Question

help please.. i need help with number 7

Answer 1

i did part a

Answer 2

part b which is a true stmt

Answer 3

i did a set up for the proof but i got stuck on the backwards direction

Answer 4

So you need help solving for a?

Answer 5

i need help with b since i was doing the bi conditional proof

Answer 6

i got stuck on the backwards direction

Answer 7

Yes, but in b it says to "solve for a" do you see that?

Answer 8

Should I circle it for you?

Answer 9

oh wait.. it says part a lol umm how do i determine if they are true?

Answer 10

You seem confused

Answer 11

yeah i am :/

Answer 12

You should identify exactly where you need help.

Answer 13

hmm, well i need help with b. Where do i start? i think my work is wrong

Answer 14

I don't know what "b" means. I don't see a "b"

Answer 15

I already asked you if you needed help solving for a but it seems you did not recognize what I was talking about

Answer 16

Okay, but I was looking at your "proofs". There is a part where it says "solve for a" and you did not solve for a. Why did you not solve for a?

Answer 17

i didnt solve it since i was lost.

Answer 18

You were lost because you could not solve the equation for a?

Answer 19

well i think i solved it which i created the bi conditional statement for it

Answer 20

I'm referring to the Backwards Direction P2

Answer 21

oh okay, sorry i got lost. Yes, i could solve for which i'll get it into factored form

Answer 22

i dont know what question you are talking about send it one more time plz

Answer 23

He already posted it twice

Answer 24

the picture is at the top and question 7

Answer 25

i am sorry i cant help

Answer 26

I am sorry i can¨t help so sorry

Answer 27

becsuse i really dont know the answer

Answer 28

because

Answer 29

i dont really now the answer i am in 8th grade

Answer 30

im completely lost tbh lol i solved for a already for i ended up factoring it , which i think its wrong

Answer 31

@xXMarcelieXx to solve for \(a\) you have to write it in vertex form by completing the square

Answer 32

oh that sounds about right

Answer 33

did you figure out part A. and part B

Answer 34

oh wait i think got it
one sec

Answer 35

Yes correct

Answer 36

okay so then, we would square each side and we up having sqroots but isnt that hard to solve?

Answer 37

i really dont know

Answer 38

In forward direction, when you solved for a, what did you do with the expression after that?

Answer 39

i plugged in a into the conclusion and it was easier to solve

Answer 40

So why not do that for the backwards direction as well?

Answer 41

i did this but i cant get it into the form of 8p though

Answer 42

One moment

Answer 43

okie

Answer 44

what grade are you in xXMarcelieXx

Answer 45

it is kinda off topic but i kinda what to know

Answer 46

I have something that might work.

Answer 47

I don't know where @xXMarcelieXx went to though

Answer 48

$$ \begin{array}{l}
a^{2} +4a-4=8r\\
a^{2} -8a+4a-4=8r\ -\ 8a\\
a^{2} -4a-4=8r-8a\\
a^{2} -4a+4-4=8r-8a+4\\
a^{2} -4a+4=8r-8a\\
( a-2)^{2} =( 8r-8a)\left(\dfrac{8r+8a}{8r+8a}\right)\\
( a\ -\ 2)^{2} =\dfrac{( 8r-8a)( 8r+8a)}{8r+8a}\\
( a\ -\ 2)^{2} =\dfrac{64r^{2} -64a^{2}}{8r+8a}\\
( a\ -\ 2)^{2} =\dfrac{64( r+a)( r-8)}{8( r+a)}\\
( a-2)^{2} =\dfrac{64}{8}( r-8)\\
a-2=\sqrt{\dfrac{64}{8}( r-8)}\\
a-2=\sqrt{64}\sqrt{\dfrac{r-8}{8}}\\
a\ -2=\ 8\sqrt{\dfrac{r-8}{8}}
\end{array}$$

Answer 49

Somehow we'd have to show that root expression is equal to \(p\)

Answer 50

And if it can't be proven then maybe the original statement is false

Answer 51

Also I just realized something is off with your work

Answer 52

oh, where is it off?

Answer 53

Basically, what I'm trying to figure out is, how did you come up with your table of values?

Answer 54

You're using two variables I think when you should just be using \(a\) as the variable

Answer 55

That's how you've always done these unless the original statement had two variables which it does not

Answer 56

Remember, when we are doing these normally, the other variable isn't introduced until later. We should keep that same consistency here

Answer 57

I'm referring to your table of true/false values btw

Answer 58

right, i messed on my algebra since i didnt complete the square right.
i thought we always used two variables for this stuff?

