Question

What are the zeroes of f(x) = x^2 - 2x - 3?

x = -3, 1

x = -3, -1

x = 3, 1

x = 3, -1

Answer 1

try to replace x with each number

Answer 2

f(x)=x^x-xx-x ?

Answer 3

What is that^ @Alexandra675 ?

Answer 4

lol he said to replace x with each number sooo...

Answer 5

there is no x^x

Answer 6

x^2

Answer 7

look at the answers:
x = -3, 1

x = -3, -1

x = 3, 1

x = 3, -1

they don't make sense do they?

Answer 8

Kinda the the 3 is from the problem I don't know where they got the 1 from...

Answer 9

A point has coordinates (x, y)

Answer 10

yea

Answer 11

What does "x = -3, 1 " mean?

Answer 12

That there are two points and those are both the x coordinats?

Answer 13

so you do \(\huge\ -3^2 - 2(-3) - 3\) and \(\huge\ -1^2 - 2(-1) - 3\)

Answer 14

Nope. It should be written as "x = -3, y = 1 "

Answer 15

oh ok so where does the 1 come from?

Answer 16

oh yes your totally right @skullpatrol

Answer 17

sorry

Answer 18

XD

Answer 19

The 1 is what y equals.

Answer 20

\(\huge\ -3^2 - 2(-3) - 3 =1\)

Answer 21

oh ok so its A then?

Answer 22

Sorry, that is wrong :(

Answer 23

huh what are you talking about? @skullpatrol

Zeros of f(x) is x = -3, 1 means f(x) = 0 at x= -3 or x=1

Answer 24

The equation has 2 solutions for x

Answer 25

that "1" part is not y.

Answer 26

Can we start again @Alexandra675 ?

Answer 27

Like do the problem over? Sure

Answer 28

Yes, I confused it with your first question, sorry about that.

Answer 29

What are the zeroes of f(x) = x^2 - 2x - 3?

x = -3, 1

x = -3, -1

x = 3, 1

x = 3, -1

Answer 30

lol y did I do tht XD

Answer 31

start by factoring x^2 - 2x - 3

Answer 32

\[(x-3)(x+1)\]

Answer 33

Correct! now set it equal to 0

Answer 34

x=3,-1

Answer 35

and tell me for what values of x will it be true

Answer 36

???

Answer 37

Yes for x=3 or x=-1, x^2 - 2x - 3 = 0

Answer 38

those are the "zeros"

Answer 39

Hold on let me work this out

Answer 40

please do :)

Answer 41

The first one works

Answer 42

I think...

Answer 43

If you factored them correctly, then they should be right, just watch out for the extraneous solutions.

Answer 44

The second one equals -2 tho

Answer 45

No what is it still correct?

Answer 46

...

Answer 47

@skullpatrol

Answer 48

If you factor \(x^2-2x-3\) you get \((x-3)(x+1)\) right?
Now solve them \(0=(x-3), 0=(x+1)\)
You will get \(x=3, -1\)

If you plug them in:
\(0=(3)^2-2(3)-3\) you get \(0=9-6-3\) which is just \(0=0\) so that works.

Plug in -1
\(0=(-1)^2-2(-1)-3\) you get \(0=1+2-3\) which is also \(0=0\)

So those points work, as @skullpatrol said

Answer 49

Oh ok I guess I did my calculations wrong for -1 thank you!