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OleMash [197]
4 years ago
12

PLEASE HELP

Mathematics
2 answers:
lara31 [8.8K]4 years ago
5 0

Answer: The correct solution is option third.

Explanation:

The given equation is,

-3x^2-4x-4=0

The quadratic formula to find the value of x for the equation  ax^2+bx+c=0  is,

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

The values are,

a=-3,b=-4,c=-4

x=\frac{-(-4)\pm \sqrt{(-4)^2-4(-3)(-4)}}{2(-3)}

x=\frac{4\pm \sqrt{16-48}}{-6}

x=\frac{4\pm \sqrt{-32}}{-6}

Since \sqrt{-1}=i,

x=\frac{4\pm 4i\sqrt{2}}{-6}

x=\frac{2(2\pm 2i\sqrt{2})}{-6}

x=\frac{2\pm 2i\sqrt{2}}{-3}

x=\frac{-2\pm 2i\sqrt{2}}{3}

Therefore third option is correct.

IceJOKER [234]4 years ago
4 0

Answer:

x equals quantity of negative 2 plus or minus 2i square root of 2 all over 3

Step-by-step 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(+/-)\sqrt{-32}} {6}

remember that

i=\sqrt{-1}

x=\frac{-4(+/-)4i\sqrt{2}} {6}

Simplify

x=\frac{-2(+/-)2i\sqrt{2}} {3}

x1=\frac{-2(+)2i\sqrt{2}} {3}

x2=\frac{-2(-)2i\sqrt{2}} {3}



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