Answer 59

Negative.

Answer 60

Not for verifying the original statement

Answer 61

The original statement only has one variable to use for verifying it which is variable \(a\)

Answer 62

Do you get what I'm saying?

Answer 63

As I said, when doing a proof, you only introduce another variable AFTER you have done verifying with tables

Answer 64

You're still more confused than you should be

Answer 65

oh okay, i think i get what you are saying

Answer 66

how did you get this second step ?

Answer 67

I obviously subtracted 8a from both sides

Answer 68

Which you are allowed to do

Answer 69

Actually I realize the statement should be
\((a - 2)^2 = 8r - 8a + 8\)
\((a - 2)^2 = 8(r - a + 1)\)

Answer 70

One moment

Answer 71

okie

Answer 72

\(\begin{array}{l}
a^{2} +4a-4=8r\\
a^{2} -8a+4a-4=8r\ -\ 8a\\
a^{2} -4a-4=8r-8a\\
a^{2} -4a+4-4=8r-8a+4\\
a^{2} -4a+4=8r-8a\ +\ 8\\
( a\ -\ 2)^{2} =8r\ +\ 8\ -\ 8a\left(\dfrac{8r+8+8a)}{8r+8+8a}\right)\\
( a\ -\ 2)^{2} =\dfrac{( 8r\ +\ 8\ -8a)( 8r+8+8a)}{8r+8+8a}\\
( a\ -\ 2)^{2} =\dfrac{(( 8r\ +\ 8) -8a)(( 8r+8) +8a)}{8r+8+8a}\\
( a\ -\ 2)^{2} =\dfrac{( 8r+8)^{2} -( 8a)^{2}}{8r+8+8a}\\
( a\ -\ 2)^{2} =\dfrac{( 8( r+1))^{2} -( 8a)^{2}}{8r+8+8a}\\
( a\ -\ 2)^{2} =\dfrac{64( r+1)^{2} -64a^{2}}{8( r\ +\ 1+a)}\\
( a-2)^{2} =\dfrac{64\left(( r+1)^{2} -a^{2}\right)}{8( r+1+a)}\\
( a-2)^{2} =\dfrac{64}{8} \cdot \dfrac{( r+1+a)( r+1-a)}{( r+1+a)}\\
( a-2)^{2} =\dfrac{64}{8}( r+1-a)\\
a-2=\sqrt{\dfrac{64}{8}( r+1-a)}\\
a-2=\sqrt{64}\sqrt{\dfrac{r+1-a}{8}}\\
a-2=8\sqrt{\dfrac{r+1-a}{8}}
\end{array}\)

Answer 73

So you'd have to prove that \(p\) is equal to that root expression

Answer 74

But the expression has an a in it which is not good

Answer 75

I'm leaning towards the statement is false

Answer 76

ohhh, so then does that mean that the whole statement is false?

Answer 77

Hasn't exactly been verified whether true or false

Answer 78

But what you can do is select an \(a\), then see if you can find an \(r\) such that \(p\) is an integer.

Answer 79

The root expression must be equal to \(p\)

Answer 80

Hang on a min ...

Answer 81

Basically the root expression is equal to \(p\)

Answer 82

So it is proven

Answer 83

I did it by accident

Answer 84

\(p = \sqrt{\dfrac{r + 1 -a}{8}}\)

Answer 85

Not sure if your teacher will accept it though since \(a\) is included in the expression for \(p\)

Answer 86

But either way, its a formula that takes values \(a\) and \(r\) and returns a \(p\)

Answer 87

where \(a\) , \(r\) and \(p\) are integers that satisfy the given condition

Answer 88

For example \(a = 10\) and \(r = 17\) produces \(p = 1\)

Answer 89

You can produce infinite of these values

Answer 90

You can always find an \(a\) and and \(r\) that produces a \(p\)

Answer 91

Do you follow @xXMarcelieXx

Answer 92

oh okay, got it :)

Answer 93

@xXMarcelieXx have you submitted this to your teacher yet?

Answer 94

i got confused what she wrote after the or stuff

Answer 95

I'd have to review mods in order to continue helping you with them. I don't know how to manipulate them like that.

Answer 96

oh okay, i remember i did the the first part of (p->Q) with mods but the second part was too much