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3 years ago

What are the possible values of x in 6x^2 + 432 = 0

Mathematics

1 answer:

coldgirl [10]3 years ago

7 0

Answer:

x = ±6i sqrt(2)

Step-by-step explanation:

6x^2 + 432 = 0

Subtract 432 from each side

6x^2 =- 432

Divide by 6

6x^2 /6 =- 432/6

x^2 =-72

Take the square root of each side

sqrt(x^2) = ±sqrt( -72)

x = ±sqrt(-1) sqrt(2*36)

We know the square root of -1 = i

x = ±i sqrt(2*36)

x = ±i sqrt(36) sqrt(2)

x = ±6i sqrt(2)

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