# Quadratic formula to find zero

• May 2nd 2010, 04:55 PM
Jackie10
Use the quadratic formula to find the zeros of the function

f(x)=x^2-6x+18

i get

x=6 +- Square r0ot of -6^2-4(1)(18)

x=6 +- squareroot of 36-72

which = -36

then
6 +_ 6i
__________

2(1)

=

3+- 3i

right...

Now iT states to put it in factored form

f(x)= a(x-r1)(x-r2)

where a= ?

r1= ?

r2=?

i tried putting a=3

r1=3i
r2=3i

but i guess im wr0ng help please...

=[

thank you so much to everyone on the forum!
• May 2nd 2010, 05:12 PM
masters
Quote:

Originally Posted by Jackie10
Use the quadratic formula to find the zeros of the function

f(x)=x^2-6x+18

i get

x=6 +- Square r0ot of -6^2-4(1)(18)

x=6 +- squareroot of 36-72

which = -36

then
6 +_ 6i
__________

2(1)

=

3+- 3i

right...

Now iT states to put it in factored form

f(x)= a(x-r1)(x-r2)

Hi Jackie10,

I would do it this way:

Since the roots are 3+3i and 3-3i, the factored form would be:

f(x) = [x-(3+3i)][x-(3-3i)] = (x-3-3i)(x-3+3i)
• May 2nd 2010, 05:13 PM
Jackie10
=] Thank you so then the top part is correct right...

and what do i put for a in the formula?
• May 2nd 2010, 06:39 PM
Debsta
Quote:

Originally Posted by Jackie10
=] Thank you so then the top part is correct right...

and what do i put for a in the formula?

"a" is the coefficient of x^2, so a=1

r1 doesn't equal -3i
r1=3-3i