Question

I'm going to cry help :c
Solve each of the quadratic equations below and describe what the solution(s) represent to the graph of each. Show your work to receive full credit.

y = x2 + 3x + 2
y = x2 + 2x + 1

Answer 1

no crying over math

Answer 2

Math is hard.

Answer 3

yes that is true

Answer 4

here is a graph of the first one
http://www.wolframalpha.com/input/?i=y+%3D+x^2+%2B+3x+%2B+2+

Answer 5

you can see that
\[y=x^2 + 3x + 2 \] factors as \[y=(x+2)(x+1)\]

Answer 6

this means if you want to find out where the graph crosses the \(x\) axis it is relatively easy. it crosses the \(x\) axis where \(y=0\) so set
\[(x+2)(x+1)=0\] and solve for \(x\)

Answer 7

you get \(x+2=0\implies x=-2\) or \(x+1=0\implies x=-1\)
so you know where it crosses the \(x\) axis at the points \((-2,0)\) and \((-1,0)\)
you can see it from the picture i sent

Answer 8

does this make any sense to you?

Answer 9

No LOOOL :c

Answer 10

any question i can answer? that was my best explanation
also a nice picture

Answer 11

Can you show me how to like factor it out? :c Sorry I didn't take algebra last year so that's why I'm all confused on it.

Answer 12

ok

Answer 13

we start with
\[y=x^2+3x+2\] and we want to make it look like
\[y=(x+a)(x+b)\]

Answer 14

so we need two numbers, \(a\) and \(b\) with \(a\times b=2\) and also \(a+b=3\)

Answer 15

it is pretty clear that the two numbers that work are 1 and 2 because \(1\times 2=2\) and \(1+2=3\)

Answer 16

so we can see that
\[x^2+3x+2=(x+1)(x+2)\] and we can check that it is correct by multiplying out on the right and see that we get what we want

Answer 17

that is, if you multiply
\[(x+1)(x+2)\] you get
\[x^2+x+2x+2=x^2+3x+2\] so we know it is right

Answer 18

so far so good?

Answer 19

OH!! Thanks so much ^__^

Answer 20

Yeah I get it c:

Answer 21

good
can you do the next one?