Question

How do you solve `w^ { 2} = - 2w - 12`?

Answer 1

`w=sqrt(2(-w-6)`

Explanation:

`w^2=-2w-12`

`:.w^2=2(-w-6)`

`:.w=sqrt(2(-w-6))`

Answer 2

`w=-1+-isqrt11`

Explanation:

Rearrange the quadratic and equate to zero.

`rArrw^2+2w+12=0`

This does not factorise, so check the `color(blue)"discriminant"`

For this quadratic `a=1,b=2" and " c=12`

`rArrb^2-4ac=2^2-(4xx1xx12)=-44`

`b^2-4ac<0rArr" no real roots"`

The roots are complex. We can find them using the
`color(blue)"quadratic formula"`

`w=(-b+-sqrt(b^2-4ac))/(2a)`

`rArrw=(-2+-sqrt(-44))/2`

`color(white)(xxxx)=(-2+-2isqrt11)/2`

`rArrw=-1+-isqrt11`