Question

Explain, what do you mean by complex roots and real roots?

Answer 1

Real roots may have positive, rational, irrational, non-integer roots for the respective equation where imaginary unit that is $i = \sqrt { - 1}$ is always zero where the complex roots are the vice-versa of the real roots containing more number of roots.

Complete step-by-step solution:
$\Rightarrow$ Real roots
As per as the mathematical part is concerned, real roots are the algebraic representation of polynomial of having at least one or maximum of two roots (in complex number system remember that real roots exists if only if imaginary part i.e. $i\sqrt { - 1}$ is zero as per as complex equations are concerned)!
We know that, from the factorization method we can get the real roots.
Let’s take an example of the equation for more clarifications,
${x^2} + 3x + 2 = 0$ Which is the quadratic equation of degree two respectively.
As a result, simplifying the equation by factorising the terms that is expanding the equation given above, we get
$\Rightarrow {x^2} + 2x + x + 2 = 0$
Taking the terms common in one bracket, we get
$\Rightarrow x(x + 2) + 1(x + 2) = 0 \\\ \Rightarrow (x + 2)(x + 1) = 0 \\\$
Now, the equation may have one or two real roots, solving it, we get
$\Rightarrow x + 2 = 0$ Or, $x + 1 = 0$
$\Rightarrow x = - 2$ Or, $x = - 1$
Hence, the equation we have considered has two real roots i.e. $- 2, - 1$ respectively.
$\Rightarrow$Complex roots (vice-versa of the real roots)
As a result of the real roots, complex roots seems to be the most unrealistic roots of the equation containing non-integer, unreal, etc… in this case the equation can be solved by another factorization formula that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ which has drastic two unreal roots of the equation $a{x^2} + bx + c = 0$. The conditions exist that when ${b^2} - 4ac < 0$ the roots are imaginary, when ${b^2} - 4ac > 0$ are real and ${b^2} - 4ac = 0$ respectively.
$\Rightarrow$ Complex number system in the form of $z = a + ib$ where, $a,b$ are the real roots and $i$ is the imaginary unit or an imaginary constant that is $i = \sqrt 1$ which is also the complex unreal root of the system.
Let’s consider the five degree polynomial equation, to find the roots which may contain complex unreal roots of the equation ${x^5} - {x^4} + {x^3} - {x^2} - 6x + 6 = 0$
Expanding the equation, we get
$\Rightarrow (x - 1)({x^4} + {x^2} - 6) = 0$
Further simplifying it, we get

\Rightarrow (x - 1)({x^2} - 2)({x^2} + 3) = 0 \\\ \Rightarrow (x - 1)(x - \sqrt 2 )(x + \sqrt 2 )({x^2} + 3) = 0 \\\ \Rightarrow (x - 1)(x - \sqrt 2 )(x + \sqrt 2 )(x + 3i)(x - 3i) = 0 $$ **Thus, the equation contains total five roots out of which three of them are real roots (i.e.$(x - 1),(x - \sqrt 2 ),(x + \sqrt 2 )$) and two of them (i.e.$(x + 3i),(x - 3i)$) are complex roots respectively.** **Note:** Remember real roots and complex roots are the two different groups or systems in the field of mathematics. One must be keen to observe the desired solution of the equation to determine respective roots whether they are real or complex!