Which is a zero of the quadratic function f(x) = 9x2 – 54x – 19?

The zeros of the quadratic function f(x) = 9x2 – 54x – 19 is

x = -1/3   or   x= 6

Step-by-step explanation:

To find the zeros of the quadratic function, we equate the equation to zero and then solve for x

9x2 – 54x – 19  = 0   -----------------------(1)

comparing this equation with a standard form equation 'ax² + bx +c = 0

a =9    b= -54   and    c=-19

9 × -19 = -171

Find two factors such that their sum will give -54 and their products will give -171

The two factors are -57 and 3

-57 × 3 = -171

-57 + 3 = -54

So we will replace -54x by  -57x and 3x in equation (1)

9x2 + 3x - 57x – 19 =0

(9x2 + 3x) (- 57x  – 19)= 0

In the first parenthesis, 3x is common, so we will factor out 3x while in the second parenthesis -19 is common, so we will factor out -19

3x(3x+ 1) - 19 (3x + 1) = 0

(3x+1)(3x - 19) =0

Either  3x + 1 = 0

            subtract 1 from both-side of the equation

           3x + 1 - 1= 0-1

            3x = -1

Divide both-side of the equation by 3

          3x/3  = -1/3

        x=-1/3

OR

3x - 19= 0

Add 19 to both-side of the equation

3x - 19 + 19 = 0 + 19

3x = 19

Divide both-side of the equation by 3

3x/3 = 19/3

x = 6

Therefore, the zeros of the quadratic function f(x) = 9x2 – 54x – 19 is

x = -1/3   or   x= 6

i believe

step-by-step explanation:

from my conclusion c