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.