some.random.cool.kid
  • some.random.cool.kid
can someone help me with quadratics please? you can make your own examples and guide me to the the answer.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
phi
  • phi
can you be more specific?
some.random.cool.kid
  • 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
  • phi
can you post a problem ?

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More answers

some.random.cool.kid
  • some.random.cool.kid
ok
some.random.cool.kid
  • some.random.cool.kid
y = -16x^2 + 96x for instance this one : above
phi
  • phi
and what is the question?
some.random.cool.kid
  • 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
  • some.random.cool.kid
if this doesn't help I can try and find some other one ...
some.random.cool.kid
  • some.random.cool.kid
i have a perfect one 𝑔𝑔 − 8𝑔𝑔 − 20.
phi
  • 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
  • some.random.cool.kid
yeah negative and positive ...
some.random.cool.kid
  • some.random.cool.kid
those were whats on the graph ...
some.random.cool.kid
  • some.random.cool.kid
what about factoring quad's though?
some.random.cool.kid
  • some.random.cool.kid
lets say I want to use this .. Factor 𝑦𝑦 + 11𝑦𝑦 + 24.
some.random.cool.kid
  • some.random.cool.kid
thats one y each not two
phi
  • 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
  • some.random.cool.kid
/uh well I can multiply x and x by distributing right?
phi
  • 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
  • some.random.cool.kid
uh well i see that you can do x^a'b?
some.random.cool.kid
  • some.random.cool.kid
imlost
phi
  • phi
B*(x+a) now multiply B times each term inside the parens (this is called "distributing")
some.random.cool.kid
  • some.random.cool.kid
yes but I'm not sure whether its x^b + a^b or ...
some.random.cool.kid
  • some.random.cool.kid
theres no variable for B so idk what I'm multiplying.
phi
  • 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
  • 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
  • some.random.cool.kid
/wasnt*
phi
  • 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
  • some.random.cool.kid
well I write that earlier I thought you disagreed with it...
some.random.cool.kid
  • some.random.cool.kid
wrote*
phi
  • phi
try B*(x+a)
some.random.cool.kid
  • some.random.cool.kid
so this is bx + ab?
some.random.cool.kid
  • some.random.cool.kid
or xb*
phi
  • 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
  • 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
  • phi
*B is short for (x+b) (not x+a )
some.random.cool.kid
  • 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
  • some.random.cool.kid
gave*
some.random.cool.kid
  • some.random.cool.kid
though*
phi
  • 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
  • some.random.cool.kid
I can certainly try.
some.random.cool.kid
  • some.random.cool.kid
uh so i think its x^2 + b and ax + b?
some.random.cool.kid
  • some.random.cool.kid
or am i missing something?
phi
  • phi
if the problem was 2(2+3) you did 2*2 + 3
some.random.cool.kid
  • some.random.cool.kid
yeah? thats 4 + 3? but Im not sure what I missed in the first one?
phi
  • phi
2(2+3) is 10 2*5=10 or distributing: 2*2 + 2*3 = 4+6 = 10
some.random.cool.kid
  • some.random.cool.kid
oh distribute both...
phi
  • phi
if we had x(x+b) we multiply x times *each* term inside the parens
some.random.cool.kid
  • some.random.cool.kid
so X^2 and xb?
phi
  • phi
yes
some.random.cool.kid
  • some.random.cool.kid
ok
phi
  • phi
(x+b)(x+a)= x(x+b) +a(x+b) (x+b)(x+a)= x^2 +bx + ax + ab
some.random.cool.kid
  • some.random.cool.kid
so what about both numbers and letters can you perhaps explain those?
phi
  • 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
  • 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
  • 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
  • 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
  • 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
  • some.random.cool.kid
alright... that was a bit of a long shot does it normally take that much steps?
phi
  • 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
  • phi
try factoring x^2+4x+4
some.random.cool.kid
  • some.random.cool.kid
well uh ... I assume its 4^2x? for the first part
some.random.cool.kid
  • some.random.cool.kid
then add the 4?
phi
  • phi
factoring x^2+4x+4 means find a and b so that (x+a)(x+b)= x^2+4x+4
phi
  • 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
  • phi
try factoring x^2 +4x + 3
some.random.cool.kid
  • some.random.cool.kid
back
some.random.cool.kid
  • some.random.cool.kid
ok
some.random.cool.kid
  • some.random.cool.kid
wait idk how to factor this my teacher has asked me this before and i failed
phi
  • phi
x^2 +4x + 3 what is the first step? (look what I posted up above)
some.random.cool.kid
  • some.random.cool.kid
except she did it with a minus
some.random.cool.kid
  • some.random.cool.kid
i really don't know please walk me through ...
some.random.cool.kid
  • some.random.cool.kid
all i know is that I factor it ..
phi
  • phi
the first step is look at the last number:
some.random.cool.kid
  • some.random.cool.kid
3?
phi
  • phi
find all pairs of numbers that multiply to give you 3
some.random.cool.kid
  • some.random.cool.kid
3 and 1
phi
  • phi
now add them. if they add to the "middle number" 4, you found the correct pair
some.random.cool.kid
  • some.random.cool.kid
k
phi
  • 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
  • some.random.cool.kid
uh ok
phi
  • phi
try y^2 + 11y + 24 which you posted at the start
some.random.cool.kid
  • some.random.cool.kid
do I start with 24
phi
  • phi
yes
some.random.cool.kid
  • some.random.cool.kid
in that case 12 and 2 can work..
some.random.cool.kid
  • some.random.cool.kid
or 24 ad 1
phi
  • phi
I always start with 1, 2, 3, etc up to the square root of 21 (less than 5)
phi
  • phi
1,24 2,12 3,8 4,6
some.random.cool.kid
  • some.random.cool.kid
ok
some.random.cool.kid
  • some.random.cool.kid
so i pick 12 and 2
phi
  • phi
do any of the pairs add up to the middle number in y^2 +11y + 24
some.random.cool.kid
  • some.random.cool.kid
no
some.random.cool.kid
  • some.random.cool.kid
so is it done already
some.random.cool.kid
  • some.random.cool.kid
/thats been simplified?
phi
  • phi
not yet. recheck each pair
some.random.cool.kid
  • some.random.cool.kid
ok im lost Im not sure I see that that pairs except 12 and 2 equally the 24
phi
  • phi
add each pair 1,24 2,12 3,8 4,6
some.random.cool.kid
  • some.random.cool.kid
ill be back in two hours
phi
  • phi
1+24= 25 2+12= 14 3+8= 11 4+6= 10

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