Answer: The correct solution is option third.
Explanation:
The given equation is,
[tex]-3x^2-4x-4=0[/tex]
The quadratic formula to find the value of x for the equation [tex]ax^2+bx+c=0[/tex] is,
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
The values are,
[tex]a=-3,b=-4,c=-4[/tex]
[tex]x=\frac{-(-4)\pm \sqrt{(-4)^2-4(-3)(-4)}}{2(-3)}[/tex]
[tex]x=\frac{4\pm \sqrt{16-48}}{-6}[/tex]
[tex]x=\frac{4\pm \sqrt{-32}}{-6}[/tex]
Since [tex]\sqrt{-1}=i[/tex],
[tex]x=\frac{4\pm 4i\sqrt{2}}{-6}[/tex]
[tex]x=\frac{2(2\pm 2i\sqrt{2})}{-6}[/tex]
[tex]x=\frac{2\pm 2i\sqrt{2}}{-3}[/tex]
[tex]x=\frac{-2\pm 2i\sqrt{2}}{3}[/tex]
Therefore third option is correct.
Answer:
x equals quantity of negative [tex]2[/tex] plus or minus [tex]2i[/tex] square root of [tex]2[/tex] all over [tex]3[/tex]
Step-by-step explanation:
we have
[tex]-3x^{2} -4x-4=0[/tex]
Rewrite (Multiply by [tex]-1[/tex] both sides)
[tex]3x^{2}+4x+4=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]3x^{2}+4x+4=0[/tex]
so
[tex]a=3\\b=4\\c=4[/tex]
substitute
[tex]x=\frac{-4(+/-)\sqrt{4^{2}-4(3)(4)}} {2(3)}[/tex]
[tex]x=\frac{-4(+/-)\sqrt{-32}} {6}[/tex]
remember that
[tex]i=\sqrt{-1}[/tex]
[tex]x=\frac{-4(+/-)4i\sqrt{2}} {6}[/tex]
Simplify
[tex]x=\frac{-2(+/-)2i\sqrt{2}} {3}[/tex]
[tex]x1=\frac{-2(+)2i\sqrt{2}} {3}[/tex]
[tex]x2=\frac{-2(-)2i\sqrt{2}} {3}[/tex]
