Question

How do you solve `12x ^ { 2} + 2x - 1= - 3`?

Answer 1

Solution is `x=-1/12+-isqrt23/12`

Explanation:

`12x^2+2x-1=-3`

`hArr12x^2+2x+2=0` and dividing each term by `2` we get

`6x^2+x+1=0`

Using quadratic formula `x=(-1+-sqrt(1^2-4xx1xx6))/(2xx6)`

= `(-1+-sqrt(-23))/12`

= `-1/12+-isqrt23/12`

Answer 2

`x = -1/12 +-(sqrt(23)i)/12`

Explanation:

Put in the form `Ax^2 + Bx + C = 0: " "12x^2 + 2x +2 = 0`

From the graph you can see that the equation does not cross the `x`-axis. This means that the values of `x` are complex.
graph{12x^2+2x+2 [-10.33, 9.67, -1.12, 8.88]}

Use the quadratic equation to find the complex roots:
`x = (-B +- sqrt(B^2 -4AC))/(2A) = (-2 +- sqrt(4-4(12)(2)))/(2*12)`

`x = -1/12 +- sqrt(-92)/24 = -1/12 +- (sqrt(4 * 23)i)/24 = -1/12 +-(sqrt(23)i)/12`