HD Homework Help Thread!

[StS]

Words

Ah, back to the basics. Many thanks to both you and myndzi; I feel like these are things I should know in-depth, yet my brain just won't supply the right words.  

[Destiny]

How to partial factor y=x^2+3x-10 and how to complete the square of  x^2 + 12x + 4 = 0. The finding max rev and profit i'll find later.

[StS]

How to partial factor y=x^2+3x-10 and how to complete the square of  x^2 + 12x + 4 = 0. The finding max rev and profit i'll find later.

I'll let someone else take care of partial factorisation, but completing the square goes as follows:

x2 + 12x + 4 = 0

You want to get the left hand side to be of the form (ax)^2 + 2abx + b^2, so that you can factorise it to (ax+b)^2. You therefore need to find a and b by comparing them:

a = 1 (because of the x^2 term)
and from the 12x term, you find: 2*1*b*x = 12*x
so b = 6

You now know that

x2 + 12x +36 = (x + 6)^2

Thus:

x2 + 12x + 4 = 0
x2 + 12x + 36 - 32 = 0
(x + 6)^2 = 32

Which you can now solve for x by taking the square root on both sides.

Hope this helps; if you've got any problem with any step, let me know.

[Rosti_LFC]

How to partial factor y=x^2+3x-10 and how to complete the square of  x^2 + 12x + 4 = 0. The finding max rev and profit i'll find later.

Completing the square is the easiest one, so I'll do that first.

For completing the square you basically follow a couple of steps:

Where you have the equation ax^2 + bx +c = 0

1) Start off with a(x+b/2a)^2. This is the key component that you're always going to get in any answer, and it's your focus point for finding the solution. Later on this becomes a bit intuitive but to start with you just have to remember that you need to divide b by 2a, and take the "a" co-efficient for x^2 and put it at the front of the brackets.

In this case it gives you (x+6)^2, as a=1 and b=12 from the original equation you're given

2) Take your bit for step 1) and expand the squared brackets

In this case (x+6)^2 = x^2 + 12x + 36

Note here that the "x^2 +12x" is the first part of the original equation - this should always be the case because of the way step 1) works.  If it's not, you've fucked up somewhere. What you're interested with here though is the +36 bit, because that's different.

3) So the last issue is that your squared brackets give you +36 instead of +4 at the end in the original equation, and you need to fix this. This is easy: to go from +36 to +4 you need to subtract 32. Add the -32 onto your step 1) and you're done:

(x+6)^2 - 32

To check:
(x+6)^2 - 32 = [x^2 + 12x + 36] - 32 = x^2 + 12x + 4 = the original equation (if it doesn't, you've made a mistake somewhere)

So why the hell go to the trouble of doing this? It's because completing the square gives you the same equation in a format that's a lot easier to interpret. x^2 + 12x + 4 is a quadratic equation and it's not obvious what the graph looks like. On the other hand (x+6)^2 - 32 is a much simpler form of the same equation to interpret. It represents the graph y = x^2, except with the curve shifted -6 on the x-axis (from the x+6 instead of x transformation) and shifted -32 on the y-axis. This is what x^2 + 12x + 4 will look like on a graph as well, but you can't just pull that out directly from the equation just by looking at it.

It also makes the solution a lot easier, because you're now trying to solve (x+6)^2 - 32 = 0

Which gives:

(x+6)^2 = 32
(x+6) = ±√32
x = ±√32 - 6

You can check on a calculator that your answer is right by substituting "x = √32 - 6" and "x = -√32 - 6" into "x^2 + 12x + 4" on a calculator and finding that they do indeed give 0 as the answer.


I'll do partial factorisation tomorrow because it's late and I'm a bit more rusty on it, but it basically works along similar kind of lines.

EDIT: StS expresses the first bit in a slightly different way but it's basically the same thing just with "a" and "b" taking different forms.

[Destiny]

D'awww thanks guys!

[FireTstar]

Ele just pm me via fb and I will help you will your algebra ii stuff if you need anymore help that is

[Destiny]

Now on the English side of hw :3 Anyone here majored in it, plans on majoring in it or is really good at writing essays?

[Extruderx]

Hmm... Me (I think), except English is obviously not a main language in my country, so I'd probably do many grammar (and cultural) errors.

[Dark8Dragon]

... and how to complete the square of  x^2 + 12x + 4 = 0. The finding max rev and profit i'll find later.

Hi Destiny, if you want a very easy and fast solution you can also use this:

Whenever you have an equation of ax²+bx+c=0 you can use the formula (-b+-sqrt(b²-4ac)) / 2a which would mean in this example:

(-12+-(sqrt(12² -4*1*4)) / 2*1
= (-12+-(sqrt(144 - 16)) / 2
= (-12+-(sqrt(128)) / 2
= (-12+-11,31) / 2
==> x1 = -0,34; x2 = -11,66;

Takes a half minute to get the result  

How to partial factor y=x^2+3x-10...

x²+3x-10 ==> (x+5)(x-2) = x*x - 2x + 5x - 10

1. It helps if you just take a look at the c=-10, only a=5;10*2;1=b fits into (x+a)(x-b).
2. You have either +-{10,1} or +-{5,2}. The one where you can build a sum of 3 (-2+5) is the b.

