Answer:
[tex]x = \frac{1 + \sqrt{(-219)} }{22} , \space x = \frac{1 - \sqrt{(-219)} }{22}[/tex]
Step-by-step explanation:
f(x)=11x^2+x+5 is the given equation,
now comparing it with the standard equation [tex]ax^{2} + bx + c = 0[/tex], we get
[tex]11x^{2} + x + 5 = 0[/tex]
Here, a = 11, b = 1 and c = 5
Now by QUADRATIC FORMULA
x = [tex]\frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}[/tex]
Now, [tex]b^{2} - 4ac = 1^{2} - 4 (11) (5) = 1 - 220 = -219[/tex]
Now as discriminant D < 0, then the roots are imaginary and distinct.
So, roots are [tex]x = \frac{-1 + \sqrt{(-219)} }{22} , \space x = \frac{-1 - \sqrt{(-219)} }{22}[/tex]
These are the two roots of the given equation.
