Question

Please help me :( http://prntscr.com/b0fgrr

Answer 1

@myininaya @ganeshie8 pls help :(

Answer 2

they made this problem grosser
I don't know why they left that 2 there...
since you can divide both sides by 2 first
and just have x^2+6x-7=0
but whatever
I guess we will come back to play with the factor 2 later
even though I think that is dumb

Answer 3

so what number do we need to add to both sides in order to write x^2+6x+something as a square ?

Answer 4

hint:
\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2\]

Answer 5

compare your x^2+6x+....
to x^2+kx...
what does k need to be?

Answer 6

oh i went to eat yogurt sorry

Answer 7

lemme read it

Answer 8

@myininaya im not really sure :( I am like extremely confused on this subject

Answer 9

can you tell me what value of k we need so that x^2+kx appears the same as x^2+6x ?

Answer 10

6

Answer 11

right k is 6

Answer 12

so let me input 6 into our formula up above
\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \\ x^2+6x+(\frac{6}{2})^2=(x+\frac{6}{2})^2\]

Answer 13

ok so we now know what we need to add to both sides

Answer 14

yes we have that thing, do we need to make the 2 sides equal?

Answer 15

let's go back up to our problem
\[x^2+6x-7=0\]
add 7 to both sides
and then add the number in our formula above to both sides that will complete the square on the left hand side of our expression

Answer 16

the number in our formula? wait so we would've had \[x^2+6x=7\]

Answer 17

yes that is what you have after adding 7 to both sides

Answer 18

yes ma'am

Answer 19

now what is the number we need to add to both sides so that we can complete the square on the left hand side ?

Answer 20

do we need either like a 36 or like a 3 or something

Answer 21

you remember this right:
\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \\ x^2+6x+(\frac{6}{2})^2=(x+\frac{6}{2})^2 \]?

Answer 22

yes, but it never showed an x, would it be 6?

Answer 23

we are aiming to write the left hand side as (something)^2

Answer 24

\[x^2+6x+?=7+? \\ \text{ we need to figure out what that ? is } \\ \text{ so that we can write the left hand side as a square } \\ \text{ using } x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \\ x^2+6x+\color{red}{(\frac{6}{2})^2}=(x+\frac{6}{2})^2 \]

Answer 25

what do u mean by making it a square, i remember i did a problem like this earlier, it was simpler, i got to this part before i just don't know what it is

Answer 26

i thought it was either 3 or 9 at first

Answer 27

(6/2)^2 is 3^2 aka 9

Answer 28

yes

Answer 29

that is what you need to add to both sides

Answer 30

oh we just had to solve that little thingy

Answer 31

\[x^2+6x+(\frac{6}{2})^2=7+(\frac{6}{2})^2\]

Answer 32

now you can write the left hand side as a square

Answer 33

\[x^2+6x+9=16\]

Answer 34

right
but I would leave that 9 on the left as 3^2
It makes easier for me to remember how to write the square

Answer 35

oh okay, it would have to be a (x+3)^2 right?

Answer 36

\[\text{ examples } \\ x^2+2x+1^2=(x+1)^2 \\ x^2+4x+2^2=(x+2)^2 \\ x^2+6x+3^2=(x+3)^2 \\ x^2+8x+4^2=(x+4)^2 \\ x^2+10x+5^2=(x+5)^2 \\ x^2+12x+6^2=(x+6)^2 \\ x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2\]

Answer 37

correct

Answer 38

\[(x+3)^2=16 \]

Answer 39

should be a 2 at the front of that?

Answer 40

now if you want to get one of their equations
just multiply both sides by 2

Answer 41

the equation still holds if you multiply the same number on both sides (just don't multiply both sides by 0 :p)

Answer 42

well its either option C or D, we just need to figure out the X's

Answer 43

because of the math property multiplying thing

Answer 44

anyways multiplying by 2 goes backwards
we have
(x+3)^2=16

take the square root of both sides

Answer 45

square root of 16 is 4

Answer 46

that means x+3=4 or x+3=-4

Answer 47

and the left side is (x+3)?

