Question

I am doing solving Quadratic equations by factoring and Applications of quadratic equations. I have a test TODAY! I dont understand it at all I am doing solving Quadratic equations by factoring and Applications of quadratic equations. I have a test TODAY! I dont understand it at all @Mathematics

Answer 1

What is a quadratic equation?

Answer 2

hold on I will give you one to look at

Answer 3

y^2 + 3y +2 = 0

Answer 4

\[Ax^2 + Bx + C = 0\]

Answer 5

It says solve the equation and check your solutions
I just dont understand this

Answer 6

There are so many applications of quadratic equations, but solving them by simply factoring might require a bit of.... 'intuition'.

can you explain how did I rewrite \[y^2 + 3y + 2 = 0\]as\[(y+2)(y+1)=0\]

Answer 7

I am just so confused on how to solve this and my teacher goes so fast that my brain can't absorb it

Answer 8

but she brakes it all down and says y +2= something and y - 1= 0

Answer 9

do you know how to factor quadratic expressions?

Answer 10

not really lol. Im sorry. I know I am confusing but its bc I am so confused

Answer 11

I just need someone to break down the steps to solve the problems. Like dummify them I guess

Answer 12

first step to solving a quadratic equation 'by factoring' is to factor the quadratic expression \[Ax^2 + Bx + C\]
Look for two numbers that multiply together to equal C, and add up to equal B.

for example, for \[y^2 + 3y + 2\]
notice that the numbers 2 and 1 multiply to 2 and add up to 3

Answer 13

once you've found those 2 numbers, you can factor the quadratic expression \[y^2+3y+2\]
into
\[(y+2)(y+1)\]

Answer 14

Basically, factorise the equation you have, then make it equal to 0:
(y+2)(y+1)=0
Then take each equation and work out x, so:
(y+2)=0, so y must equal -2 and (y+1)=0, so y must equal -1. Those are your two x-coordinates.

Answer 15

Ok. I got that part down. so I guess I do know the factoring part lol

Answer 16

ok so basically what you are saying is change the signs like (y + 2) to -2. and +1 to -1 and Im done?

Answer 17

sorry had a phone call

once you've factored the quadratic expression, it becomes very easy to solve the quadratic equation
\[(y+2)(y+1)=0\]

the equation is now telling you that either factor, y +2, or y +1, can equal to zero, so you can consider 2 possible solutions

\[y+2 = 0\]
\[y + 1 = 0\]

solve those 2 equations to get

\[y = -2\ \ \ or\ \ \ y = -1 \]

Answer 18

Yes, in a sense. You've just got to make y+2=0 and y+1=0

Answer 19

ok heres another one. (6k + 5) (k+4) = 0 some how she made it into a fraction 6k/6 = 5/6 and the answer ended up being k+4=0 k =-4 How did that happen? lol

Answer 20

it's the same thing\[(6k+5)(k+4)=0\]

you can break the above equation down to two equations:\[6k+5 = 0\]\[k + 4 = 0\]

solving both of those equations gives us (I'll start with the first one)\[6k = -5\]\[k = -\frac{5}{6}\]
and solving the second equation gives us \[k = -4\]
so our solutions are y = -5/6 or y = -4

Answer 21

Also I have a big problem with knowing if its supposed to be a positive or negative, Like when your doing the foil method on a problem with a positive and 2 negs.

Answer 22

what's the foil method? :(

Answer 23

First Outter Inner Last

Answer 24

I've never heard of that method. I might have used it before though :-P

Answer 25

but how did you get -5 from 6k?

Answer 26

solve this equation \[6k + 5 = 0\]

Answer 27

well basically you muliply the first of the equation times the first of the second... (y + 2) (4y- 1) you would multiply y * 4

Answer 28

would it be -1?

Answer 29

k= -1 i mean

Answer 30

how did you get k = -1?

Answer 31

6k - 5? was that wrong?

Answer 32

For that one you just wrote, @bhunsuck, the foiled answer is: 4y^2+-1y+8y-2. This'll give you:
4y^2+9y-2

Answer 33

oh! you mean expanding the expression :-P

Answer 34

but is the problem done?

Answer 35

I'm sure it is :-D

Answer 36

so the answer will be 4y^2+9y-2=0

Answer 37

For (y + 2) (4y- 1), yes it is.

