How to find real number solutions of a polynomial equation

Asked by admin @ 10/11/2021 in Mathematics viewed by 280 People

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

Answered by admin @ 10/11/2021

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

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