Question

Solve the equation $ {x^5} - {x^4} + 8{x^2} - 9x - 15 = 0 $ , one root being $ \sqrt 3 $ and another $ 1 - 2\sqrt { - 1} $ .

Answer 1

Hint: The roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. The roots of an equation can be found out by factorizing the equation. When a polynomial equation with rational coefficients has an irrational root, then its conjugate is also the root of that equation and the same is true for a complex root. So from the given two roots, we can find the other two roots and then find out its remaining roots.

Complete step-by-step answer:
We are given that $ {x^5} - {x^4} + 8{x^2} - 9x - 15 = 0 $ , this is a polynomial equation of degree five, so the number of solutions of this equation is also five, one of the roots is given as $ \sqrt 3 $ .
So, $ - \sqrt 3 $ is another root of the given equation.
 $ \sqrt 3 $ and $ - \sqrt 3 $ are the two roots of the given equation, so
$ (x + \sqrt 3 )(x - \sqrt 3 ) = {x^2} - 3 $ is the factor of the given equation.
Dividing
$ {x^5} - {x^4} + 8{x^2} - 9x - 15 = 0 $ by $ {x^2} - 3 $ we get –
 $
  {x^5} - {x^4} + 8{x^2} - 9x - 15 = ({x^2} - 3)({x^3} - {x^2} + 3x + 5) = 0 \\
   \Rightarrow {x^3} - {x^2} + 3x + 5 = 0 \;
  $
 $ 1 - 2\sqrt { - 1} $ can be written as $ 1 - 2i $ where $ i = iota $ is an imaginary number that is $ 1 - 2\sqrt { - 1} $ is a complex number, thus another root of the equation
$ {x^3} - {x^2} + 3x + 5 = 0 $ is its conjugate that is
$ 1 + 2i = 1 + 2\sqrt { - 1} $ .
 $ 1 - 2i $ and $ 1 + 2i $ are the two roots of the above equation, so
$ (x - 1 + 2i)(x - 1 - 2i) = {(x - 1)^2} + {(2i)^2} = {x^2} + 1 - 2x + 4 = {x^2} - 2x + 5 $ is the factor of the above equation.
Dividing $ {x^3} - {x^2} + 3x + 5 = 0 $ by $ {x^2} - 2x + 5 $ , we get –
 $
  {x^3} - {x^2} + 3x + 5 = (x + 1)({x^2} - 2x + 5) = 0 \\
   \Rightarrow x + 1 = 0 \\
   \Rightarrow x = - 1 \;
  $
Thus, $ x = - 1 $ is another root of the equation $ {x^5} - {x^4} + 8{x^2} - 9x - 15 = 0 $ .
Hence the five roots of the equation $ {x^5} - {x^4} + 8{x^2} - 9x - 15 = 0 $ are $ \sqrt 3 ,\, - \sqrt 3 ,\,1 - 2\sqrt { - 1} ,\,1 + 2\sqrt { - 1} \,and\, - 1 $
So, the correct answer is “ $ \sqrt 3 ,\, - \sqrt 3 ,\,1 - 2\sqrt { - 1} ,\,1 + 2\sqrt { - 1} \,and\, - 1 $ ”.

Note: The roots of an equation are the points on the x-axis at which the y-coordinate of the function is zero. In a polynomial equation, the highest exponent of the polynomial is called its degree. And according to the Fundamental Theorem of Algebra, a polynomial equation has exactly as many roots as its degree.