Answer 38

huh.. cool... do you mind working a few more problems? I do better with repetition

Answer 39

Yeah, sure. Fire away. :)

Answer 40

ty :D

Answer 41

6\[x^2\]= 4+\[5x\]

Answer 42

ooops sorry, im not very good with this thing yet lol

Answer 43

you will get better :-D

Answer 44

Is that \[6x ^{2}= 4+ 5x\]?

Answer 45

i know you have to change te signs and go from squares, variable then numbers such as 6x^2 - 5x-4=0

Answer 46

yes it is damned

Answer 47

hello lol

Answer 48

\[6x ^{2}-5x-4=0\]
Then, factorise:
(\[\left( -2x-1 \right)\left( -3x+4 \right)\]
Then, make both equations equal to zero and figure them out, so:
(-2x-1)=0 and (-3x+4)=0
Finally, find out x, which seems pretty difficult. I think it's:
x=1/-2 and x=-4/-3

Answer 49

...

Answer 50

ok so how did u get the -2x -1 and -3 +4?

Answer 51

Sorry I took so long. Was trying to figure out the last bit. Just factorise. It's basically trial and error and you have to try different numbers until you figure it out. For this one, I knew the two x's would either be 6x and x or 3x and 2x because they multiply to make 6x^2. Then I had to find two other numbers that multiplied to make the last number but added up to make the middle number, but taking the x number's into account. This may be a slightly terrible way of describing it to you XD

Answer 52

yeah kinda is hard to understand lol

Answer 53

ur answer x = -1/2 and x= 4/3 is right btw

Answer 54

Okay, what don't you get, specifically? Thank god for that XD

Answer 55

Oh jesus... I know how to factor.. I know I know how to factor... But when I look at these problems it all goes out of my head! lol hold on a sec. I am trying to soak this in and figure it out.

Answer 56

I know how it feels, doing revision just before a test. Didn't go to badly, luckily.

Answer 57

OMG, ok lol... My husband is soooooo right.. (shhh I didn't say that) I over think things and then I confuse myself! Ok I got the first part. I feel so silly lol

Answer 58

I was lost when I first did this stuff too :-P

Answer 59

hi agdg, wb :)

Answer 60

Haha! You have a sensible husband then. Okay, what other bit don't you get?

Answer 61

ok so let me get this straight... U got -2x -1 and -3 +4 bc 2x 3= the 6 and added together makes the 5. and 1 x 4 = 4 and added together makes the 5 right?

Answer 62

How do you know to turn it into a fraction? when theres more than one number in the answer?

Answer 63

XD Kinda. I got the -2x and -3x because yes they do make 6 but then I had to pick numbers that added up to 5 but multiplied to make 4, taking into account that those two numbers had to be multiplied by the 3x and 2x.
Well, when I got the two brackets, all I did was rearrange them to make it equal to x, so:
(-2x-1)=0
-2x=1
x=1/-2. You remove the brackets by the way because they're not really serving any purpose.

Answer 64

cool

Answer 65

can u explain x^2=7x? The way I think is just take the x-7=0 x=7

Answer 66

Divide everything by x and you have your answer for that one. So, yes, x=7

Answer 67

so y in gods name dont they just say take out the x's and that will be your answer? lol

Answer 68

be careful about dividing by x: what if x = 0?

Answer 69

well how would i know?
My main goal at this point is to have to learn as LITTLE steps to solve as possible bc my brain can't hold all the different crap

Answer 70

instead of dividing, try taking the square root of both sides :-D

Answer 71

i dont know how to find the square roots. thats one of mt problems

Answer 72

you're right; the square root won't work for that particular equation :(

go ahead and divide by x, but keep in mind that one of the possible solutions is x = 0

Answer 73

But you can't really square root 0, can you?

Answer 74

i can find them in the smaller numbers but not the bigger ones.

Answer 75

the square root of zero is zero.

still doesn't work with the equation x^2 = 7x though :(

I would divide by x

x = 0 or x = 7

Answer 76

ok here one where y = 0 but I dont understand how she got that answer

Answer 77

what's the equation?

Answer 78

10y^2 = 5y

Answer 79

same reason: one possible solution to 10y^2 = 5y is

y = 0

Answer 80

well we are supposed to list both solutions. y=0 y= 1/2

Answer 81

I have like 30 problems to do before class in like 2 hours. So i guess I better go. THANK YOU SO MUCH for your help. Do you come on here alot? I know I will need more help lol