Solving Quadratics by Factoring and Graphing

zee-f · 2011-07-25 02:26:48

1. x^2 49 = 0

so for this one this is what i did
(x-7)(x+7)=0
x-7=0
x+7=0
x=7 x= -7

2. 3x^3 12x = 0

2- 3x(x^2-4)=0
3x=0 x=0
x^2-4=0
x^2=4

3. 12x^2 + 14x + 12 = 18

3- -18+12x^2+14x+12=0

12x^2+14x-6=0
2(6x^2+7x-3)=0

then i just dont know what to do

4. x^3 + 22x^2 121x = 0

(x+11)(x-11)=0

x=-11 or x=11

5. x^2 4x = 5

so subtract 5 from both sides

x^2-4x-5=0
(4+x)(1+x)=0

x= -4 or x=-1

anonimnystefy · 2011-07-25 02:34:42

hi zee-f

the first one is correct,but the second one isn't finished.

you didn't solve x^2-4=0.

bobbym · 2011-07-25 02:38:24

Hi;

#5 is incorrect.

anonimnystefy · 2011-07-25 02:44:22

for the third yoiu can do this:
this is what you got:
2(6x^2+7x-3)=0
now you can do this:
6x^2+7x-3=0/2
6x^2+7x-3=0
6x^2-2x+9x-3=0
2x(3x-1)+3(3x-1)=0
(2x+3)(3x-1)=0
2x+3=0
2x=-3
x=-3/2

3x-1=0
3x=1
x=1/3

anonimnystefy · 2011-07-25 02:46:34

also check 4 again you threw out the GCF.

Bob · 2011-07-25 06:15:20

hi zee-f

number 5 see diagram.

You've got to put in numbers where I've highlighted.

Two numbers that multiply to make minus 5 .... so one of them must be minus. (which will be fixed by the next step)

Then fill in the x bits to make -4x. If you get +4x you've picked the wrong number to be minus.

Bob

fractions-helper · 2011-07-25 08:52:03

let's do # 4

x^3 + 22x^2 121x = 0

Take "- x" out as gcf

- x (x^2 - 22x + 121) = 0

- x (x - 11)^2 = 0

So, -x = 0 which is x = 0 and (x - 11) = 0 which is x = 11

Hence x = 0, 11

bobbym · 2011-07-25 13:14:48

Hi zee-f;

I have been watching you post zillions of answers, I know you are trying, let's take #5 for instance:

5)

x^2-4x-5=0
(4+x)(1+x)=0

x= -4 or x=-1

Okay, you factored it wrong, that happens. Haven't you been taught how to check your answers by plugging in? The -1 is okay

(-1)^2 - 4(-1) - 5

1 + 4 - 5 = 0 Checks!

Let's look at -4 as an answer.

(-4)^2 - 4(-4) - 5 ?= 0

16 + 16 - 5 = 27

27 is not 0 so - 4 is not a root.

My point is most of the work you are doing can be checked for validity by you. Also when you factor something you do not multiply back to see if it is right.

I love posting to you but I think you will never be able to stand on your own two feet if you can not check your work.

If you want I will start a new thread on plugging in and multiplying binomials, just for you.

zee-f · 2011-07-26 01:42:46

i would love that bobbym am ganna try to do all of them again by my own just to try again

bobbym · 2011-07-26 02:10:00

Hi;

When you are ready, just start a new thread and ask how to check whether answers are right. Then copy problem #5.

zee-f · 2011-07-27 05:43:17

hey so i dont want to get confused lets start with number 2

this is what i did

3x^3-12x=0

x(3x^2-12)=0

x=0

bobbym · 2011-07-27 05:59:38

What about the 3x^2 - 12 ?

anonimnystefy · 2011-07-27 06:00:25

hi zee-f

you didn't get the whole GCF.look if you can get anything else out of the brackets.

zee-f · 2011-07-27 06:59:26

3x-12x=0

3x(x^2-4)=0

3x=0
3x/3 0/3

x=0

x^2-4=0
-4 -4

x^2=-4

bobbym · 2011-07-27 07:02:09

x^2-4=0
-4 -4

-4 - 4 does not equal 0

zee-f · 2011-07-27 07:07:38

no i subtracted 4 from both sides to get the x^2 by its self

zee-f · 2011-07-27 07:09:09

its ganna solve the x^2 for me when i do that

bobbym · 2011-07-27 07:10:13

That is what I am saying. Subtracting 4 from the left side does not get the x^2 by itself.

-4 - 4 = - 8 so you did not get rid of the - 4

anonimnystefy · 2011-07-27 07:10:48

hi zee-f

if you have x^2-4 and then subtract 4 you get
x^2-4-4=x^2-8

zee-f · 2011-07-27 07:13:29

thats confusing because x^2-4=0

so dont i subtract the the four to get rid of it ?

zee-f · 2011-07-27 07:14:42

so subtract a four not a negative four

bobbym · 2011-07-27 07:16:31

No add a 4.

-4 + 4 = 0

anonimnystefy · 2011-07-27 07:17:07

hi zee-f

you can subtract a negative four which would be like adding a four,but what you did is subtract a four.

zee-f · 2011-07-27 07:19:47

so x^2 would equal a +4

bobbym · 2011-07-27 07:20:11

yes, once you do the right thing.

Math Is Fun Forum

#1 2011-07-25 02:26:48

Solving Quadratics by Factoring and Graphing

#2 2011-07-25 02:34:42

Re: Solving Quadratics by Factoring and Graphing

#3 2011-07-25 02:38:24

Re: Solving Quadratics by Factoring and Graphing

#4 2011-07-25 02:44:22

Re: Solving Quadratics by Factoring and Graphing

#5 2011-07-25 02:46:34

Re: Solving Quadratics by Factoring and Graphing

#6 2011-07-25 06:15:20

Re: Solving Quadratics by Factoring and Graphing

#7 2011-07-25 08:52:03

Re: Solving Quadratics by Factoring and Graphing

#8 2011-07-25 13:14:48

Re: Solving Quadratics by Factoring and Graphing

#9 2011-07-26 01:42:46

Re: Solving Quadratics by Factoring and Graphing

#10 2011-07-26 02:10:00

Re: Solving Quadratics by Factoring and Graphing

#11 2011-07-27 05:43:17

Re: Solving Quadratics by Factoring and Graphing

#12 2011-07-27 05:59:38

Re: Solving Quadratics by Factoring and Graphing

#13 2011-07-27 06:00:25

Re: Solving Quadratics by Factoring and Graphing

#14 2011-07-27 06:59:26

Re: Solving Quadratics by Factoring and Graphing

#15 2011-07-27 07:02:09

Re: Solving Quadratics by Factoring and Graphing

#16 2011-07-27 07:07:38

Re: Solving Quadratics by Factoring and Graphing

#17 2011-07-27 07:09:09

Re: Solving Quadratics by Factoring and Graphing

#18 2011-07-27 07:10:13

Re: Solving Quadratics by Factoring and Graphing

#19 2011-07-27 07:10:48

Re: Solving Quadratics by Factoring and Graphing

#20 2011-07-27 07:13:29

Re: Solving Quadratics by Factoring and Graphing

#21 2011-07-27 07:14:42

Re: Solving Quadratics by Factoring and Graphing

#22 2011-07-27 07:16:31

Re: Solving Quadratics by Factoring and Graphing

#23 2011-07-27 07:17:07

Re: Solving Quadratics by Factoring and Graphing

#24 2011-07-27 07:19:47

Re: Solving Quadratics by Factoring and Graphing

#25 2011-07-27 07:20:11

Re: Solving Quadratics by Factoring and Graphing

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