Here's the question you clicked on:
Emily778
Solve the quadratic equation by completing the square. x^2-12x+7=0
add 29 to both sides.
can you add 29 to both sides?
Both. \[x^2-12+7+29=0+29\]
To make the left side into perfect square because you have to have (-6) (-6) which equals 36.
x^2 - 12x + 7 = 0 x^2 - 12x = -7 x^2 - 12x + (-12/2)^2 = (-12/2)^2 - 7 x^2 - 12x + 36 = 36 - 7 (x - 6)^2 = 29 Take square root of both sides to finish solving for x
Oh. Okay. I still don't quite get it, but lets move on. What shall I do next? :D
Uhh... which one is right?
x^2-2*x*6+7=0 x^2-2*x*6+36-36+7=0 (x^2-2*x*6+36)-29=0 (x-6)^2-29=0 (x-6)^2=29 x-6=29^1/2 x=(29)^1/2+6
Are all of these right, just in different ways?
You add 36 to both sides, not 29
@Hero, no you add 29, because that makes the left side into perfect square. @Emily778, I know what you are confused about, lets learn how to complete the square.
I'll just do this x^2 - 12x + 7 = 0 x^2 - 12x = -7 x^2 - 12x + (-12/2)^2 = (-12/2)^2 - 7 x^2 - 12x + 36 = 36 - 7 (x - 6)^2 = 29
@SolomanZelman, if that's true, then explain how I got my result
actually in (a+b)^2=a^2+b^2+2ab here 2ab is 12x=2*x*6 so u need to make b^2 that is 36
I know exactly how the process works and what to add to both sides.
@Emily778, do you get how I came up with 29? lets do a different example. GUYS PLEASE, LETS TEACH HER, NOT DO HER WORK> She needs to understand.
@hero if u will add 29 both side than on x^2-12x+7+29=29 x^2-12x+36=29
She'll only understand if those teaching know what they are talking about
@hero plzz plzzzz plzzzz for god sake use little of ur mind
@Emily778, do you get how I came up with 29? Do you want to learn how to complete the square?
I think @Emily778 already has it figured out lol
MODERATOR, please don't ruin the thread, and confuse teh asker, lets teach her!
@Hero What do I do after this? Or is it already done? x^2 - 12x + 7 = 0 x^2 - 12x = -7 x^2 - 12x + (-12/2)^2 = (-12/2)^2 - 7 x^2 - 12x + 36 = 36 - 7 (x - 6)^2 = 29
Sorry If I was offensive, @Hero .
Finish solving for x @Emily778
Take the square root of both sides
Can you show me how? I am so confused:(
@Emily778 PLEASE read this CAREFULLY. Do you know how to come up with the number that you have to add to both sides to ake the left side into a perfect square?
to make the left side...typo
Emily, do you get it conceptually?
Add 6 to both sides to finish solving for x
so, like this @Hero (x-6)^2+6=29+6?
@Emily778, please observe what I did above first
I think you missed that last bit I posted in a drawing
oh, you have to square root first?
Yes you can't do anything until you take the square root of both sides.
@SolomonZelman, I assure you that we're supposed to add 36 to both sides. I wouldn't tell you anything wrong here.
If you don't believe me, I have ways of convincing you.
For one, I know you would never argue with WolframAlpha.
Hero, how does 36 make the left side into a perfect square, it is \[x^2+12x+7=0\]
Hang on, @SolomonZelman, WolframAlpha is undefeated.
I meant\[x^2-12x+7=0\]
okay, so it's actually x-6+6=+=sqrt29=6?
Well yes it is then.
omg don't argue with @Hero
@Hero, I am sorry to say this, but you have to rvw over some staff.
he is correct and I am telling you it
@hero okay, so it's actually x-6+6=+=sqrt29=6?
Do you guys know how to complete the square, you don't need any online calculator for this. Idk how to even tell you this. you help with hard staff that I have no clue about, and make a simple error, in completing a square.
Don't tell me 1+1=4
I know I do not need a calculator and I know he is correct
Notice that wolframalpha says to add 36 to both sides.
In this here link: http://tr.im/4klyc
maybe you put in -7, instead of +7.
OK, when you subtract 7, but you can just add 29, without subtracting 7.
Thankyou! Is that it @Hero
There's only one correct way to do it @SolomonZelman. Wolframalpha shows the steps. Go review them.
@Emily778, I think you have it but review your symbology for mistakes and check the link on WolframAlpha
@Hero it wastes a step on subtracting 7 from both sides, you still get the same thing, \[x^2-12+36=29\] Do you see my point?
Do you guys just argue, or actually READ MY REPLIES?
solomon, study more before you question
@nincompoop, you obviously didn't understand what i said.
Do you understand what I said, I don't need to study, I am fine.
@SolomonZelman, there's a difference between wasting steps and skipping steps.
Skipping steps leads to confusion
I am not saying the Wolframalpha is wrong, it just has a different way of showing... Yes, my bad, perhaps I skipped it, not wasted.... My teacher told me that both ways are valid, and to me that makes perfect sense. And YES, instead of arguing lets continue studying.
I still think though, that emily didn't get it conceptually.
That was quite offensive.
Yes, but some things can be figured out through self-study, observation and understanding. One thing is for sure. She surely won't get it if she is fed incorrect info.
True, and I was trying to ask emily if she figured out the first step(s) in completing the square.
Yes, me too, that's why I kept on asking that.
By the way @SolomonZelman, I'm fully aware of your method of doing it. But I abandoned it after I realized that steps were being skipped. It is wrong for teachers to teach it that way.
Well, not that they taught me that, but didn't get any points off.
it's cool can't teach someone about skipping when that person barely knows anything to begin with
I don't really see the difference between subtracting seven at first or not. To me it makes perfect sense to do my way.
I see your point though....
And she still hasn't figure it out conceptually, just got the answer and left.
@nincompoop, actually in biology that could be true, like when parents (or future parents) combine or whatever the word for that is... LIKE twins.
it's called precision
\(\huge {\color{green} {Yup,} }\) forgot the name for this, I learned that last year.
\(\huge {\color{green} {H} }\)\(\huge {\color{blue} {a} }\)\(\huge {\color{red} {v} }\)\(\huge {\color{orange} {e} }\) \(\huge {\color{purple} {a} }\) \(\huge {\color{lightblue} {n} }\)\(\huge {\color{grey} {i} }\)\(\huge {\color{brown} {g} }\)\(\huge {\color{lightgreen} {h} }\)\(\huge {\color{yellow }{t} }\)\(\huge {\color{darkgreen } {!} }\)
The difference between the methods is, one is systematic, and the other is not. The non-systematic method is an approach that some would refer to as improper methodology.
\[\(\huge {\color{green} {B^{Y^{E ^{!}}}} }\)\]
Didn't delete the parenthesis....
The method I use can be generalized to this: You begin with x^2 + bx - c = 0 add c to both sides: x^2 + bx = c Then add (b/2)^2 to both sides: x^2 + bx + (b/2)^2 = (b/2)^2 + c
This is a systematic approach and you end up with (x + b/2)^2 = b^2/4 + c
Well, then improper methodology just shows that the person who did the steps is SMART.
Now let's see @SolomonZelman explain his method in a systematic, generalized manner.
I would just use an example similar to the one we solved just now.
Examples are not generalized. And they don't show the systematic methodology.
Systematic methods have a better chance of being understood than non-systematic methods.
OK, I already said that I agree that to start teaching, systematic.... but not using it still shows that the solver is smarter.