[Rosti_LFC]

Whenever you have an equation of ax²+bx+c=0 you can use the formula (-b+-sqrt(b²-4ac)) / 2a which would mean in this example:

Except doing it this way in actual an exam/homework would give you zero marks, or at the very least zero redundancy if you fucked up. Whole point of them stating a technique is to make you practice/demonstrate the technique. In practical situations and jobs you just use Wolfram Alpha for this kind of thing anyway.

[Destiny]

Oh...i think i gave the wrong equation for partial factoring. I think the partial factoring equations I meant were something like 13x+7x+4=0

[Dark8Dragon]

Except doing it this way in actual an exam/homework would give you zero marks, or at the very least zero redundancy if you fucked up. Whole point of them stating a technique is to make you practice/demonstrate the technique. In practical situations and jobs you just use Wolfram Alpha for this kind of thing anyway.

Hm ok that sounds bad. It was no problem for me in school, it was even the best way to solve this problem. Realizing that the term has the form ax² + bx + c =0 and then take the most efficient way of solving. Even in the university where I am studying Computer Science and Business/Economics we used this "Midnight-Formula". Wolfram Alpha is awesome btw.

Oh...i think i gave the wrong equation for partial factoring. I think the partial factoring equations I meant were something like 13x+7x+4=0

If you have no equation of a 2nd grade (x²) the form (x-a)(x+b) doesnt work. Its even easier:
13x+7x+4 = 20x + 4 ==> both numbers are dividable by 4 so you can make a 4(5x+1) of it.

And if you want to know what x is, simply put the 4 on the other side and you will get 20x = -4 , divide by 20 on both sides to get the x ==>  20x/20 = -4/20 ==> x = -1/5 or -0,2

[Rosti_LFC]

Hm ok that sounds bad. It was no problem for me in school, it was even the best way to solve this problem. Realizing that the term has the form ax² + bx + c =0 and then take the most efficient way of solving. Even in the university where I am studying Computer Science and Business/Economics we used this "Midnight-Formula". Wolfram Alpha is awesome btw.

Except typically in school (or most mathematics education really), you're learning various ways of doing things to pick up useful tools. And whilst the quadratic formula is fairly quick and works on all quadratic equations, it's not always the fastest way to find the solutions - it's completely overkill for any equations where completing the square or just basic factorisation can be done fast by inspection - typically you can just do those completely in your head just by looking at the equation.

Also the quadratic formula will only ever tell you the solutions where the equation equals zero, whilst other forms of algebraic manipulation will tell you other things about the nature of the function. And the quadratic formula has almost zero use outside of second order quadratic equations, whilst completing the square can be useful outside of that.


Anyway, Destiny, partial factorisation:

Typically for partial factorisation you're doing it to find out the maximum/minimum point on the quadratic. What you basically do is find two points where the y-value for the curve will be the same, and because ax^2+bx+c curves are always symmetrical, finding two points with the same y-value lets you find the maximum or minimum point because it will always be directly in between.

So for 13x^2+7x+4 (assuming you missed the ^2 because it makes no sense otherwise):

Step 1) Drop any term that doesn't have an x with it (the +c term). In this case remove the +4, so you just get 13x^2 + 7x

Step 2) Factorise the rest of the equation. In this case there are no common factors between 13 and 7, so you just end up with x(13x+7)

Technically just these two steps are the partial factorisation bit. But usually the point of doing it is to find the max/minimum, which is the next few steps:

Step 3) Set each side to zero, so you get:

x= 0
13x + 7 = 0   --> x = -7/13

What you've done here is found the two points where x(13x-7) = 0. This means for the original equation, you've found two points where 13x^2 + 7x + 4 = x(13x-7) + 4 = 4, because the left bit will multiply to zero when either x = 0 or x = -7/13. So in this case for those two x-values you get the same y-value, which is 4.

Note that these are *not* the solutions to 13x^2+7x+4 = 0, you're doing something else because you ignored the +4 at the start.

Step 4) Take the average of the two x-values to find the line of symmetry and therefore the value of x at the maximum/minimum. (0 - 7/13)/2 = -7/26, so -7/26 is the x-coordinate.

Step 5) Insert the x-coordinate back into the original equation to get the y-coordinate for the maximum/minimum point. In this case 13(7/26)^2 + 7(7/26) + 4 gives you 159/52 or 3.06

As the first term in the equation is positive you know that the vertex at (-7/26, 159/52) is a minimum point, though you can also figure this out from the fact that 3.06 is less than the +4 you were ignoring at the start.

Feel free to ask questions if you want something I've said explaining, or help on something else.

[Destiny]

Yay I understand the partial factoring and completing the square. And yes i didnt forget the ^2  thanks everyone who replied. I do have another question but it's just for advice. Generally when im faved with wordproblems, drawing a picture helps me alot. I have trouvle knowing where to start though. Any things i should look out for?

[Dark8Dragon]

When you need some further practice with different mathtopics till Grade 8 on U.S. Standards you can take a look into the site xpmath.com. The good thing is that you play it as a game, even with highscores!
These are some maybe useful games:

Same Function in different forms (with pictures): http://www.xpmath.com/forums/arcade.php?do...&gameid=117
Order of Operations: http://www.xpmath.com/forums/arcade.php?do...y&gameid=99
2-step equations: http://www.xpmath.com/forums/arcade.php?do...y&gameid=64