Question

What are the real and complex solutions of the polynomial equation? x^3-8=0. with imaginary numbers

Answer 1

Answer: The solutions are : $2, -1 + \sqrt{3}i , -1 - \sqrt{3}i$

Our equation is $x^{3} - 8 = 0$

and we know that: $8 = 2^{3}$

then we can write our equation as: $x^{3} - 2^{3} = 0$

And using the identity: $a^{3} - b^{3} = (a-b)*(a^{2} +ab+ b^{2} )$

where a = x and b = 2, then our equation is:

$x^{3} - 2^{3} = (x-2)*(x^{2} +2x+ 2^{2} )= 0$

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.

$x^{2} + 2x + 4 = 0$

Here we need to use Bhaskara:

if $ax^{2}+ bx + c=0$ for a, b and c constants, then:

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

in our problem we have:

$x = \frac{-2 +-\sqrt{4 - 16} }{2} = -1 +-\sqrt{\frac{4 - 16}{4} } = -1 +- \sqrt{-3} = -1 +-\sqrt{3} i$

So here we have two complex solutions: $x = -1 + \sqrt{3}i$ and $x = -1 - \sqrt{3} i$.

Answer 2

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