Question

How do you find all the real and complex root of the ${x^3} + 8 = 0$

Answer 1

Hint: In order to solve this question we will first find one root by hit and trial then we will use long division method to find the remaining equation as we will be knowing one of the root then we will be finding the roots by the method of factorization or long division method like this we will get all the all the possible real and complex roots of the equation.

Complete Step by step solution:
For solving this question we need to find the first root so as the cubic equation given to us is $ \Rightarrow {x^3} + 8 = 0$
So by substituting 8 on another side:
$ \Rightarrow {x^3} = - 8$
Putting x=-2 will satisfy this equation we will knowing that this is one of the root of this equation; Now;
x-2=0 we will be dividing the original equation by this to get the remaining quadratic equation;
$ \Rightarrow \dfrac{{{x^3} - 8}}{{x - 2}} = {x^2} - 2x + 4$
The new quadratic equation we found that is ${x^2} - 2x + 4 = 0$
For solving this we will be using the famous Dharacharya formula for finding the roots of this quadratic equation;
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
For this question the value of a = 1, b = -2 , c = 4.
Putting all these values in this formula we get the values of x which are;
$ \Rightarrow x = 1 \pm \sqrt 3 i$
These two values and the x=-2 are the three values for this cubic equation.

Note: The quadratic equation forms parabolic graph; According to the graph it was seen that when we make the graph of the quadratic equation in x then the roots are the intercepts of the x axis where the graph is passing in x-axis; Now we will discuss an important thing that is if the graph is cutting two times the x-axis so there will be two real roots of that quadratic equation, while if the graph is touching at the x-axis where it is touching will be only the real root of that quadratic equation. And if the graph is neither touching nor cutting there will not be any real root of that quadratic equation; all the roots of that equation will be complex.