can someone help me with quadratics please? you can make your own examples and guide me to the the answer.

- some.random.cool.kid

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- schrodinger

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- phi

can you be more specific?

- some.random.cool.kid

ok well I'm working on quadratics functions and i need help solving them. I don't have any specific examples I just want to practice them.

- phi

can you post a problem ?

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## More answers

- some.random.cool.kid

ok

- some.random.cool.kid

y = -16x^2 + 96x
for instance this one : above

- phi

and what is the question?

- some.random.cool.kid

kimberly tracks free fall of object she models a graph I don't have the graph but...

- some.random.cool.kid

if this doesn't help I can try and find some other one ...

- some.random.cool.kid

i have a perfect one 𝑔𝑔 − 8𝑔𝑔 − 20.

- phi

some background:
a quadratic is the name for a "second order polynomial", which means the highest exponent is 2
the simplest quadratic is
y= x^2
the most general form is
\[ a x^2 + b x +c \]
where a, b, and c are numbers.
If you graph a quadratic, you get a parabola
A parabola looks like a U or a \( \cap\)

- some.random.cool.kid

yeah negative and positive ...

- some.random.cool.kid

those were whats on the graph ...

- some.random.cool.kid

what about factoring quad's though?

- some.random.cool.kid

lets say I want to use this .. Factor 𝑦𝑦 + 11𝑦𝑦 + 24.

- some.random.cool.kid

thats one y each not two

- phi

to factor quadratics, and see how to do it, it is better to start backwards
say you have
(x+a)(x+b)
can you multiply that out ?

- some.random.cool.kid

/uh well I can multiply x and x by distributing right?

- phi

use the distributive property
to see it more clearly, temporarily rename (x+b) as B
(x+a)*(x+b) becomes (x+a)*B
now distribute the B
what do you get ?

- some.random.cool.kid

uh well i see that you can do x^a'b?

- some.random.cool.kid

imlost

- phi

B*(x+a)
now multiply B times each term inside the parens (this is called "distributing")

- some.random.cool.kid

yes but I'm not sure whether its x^b + a^b or ...

- some.random.cool.kid

theres no variable for B so idk what I'm multiplying.

- phi

the ^ means exponents. it is neither
try it with numbers
2*( 3+5)
multiply 2 times each number inside the parens

- some.random.cool.kid

see now that theres variables it looks more reasonable that what i was understanding ... this is 6 + 10 = 16 after you've distributed.

- some.random.cool.kid

/wasnt*

- phi

yes, and of course it is the same as 2* (3+5) = 2* 8= 16
so 16 both ways.
if you have letters, multiplying is easy... for example x times y is written xy
just like 2 times 3 is 2*3
(getting 6 is the next step, "simplifying" )
or 2 times x is written 2x

- some.random.cool.kid

well I write that earlier I thought you disagreed with it...

- some.random.cool.kid

wrote*

- phi

try
B*(x+a)

- some.random.cool.kid

so this is bx + ab?

- some.random.cool.kid

or xb*

- phi

yes. But I used B, because B is short for (x+a)
and bx= xb (order does not matter when multiplyiing: 2*3= 3*2)

- phi

so
B*(x+a)= Bx + Ba
and if we replace B with (x+b)
(x+b)(x+a)= (x+b)x +(x+b)a
or
(x+b)(x+a)= x(x+b) +a(x+b)

- phi

*B is short for (x+b) (not x+a )

- some.random.cool.kid

/so much wisdom its making my brain fall out xD. thanks tough ... so what if you have me those letters mixed with integers also what do i do then?

- some.random.cool.kid

gave*

- some.random.cool.kid

though*

- phi

Unfortunately, you have to know this stuff before tackling quadratics.
you may not get it in one session, but you will get closer.
meanwhile
(x+b)(x+a)= x(x+b) +a(x+b)
now distribute x in the first x(x+b) and distribute the a in the second
can you do that ?

- some.random.cool.kid

I can certainly try.

- some.random.cool.kid

uh so i think its x^2 + b
and ax + b?

- some.random.cool.kid

or am i missing something?

- phi

if the problem was 2(2+3) you did 2*2 + 3

- some.random.cool.kid

yeah? thats 4 + 3? but Im not sure what I missed in the first one?

