{x} = \pm(i√(2))/(√(3))\\\\\text{x} = \pm(i√(2)*√(3))/(√(3)*√(3)) \ \text{ ... rationalizing denominator}\\\\\text{x} = \pm(i√(2*3))/(√(3*3))\\\\\text{x} = \pm(√(6))/(3)i\\\\\text{x} = (√(6))/(3)i \ \text{ or } \ \text{x} = -(√(6))/(3)i\\\\\text{where } i = √(-1)`

This leads us to the answer choice A (√6/3)i

Notes:

• The 3 is not under the square root in choice A. If it was, then the 6/3 would become 2.
• You can use the Csolve command in GeoGebra (in the CAS mode) to confirm the answer is correct. The regular "solve" won't work. WolframAlpha is another option. There are many other similar tools.