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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



User Jamieb
by
7.5k points

2 Answers

6 votes

Answer:

Explanation:


\boxed{i=√(-1) }


3x^(2) +2=0


3x^(2) =-2


x^(2) =-(2)/(3)


x=\sqrt{-(2)/(3) }


=\sqrt{(2)/(3) } *√(-1)


=\sqrt{(2)/(3) } \ i

User Eloreden
by
8.0k points
6 votes

Answer:


x = \left((√(6))/(3)\right) i

Explanation:

To solve the quadratic equation 3x² + 2 = 0, we can use the quadratic formula, which is given by:


x = (-b \pm √(b^2 - 4ac))/(2a)

In the case of 3x² + 2 = 0, we have:


  • a = 3

  • b = 0

  • c = 2

Plug these values into the quadratic formula:


x = (-0 \pm √((0)^2 - 4(3)(2)))/(2(3))

Simplify:


x = (\pm √(-24))/(6)

Since the part under the square root sign is negative, the solutions will be complex numbers.

Rewrite -24 as the product of 2² · 6 · (-1):


x = (\pm √(2^2 \cdot 6 \cdot (-1)))/(6)


\textsf{Apply the radical rule:} \quad √(ab)=\sqrt{\vphantom{b}a}√(b)


x = (\pm √(2^2) √(6) √(-1))/(6)


\textsf{Apply the radical rule:} \quad √(a^2)=a, \quad a \geq 0


x = (\pm 2 √(6) √(-1))/(6)

Divide the numerator and denominator by the common factor 2:


x = (\pm √(6) √(-1))/(3)


\textsf{Apply the imaginary number rule:} \quad √(-1)=i


x = (\pm √(6) \:i)/(3)


x = \pm \left((√(6))/(3)\right) i

So, the solutions to the equation 3x² + 2 = 0 are:


x = \left((√(6))/(3)\right) i, \qquad x=-\left((√(6))/(3)\right) i

User Xsee
by
8.1k points

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