Question

Solutions? x^2+14x+45=0? Is it x+-6 and x=2?

Answer 1

x=6

Answer 2

no

Answer 3

-6

Answer 4

you need to factor first

Answer 5

\[x^2+14x+45=(x+9)(x+4)\]

Answer 6

WAIT

Answer 7

ok

Answer 8

i made a mistake anyway. it is
\[(x+9)(x+5)\]

Answer 9

(x+5)(x+9) so then it would be x

Answer 10

x=-5 and -9

Answer 11

ok got it. now you have
\[(x+5)(x+9)=0\] set each factor = 0 and solve.

Answer 12

yes exactly!

Answer 13

Okay cool!

Answer 14

got it. good!

Answer 15

but what about 4x^2+4x+1?

Answer 16

now you have to factor first. this is a perfect square like the other ones you did

Answer 17

So this is (2x+2)(2x+2) or 1

Answer 18

1

Answer 19

2x+1 2x+1

Answer 20

yes
\[(2x+1)(2x+1)\]

Answer 21

so this would be x=1/2?

Answer 22

close

Answer 23

and -1/2

Answer 24

no just the second one

Answer 25

both factors are the same so only one solution

Answer 26

and that is the answer? -1/2?

Answer 27

that is it. you have
\[2x+1=0\]
\[2x=-1\]
\[x=-\frac{1}{2}\]

Answer 28

other factor is the same so you don't need to do it again

Answer 29

Okay, I think I got that down. but what if it its not perfect squares?

Answer 30

then you will have two factors and perhaps 2 answers.

Answer 31

perfect square gives one answer. but the problem we did before this had two answers

Answer 32

so for 2x^2+7x+3 the answer would be x=-3 and -1/2?

Answer 33

i don't know but i will check. did you factor?

Answer 34

yes I did.

Answer 35

yes you are right!

Answer 36

\[(x+3)(2x+1)=0\] etc

Answer 37

Great!! so for x^2-9=0 the answer will be x=3 and -3?

Answer 38

yes. you can factor or just write
\[x^2-9=0\]
\[x^2=9\]
\[x=\pm3\]

Answer 39

if you factor you write
\[x^2-9=(x+3)(x-3)=0\] and get the same answers

Answer 40

Okay! so I just want to go over putting an equation into ax^2+bx+c=0. if it is 9-5x^2=(x+2)^2-4

Answer 41

first multiply out on the right hand side

Answer 42

okay done.

Answer 43

then collect like terms on the right hand side. you should end up with
\[x^2+4x\]

Answer 44

okay. yes. I have that.

Answer 45

you want this set = 0 so add
\[5x^2\] to both sides and subtract 9

Answer 46

you will have a 0 on the left and everything else on the rigth

Answer 47

so it is 0=5x^2=4x-9? Is that the correct answer?

Answer 48

no I mean. +

Answer 49

between the 5x^2 and the 4x

Answer 50

no

Answer 51

we have
\[9-5x^2=x^2+4x\]

Answer 52

add \[5x^2\] to both sides to get rid of it in the left,
\[9=6x^2+4x\]

Answer 53

you forgot to add the \[5x^2\] to the \[x^2\]

Answer 54

OH. okay. so then it is 0=6x^2+4x-9?

Answer 55

then you will get
\[0=6x^2+4x-9\]

Answer 56

yes

Answer 57

which of course you probably want to write as
\[6x^2+4x-9=0\] not that it makes any difference

Answer 58

Okay. Thanks!

Answer 59

yw