Question

How do I create a quadratic equation for
x=9 and x=-1
People keep trying to explain it to me but I don't remember half of what they're talking about?

Answer 1

Any quadratic formula (x-a)(x-b). Is verified, that the roots are x = a and x = b. So if you want a specific equation just plug em in!

Answer 2

lets look at it this way. suppose you wanted to solve
\[x^2+8x-9=0\] first you try to factor. and you can factor by as
\[(x-9)(x+1)=0\] so then you would say
\[x-9=0,x=9\] or
\[x+1=0,x=-1\]

Answer 3

so you know the solutions are 9 and -1, because first you factored and then you found them

Answer 4

now we are just working backwards. you know the solutions are 9 and -1. then must have come from
\[x=9, x=-1\] so
\[x-9=0, x+1=0\] and the "factored form" must have been
\[(x-9)(x+1)=0\]

Answer 5

if we multiply this out we get
\[x^2+8x-1=0\] and that is the answer you are looking for

Answer 6

(x-9)(x+1)=0
now x-9=0 and x=0
or x+1=0 and x= -1
x^2+8x-1=0

So if I write it out like this, does it make sense?

Answer 7

or am I skipping important steps?

Answer 8

all you have to state is that, for x=9 and x=-1 to be solutions, the factored expression will be (x-9)(x+1)=0. then, multiplying it out, you get x^2-8x-9=0

Answer 9

satellite73's equation is incorrectly multiplied

Answer 10

How would it look if I multiplied it out?

Answer 11

(x-9)*(x+1) just FOIL and youll get

x^2-9x+x-9=0

combine like terms and you get

x^2-8x-9=0

Answer 12

I know this has go to be a stupid question, but what's FOIL again?

Answer 13

I just has major issues with my short term memory. It's super frustrating.

Answer 14

it stands for "First, Outer, Inner, Last", it is just how expanding a polynomial is described

Answer 15

have*