Answer:
A quadratic equation [tex]ax^2+bx+c=0[/tex] ....[1] where, a, b and c are coefficient then;
the solution fore this equation we have;
[tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex] ....[2]
Given the function in the standard form:
[tex]f(x) = -2x^2+x+5[/tex]
The quadratic equation is:
[tex]0 = -2x^2+x+5[/tex]
On comparing with equation [1] we have;
a = -2 , b = 1 and c = 5.
The discriminant is given by:
[tex]\text{Discriminant} = b^2-4ac[/tex]
then;
[tex]\text{Discriminant} = (1)^2-4(-2)(5) = 1+40 = 41[/tex]
Solve for the zeros of the quadratic function:
Substitute the given values in [1] we have;
[tex]x = \frac{-1 \pm \sqrt{41}}{2(-2)} = \frac{-1 \pm \sqrt{41}}{-4}[/tex]
then;
[tex]x = \frac{-1 + \sqrt{41}}{-4}[/tex] and [tex]x = \frac{-1 - \sqrt{41}}{-4}[/tex]
⇒[tex]x = \frac{1 - \sqrt{41}}{4}[/tex] and [tex]x = \frac{1 + \sqrt{41}}{4}[/tex]
Therefore, the zeros of the given function are:
[tex]x = \frac{1 - \sqrt{41}}{4}[/tex] , [tex]\frac{1 + \sqrt{41}}{4}[/tex]
Using quadratic function concepts, it is found that the zeros are:
A quadratic function is given according to the following rule:
[tex]y = ax^2 + bx + c[/tex]
The discriminant is:
[tex]\Delta = b^2 - 4ac[/tex]
The solutions are:
[tex]x_1 = \frac{-b + \sqrt{\Delta}}{2a}[/tex]
[tex]x_2 = \frac{-b - \sqrt{\Delta}}{2a}[/tex]
In this problem, the function is:
[tex]f(x) = -2x^2 + x + 5[/tex]
Hence the discriminant is:
[tex]\Delta = b^2 - 4ac = 1^2 - 4(-2)(5) = 41[/tex]
Then, the zeros are:
[tex]x_1 = \frac{-1 + \sqrt{41}}{2(-2)} = \frac{1 - \sqrt{41}}{4}[/tex]
[tex]x_2 = \frac{-1 - \sqrt{41}}{2(-2)} = \frac{1 + \sqrt{41}}{4}[/tex]
You can learn more about quadratic function concepts at https://brainly.com/question/24737967
