Question

Find the zeros of the function.

Answer 1

f(x) = 9x^2 + 6x – 8

Answer 2

u know to solve the quadratic equation?

Answer 3

i can try

Answer 4

can you show me how to solve this?

Answer 5

proceed....
9x^2 + 6x – 8 =0
multiply 9 by -8= -72
then factorize -72= a*b such that a+b = 6

Answer 6

-72 = 12*(-6)
a=12, b=-6

Answer 7

does it make sense?

Answer 8

i dont understand what you did here? 12 and -6 are what

Answer 9

multiply 9 by -8= -72

Answer 10

it's ok?

Answer 11

9 is the coeff of x^2 and -8 is the constant term..

Answer 12

ok

Answer 13

now factorize -72 into two parts a and b such that a*b=-72 and a+b= coefficient of x

Answer 14

allrighttt

Answer 15

hello?

Answer 16

you able to factor the quadratic ?

Answer 17

if you have trouble factoring these, you can always just go to the quadratic formula to obtain the zeros
\[x = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}\]

Answer 18

where a,b,c are the coefficients of function
\[ax^{2} +bx+c\]

Answer 19

here is the quadratic formula for you