Answer:
The zeros of the quadratic function f(x) = 9x2 – 54x – 19 is
x = -1/3 or x= 6
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