Question

What are the real and complex solutions of the polynomial equation ${{x}^{3}}-8=0$?

Answer 1

Hint: For solving this question you should know about the solutions of a polynomial equation. In this question we have to convert it in a second power quadratic form and then we apply the quadratic formula for finding roots. And we get the real and imaginary roots from this formula.

Complete step-by-step answer:
According to the question we have to find real and complex solutions of the polynomial equation ${{x}^{3}}-8=0$. So, we can see that it is in a form of ${{a}^{3}}-{{b}^{3}}$ and it can be written as $\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ form where $\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ will be a quadratic equation. So, as we know that the quadratic equations of the format $\left( {{a}^{2}}+bx+c \right)$ always contain roots, which can be imaginary or real roots. But in many equations, we can’t determine roots easily. So, then we use the formula for finding the roots. If the equation is in ${{a}^{2}}+bx+c$ form, then quadratic formula is,
$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
But Discriminant $={{b}^{2}}-4ac$ . So, we can write the quadratic formula as,
$\dfrac{-b\pm \sqrt{D}}{2a}$
And here we get two roots as,
$\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$ and $\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$
If we take an example to understand it clearly then:
Example 1: Find the roots of ${{x}^{2}}-4x+6$.
The discriminant used to determine how many different solutions and what type of solutions a quadratic equation will have. So, here according to our equation,
$\begin{align}
  & a=1,b=-4,c=6 \\
 & \Rightarrow D={{\left( -4 \right)}^{2}}-4\left( 1 \right)\left( 6 \right) \\
 & =16-24 \\
 & =-8 \\
\end{align}$
This indicates that this equation contains 2 imaginary solutions (with $i$, the square root of -1). If the answer had been positive (assume 9), then the equation would have 2 real solutions (with real numbers, they might not be rational solutions but they are real). And if the answer was 0, the equation has 1 real solution (it would be the square root of something).
So, we have ${{x}^{3}}-8=\left( {{x}^{3}}-{{2}^{3}} \right)=0$
Which can be written as,
$\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)=0$
Therefore,
$x-2=0\Rightarrow x=2$
And,
$\begin{align}
  & x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\left( 1 \right)\left( 4 \right)}}{2\left( 1 \right)}=\dfrac{-2\pm i2\sqrt{3}}{2} \\
 & =-1\pm i\sqrt{3} \\
\end{align}$
So, the real solution is 2 and imaginary are $-1+i\sqrt{3}$ and $-1-i\sqrt{3}$.

Note: If we want to calculate the roots of any equation then always try to reduce that in a form of ${{a}^{2}}+bx+c$ because it will be very easy to find the roots from this. And it gives us exact roots. So, always follow this method.