Answer:
Solutions are 2, -1 + 0.5 sqrt10 i and -1 - 0.5 sqrt10 i
or 2, -1 + 1.58 i and -1 - 1.58i
(where the last 2 are equal to nearest hundredth).
Step-by-step explanation:
The real solution is x = 2:-
x^3 - 8 = 0
x^3 = 8
x = cube root of 8 = 2
Note that a cubic equation must have a total of 3 roots ( real and complex in this case). We can find the 2 complex roots by using the following identity:-
a^3 - b^3 = (a - b)(a^2 + ab + b^2).
Here a = x and b = 2 so we have
(x - 2)(x^2 + 2x + 4) = 0
To find the complex roots we solve x^2 + 2x + 4 = 0:-
Using the quadratic formula x = [-2 +/- sqrt(2^2 - 4*1*4)] / 2
= -1 +/- (sqrt( -10)) / 2
= -1 + 0.5 sqrt10 i and -1 - 0.5 sqrt10 i
Answer: The solutions are : [tex]2, -1 + \sqrt{3}i , -1 - \sqrt{3}i[/tex]
Our equation is [tex]x^{3} - 8 = 0[/tex]
and we know that: [tex]8 = 2^{3}[/tex]
then we can write our equation as: [tex]x^{3} - 2^{3} = 0[/tex]
And using the identity: [tex]a^{3} - b^{3} = (a-b)*(a^{2} +ab+ b^{2} )[/tex]
where a = x and b = 2, then our equation is:
[tex]x^{3} - 2^{3} = (x-2)*(x^{2} +2x+ 2^{2} )= 0[/tex]
them, if x = 2 the first part is zero, so x = 2 is a solution, and now we need to see the second part in order to find the complex solutions.
[tex]x^{2} + 2x + 4 = 0[/tex]
Here we need to use Bhaskara:
if [tex]ax^{2}+ bx + c=0[/tex] for a, b and c constants, then:
x = [tex]\frac{-b +-\sqrt{b^{2} - 4ac } }{2a}[/tex]
in our problem we have:
[tex]x = \frac{-2 +-\sqrt{4 - 16} }{2} = -1 +-\sqrt{\frac{4 - 16}{4} } = -1 +- \sqrt{-3} = -1 +-\sqrt{3} i[/tex]
So here we have two complex solutions: [tex]x = -1 + \sqrt{3}i[/tex] and [tex]x = -1 - \sqrt{3} i[/tex].
