Question

Find all the zeroes of the function f(x)=11x^2+x+5

Answer 1

`x = (1 + √((-219)) )/(22) , \space  x = (1 - √((-219)) )/(22)`

Explanation:

f(x)=11x^2+x+5 is the given equation,

now comparing it with the standard equation
`ax^(2) + bx + c = 0`, we get

`11x^(2) + x + 5 = 0`

Here, a = 11, b = 1 and c = 5

Now by QUADRATIC FORMULA

x =
`\frac{-b \pm \sqrt{b^(2) - 4ac}   }{2a}`

Now,
`b^(2)  - 4ac = 1^(2)  - 4 (11) (5) = 1 - 220 = -219`

Now as discriminant D < 0, then the roots are imaginary and distinct.

So, roots are
`x = (-1 + √((-219)) )/(22) , \space  x = (-1 - √((-219)) )/(22)`

These are the two roots of the given equation.