- phi

2(2+3) is 10
2*5=10
or distributing: 2*2 + 2*3 = 4+6 = 10

- some.random.cool.kid

oh distribute both...

- phi

if we had x(x+b) we multiply x times *each* term inside the parens

- some.random.cool.kid

so X^2 and xb?

- phi

yes

- some.random.cool.kid

ok

- phi

(x+b)(x+a)= x(x+b) +a(x+b)
(x+b)(x+a)= x^2 +bx + ax + ab

- some.random.cool.kid

so what about both numbers and letters can you perhaps explain those?

- phi

if we "factor out" an x from the 2 middle terms (the opposite of distribute)
we get
(x+b)(x+a)= x^2 +(b+a)x + ab

- phi

now we can do factoring of a quadratic
say we have
x^2 +7x +12
and we are trying to find a and b so (x+a)(x+b)= x^2 +7x +12

- phi

using the letters
x^2 +(b+a)x + ab
notice that (a+b) matches the 7 and a*b matches the 12 in
x^2 +7x +12

- phi

the first thing I do is list all pairs of numbers that when multiplied give us 12
1,12
2,6
3,4

- phi

then I add each pair
1+12= 13
2+6= 8
3+4= 7
if I find a pair that add up to the middle number (7 in this case) then we hit the jackpot
the factors are
(x+3)(x+4)
and if we multiply that out we will get x^2 +7x+12

- some.random.cool.kid

alright... that was a bit of a long shot does it normally take that much steps?

- phi

it is like anything you learn. You probably forgot how hard it was to learn to read.
But in school, if you work at this every day, (and not try to learn it all in 1 day as you seem to be trying to do now) , it can be done.

- phi

try factoring x^2+4x+4

- some.random.cool.kid

well uh ... I assume its 4^2x? for the first part

- some.random.cool.kid

then add the 4?

- phi

factoring x^2+4x+4 means
find a and b so that
(x+a)(x+b)= x^2+4x+4

- phi

the first step is look at the last number: +4
find all pairs of numbers that multiply to give you 4
1,4
2,2
add each pair
1+4=5
2+2= 4
choose the pair that add up to the middle number (the 4 from 4x)
2,2 are the numbers for a and b
thus
(x+2)(x+2)= x^2 + 4x + 4

- phi

try factoring
x^2 +4x + 3

- some.random.cool.kid

back

- some.random.cool.kid

ok

- some.random.cool.kid

wait idk how to factor this my teacher has asked me this before and i failed

- phi

x^2 +4x + 3
what is the first step? (look what I posted up above)

- some.random.cool.kid

except she did it with a minus

- some.random.cool.kid

i really don't know please walk me through ...

- some.random.cool.kid

all i know is that I factor it ..

- phi

the first step is look at the last number:

- some.random.cool.kid

3?

- phi

find all pairs of numbers that multiply to give you 3

- some.random.cool.kid

3 and 1

- phi

now add them. if they add to the "middle number" 4, you found the correct pair

- some.random.cool.kid

k

- phi

3+1= 4 (which matches the 4 in x^2 + 4x +3)
so 3 , 1 are the numbers we want
the factors are
(x+3)(x+1)

- some.random.cool.kid

uh ok

- phi

try
y^2 + 11y + 24
which you posted at the start

- some.random.cool.kid

do I start with 24

- phi

yes

- some.random.cool.kid

in that case 12 and 2 can work..

- some.random.cool.kid

or 24 ad 1

- phi

I always start with 1, 2, 3, etc
up to the square root of 21 (less than 5)

- phi

1,24
2,12
3,8
4,6

- some.random.cool.kid

ok

- some.random.cool.kid

so i pick 12 and 2

- phi

do any of the pairs add up to the middle number in y^2 +11y + 24

- some.random.cool.kid

no

- some.random.cool.kid

so is it done already

- some.random.cool.kid

/thats been simplified?

- phi

not yet. recheck each pair

- some.random.cool.kid

ok im lost Im not sure I see that that pairs except 12 and 2 equally the 24

- phi

add each pair
1,24
2,12
3,8
4,6

- some.random.cool.kid

ill be back in two hours

- phi

1+24= 25
2+12= 14
3+8= 11
4+6= 10

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