Question

Which equation has the solutions x = -3 ± √3i/2 ?

2x2 + 6x + 9 = 0
x2 + 3x + 12 = 0
x2 + 3x + 3 = 0
2x2 + 6x + 3 = 0

Answer 1

C

Explanation:

Answer 2

Answer:Answer is option C : [
`x^(2)` + 3x + 3 ] =0

Note: None of options matches with given question.

instead of "-3" , there should be "-
`(3)/(2)`".

Explanation:

Note: None of options matches with given question.

instead of "-3" , there should be "
`(3)/(2)`".

Here, First thing you have to observe the nature of roots.

∴ x = -
`(3)/(2)`+
`(√(3))/(2)`i and x = -
`(3)/(2)`-
`(√(3))/(2)`

∴ [ x+(
`(3)/(2)`-
`(√(3))/(2)`i) ][ x+(
`(3)/(2)`+
`(√(3))/(2)`i) ]=0

∴ [
`x^(2)` + x(
`(3)/(2)`+
`(√(3))/(2)`i)+ x(
`(3)/(2)`-
`(√(3))/(2)`i) + (
`(3)/(2)`-
`(√(3))/(2)`i)(
`(3)/(2)`+
`(√(3))/(2)`i) ]=0

∴ [
`x^(2)` +
`(3)/(2)`x +
`(√(3))/(2)`ix +
`(3)/(2)`x -
`(√(3))/(2)`ix + (3-
`(√(3))/(2)`i)(3+
`(√(3))/(2)`i) ] =0

∴ [
`x^(2)` + 3x + (
`(3)/(2)`-
`(√(3))/(2)`i)(
`(3)/(2)`+
`(√(3))/(2)`i) ] =0

∴ [
`x^(2)` + 3x +
`(9)/(4)` - (
`(√(3))/(2)`i)(
`(√(3))/(2)`i) ] =0

∴ [
`x^(2)` + 3x +
`(9)/(4)` - (
`(3)/(4)`)
`i^(2)` ] =0

∴ [
`x^(2)` + 3x +
`(9)/(4)` + (
`(3)/(4)`) ] =0

∴ [
`x^(2)` + 3x +
`(12)/(4)` ] =0

∴ [
`x^(2)` + 3x + 3 ] =0

Thus, Answer is option C : [
`x^(2)` + 3x + 3 ] =0