Answer: The roots of the given polynomial are
[tex]x=\dfrac{-5+i\sqrt{19}}{2},~~\dfrac{-5-i\sqrt{19}}{2}.[/tex]
Step-by-step explanation: We are given to find the two values of x that are the roots of the following quadratic polynomial:
[tex]P(x)=x^2+5x+11.[/tex]
To find the roots, we must have
[tex]P(x)=0\\\\\Rightarrow x^2+5x+11=0~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We know that
the solution set of quadratic equation of the form [tex]ax^2+bx+c=0,~a\neq 0[/tex] is given by
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.[/tex]
From equation (i), we have
a = 1, b = 5 and c = 11.
Therefore, the solution of equation (i) is given by
[tex]x\\\\\\=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\=\dfrac{-5\pm\sqrt{5^2-4\times1\times 11}}{2\times 1}\\\\\\=\dfrac{-5\pm\sqrt{25-44}}{2}\\\\\\=\dfrac{-5\pm\sqrt{-19}}{2}\\\\\\=\dfrac{-5\pm i\sqrt{19}}{2}.[/tex]
Thus, the roots of the given polynomial are
[tex]x=\dfrac{-5+i\sqrt{19}}{2},~~\dfrac{-5-i\sqrt{19}}{2}.[/tex]
