Question

Solve −3x^2 − 4x − 4 = 0.

Answer 1

must use quadratic formula (basically completing the square)

so

for an equation
ax^2+bx+c=0
x=
`(-b+/- √(b^2-4ac) )/(2a)`

ax^2+bx+c=0
-3x^2-4x-4=0
a=-3
b=-4
c=-4

x=
`(-(-4)+/- √((-4)^2-4(-3)(-4)) )/(2(-4))`
x=
`(4+/- √(16-48) )/(-8)`
x=
`(4+/- √(-32) )/(-8)`
x=
`(4+/- (√(-1))(√(32)) )/(-8)`
x=
`(4+/- (i)(√((4)(4)(2))) )/(-8)`
x=
`(4+/- (i)(4√(2)) )/(-8)`
x=
`(4+/- 4i√(2) )/(-8)`
x=
`(1+/- i√(2) )/(-2)`

x=
`(1+ i√(2) )/(-2)` or
`(1- i√(2) )/(-2)`
or
x=
`(-1- i√(2) )/(2)` or
`(-1+ i√(2) )/(2)`

Answer 2

`x1=\frac{-2(+)2i√(2)} {3}`

`x2=\frac{-2(-)2i√(2)} {3}`

Explanation:

we have

`-3x^(2) -4x-4=0`

Rewrite (Multiply by
`-1` both sides)

`3x^(2)+4x+4=0`

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`3x^(2)+4x+4=0`

so

`a=3\\b=4\\c=4`

substitute

`x=\frac{-4(+/-)\sqrt{4^(2)-4(3)(4)}} {2(3)}`

`x=\frac{-4(+/-)√(-32)} {6}`

remember that

`i=√(-1)`

`x=\frac{-4(+/-)4i√(2)} {6}`

Simplify

`x=\frac{-2(+/-)2i√(2)} {3}`

`x1=\frac{-2(+)2i√(2)} {3}`

`x2=\frac{-2(-)2i√(2)} {3}`