Respuesta :
x =(4-√-32)/-6=2/-3+2i/3√ 2 = -0.6667-0.9428i
x =(4+√-32)/-6=2/-3-2i/3√ 2 = -0.6667+0.9428i
x =(4+√-32)/-6=2/-3-2i/3√ 2 = -0.6667+0.9428i
For this case we have the following polynomial:
[tex] -3x^2 - 4x - 4 = 0 [/tex]
We apply the formula of the resolvent to solve the problem.
We have then:
[tex]x = \frac{-b +/- \sqrt{b^2 - 4ac}}{2a} [/tex]
Substituting values:
[tex]x = \frac{-(-4) +/- \sqrt{(-4)^2 - 4(-3)(-4)}}{2(-3)} [/tex]
Rewriting the equation we have:
[tex]x = \frac{4 +/- \sqrt{16 - 48}}{-6} [/tex]
[tex]x = \frac{4 +/- \sqrt{-32}}{-6} [/tex]
[tex]x = \frac{4 +/- \sqrt{-2*16}}{-6} [/tex]
[tex]x = \frac{4 +/- 4i\sqrt{2}}{-6} [/tex]
[tex]x = \frac{-2 +/- 2i\sqrt{2}}{3} [/tex]
Answer:
The roots of the polynomial are given by:
[tex]x = \frac{-2 +/- 2i\sqrt{2}}{3} [/tex]
C) x equals quantity of negative 2 plus or minus 2i square root of 2 all over 3
[tex] -3x^2 - 4x - 4 = 0 [/tex]
We apply the formula of the resolvent to solve the problem.
We have then:
[tex]x = \frac{-b +/- \sqrt{b^2 - 4ac}}{2a} [/tex]
Substituting values:
[tex]x = \frac{-(-4) +/- \sqrt{(-4)^2 - 4(-3)(-4)}}{2(-3)} [/tex]
Rewriting the equation we have:
[tex]x = \frac{4 +/- \sqrt{16 - 48}}{-6} [/tex]
[tex]x = \frac{4 +/- \sqrt{-32}}{-6} [/tex]
[tex]x = \frac{4 +/- \sqrt{-2*16}}{-6} [/tex]
[tex]x = \frac{4 +/- 4i\sqrt{2}}{-6} [/tex]
[tex]x = \frac{-2 +/- 2i\sqrt{2}}{3} [/tex]
Answer:
The roots of the polynomial are given by:
[tex]x = \frac{-2 +/- 2i\sqrt{2}}{3} [/tex]
C) x equals quantity of negative 2 plus or minus 2i square root of 2 all over 3