Answer 48

oh ur ahead

Answer 49

yes so the answer is 1 or -7

Answer 50

right now we followed the directions to get this answer
but if we didn't need to show any work (if we could follow our own directions that is)
we could have factored the x^2+6x-7

Answer 51

oooo shortcuts

Answer 52

2x^2+12x-14=0
divide both sides by 2
x^2+6x-7=0
(x+7)(x-1)=0

x+7=0 or x-1=0
x=-7 or x=1

Answer 53

oh, that was a lot simpler

Answer 54

thank you ma'am <3 that was helpful because I was really confused ;-; I didn't really pay attention

Answer 55

IM LEARNING STUFFZ @ganeshie8 <3

Answer 56

\[ax^2+bx+c=0 \\ \]
a not zero...

\[\text{ First step: subtract } c \text{ on both sides } \]

\[ax^2+bx+c-\color{red}{c}=0-\color{red}{c} \\ ax^2+bx+0=-c \\ ax^2+bx=-c \\ \]


\[\text{ Second step: divide } a \text{ on both sides } \]

\[\frac{a}{\color{red}{a}}x^2+\frac{b}{\color{red}{a}}x=\frac{-c}{\color{red}{a}} \\ 1x^2+\frac{b}{a}x=\frac{-c}{a} \\ x^2+\frac{b}{a}x=\frac{-c}{a}\]


\[\text{ Third step: we need to figure out what }\\ \text{ to add to both sides to complete the square on the left }\]



\[\text{ Notice the coefficient of } x \text{ then notice what I do }\]
\[x^2+\color{blue}{\frac{b}{a}}x+(\frac{\color{blue}{b}}{2\color{blue}{a}})^2=\frac{-c}{a}+(\frac{\color{blue}{b}}{2 \color{blue}{a}})^2 \\ \]

\[\text{ Now keep a close eye on the things inside the squares and how } \\ \text{ the left hand side get rewritten in this next step }\]

So keep eye on the thing in red:
\[x^2+\frac{b}{a} x+(\color{red}{\frac{b}{2a}})^2=\frac{-c}{a}+(\frac{b}{2a})^2 \\ (x+\color{red}{\frac{b}{2a}})^2=\frac{-c}{a}+(\frac{b}{2a})^2\]

Answer 57

O______O

Answer 58

so that is how we solve this thing

Answer 59

\[\text{ Now I'm going to rewrite the right hand is all }\]
\[(x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{b^2}{4a^2} \\ (x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{b^2}{4a^2} \\ \text{ find a common denominator on the \right } \\ \text{ multiply first fraction by } \frac{4a}{4a} \\ (x+\frac{b}{2a})^2=\frac{-c}{a} \cdot \color{red}{\frac{4a}{4a}} +\frac{b^2}{4a^2} \\ (x+\frac{b}{2a})^2=\frac{-4ac}{4a^2}+\frac{b^2}{4a^2} \\ (x+\frac{b}{2a})^2=\frac{-4ac+b^2}{4a^2} \\ (x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2} \\ \text{ Almost the last step is to take \square root of both sides } \]


\[x+\frac{b}{2a}= \pm \sqrt{\frac{b^2-4ac}{4a^2}} \\ \text{ Last step: subtract } \frac{b}{2a} \\ \text{ on both sides } \\ x=\frac{-b}{2a} \pm \sqrt{\frac{b^2-4ac}{4a^2}} \\ \\ \text{ now we can rewrite this a little } \\ x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}} \\ x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

\[ \text{ which you should notice is the quadratic formula }\]

Answer 60

now I know that is probably going to take some reading :p

Answer 61

i see a bunch of numbers and letters @myininaya

Answer 62

i shall keep this forever n ever

Answer 63

lol

Answer 64

your efforts shall not go unrewarded

Answer 65

as a token of appreciation <3

Answer 66

I don't need to be rewarded. I just hope maybe you can look on it later and find it helpful in someway :p

Answer 67

But thanks for the token of appreciation

Answer 68

I do find it helpful <3 @myininaya

Answer 69

cool stuff then ! :)

Answer 70

i have to go take care of my sick kitty
enjoy the rest of your night

Answer 71

you too :)