Respuesta :

apologiabiology
must use quadratic formula (basically completing the square)

so

for an equation
ax^2+bx+c=0
x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]

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

x=[tex] \frac{-(-4)+/- \sqrt{(-4)^2-4(-3)(-4)} }{2(-4)} [/tex]
x=[tex] \frac{4+/- \sqrt{16-48} }{-8} [/tex]
x=[tex] \frac{4+/- \sqrt{-32} }{-8} [/tex]
x=[tex] \frac{4+/- (\sqrt{-1})(\sqrt{32}) }{-8} [/tex]
x=[tex] \frac{4+/- (i)(\sqrt{(4)(4)(2)}) }{-8} [/tex]
x=[tex] \frac{4+/- (i)(4\sqrt{2}) }{-8} [/tex]
x=[tex] \frac{4+/- 4i\sqrt{2} }{-8} [/tex]
x=[tex] \frac{1+/- i\sqrt{2} }{-2} [/tex]

x=[tex] \frac{1+ i\sqrt{2} }{-2} [/tex] or [tex] \frac{1- i\sqrt{2} }{-2} [/tex]
or
x=[tex] \frac{-1- i\sqrt{2} }{2} [/tex] or [tex] \frac{-1+ i\sqrt{2} }{2} [/tex]
calculista

Answer:

[tex]x1=\frac{-2(+)2i\sqrt{2}} {3}[/tex]

[tex]x2=\frac{-2(-)2i\sqrt{2}} {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]



Otras preguntas