Algebra Math Problems HELP NEEDED NOW!

[FeelTheRush]

Can someone please help? I need this by tonight midnight!! DX
@Alex @Alen
@Ravallo

I will do this. Don't worry.

---------- Post added at 06:23 PM ---------- Previous post was at 06:05 PM ----------

So the first equation will be
(9x)-8=x^2
We get this since Nine times a # is 9x, eight less than that is 9x-8.
To solve this, you have to first get common terms. So...
We can factor it like a polynomial
-X^2+9X-8=0
From there we can factor the equation...
But the solutions are extraneous, It cannot be factored. (Since you can get -8 from two factors of 9).
The best possible method now is to graph the equation above and find the x-intercepts
when you graph X^2-9X+8 (I multiplied the whole equation by -1 so X^2 wouldn't be negative).
You get two intersections.. one at -10, and one at 1.
CHECK to make sure the solutions are right.
100-90+8 DOES NOT WORK
but
1-9+8 DOES work.

Is this what your teacher wanted you to do?

---------- Post added at 06:28 PM ---------- Previous post was at 06:23 PM ----------

consecutive odd integers are
7,9,11 (for example).
So what we want to know is which consecutive odd integers ( 3 of them) equal 46.
We know we only need 3 integers since in the question it says. Square of THIRD, decreased by FIRST.
Basically all we have to do is find a # (The one we are squaring... the third #)LESS THAN 9, and the same # must be greater than 5. Since 5^2 is 25, which isn't greater than 46..
So the only ODD # we are stuck with is 7 (greatest number).
So the consecutive odds are 3,5,7
7^2 =49
49-3=46
THERE YOU GO