Bewildered by this statement.

"ax2 + bx + c = 0"

x, which is the unknown quantity, is raised to a square. The highest in that equation...
 
The pictures aren't working. :/

Can you explain that again with a working picture? :?
 
I'm truly wtfing right now
I think it means Highest power is the Square
I have no idea...
 
It's very simple.

"x" only appears twice in the quadratic equation. It has a power of of 1 and 2. Since, 2 is higher than 1, 2 is the highest power. Also known as "squared"
 
x only appears once in the quadratic equation? Who said anything about x having a power of 1 and 2? I get it if that statement(s) is true, but where did you get the first part from?
 
eXero said:
x only appears once in the quadratic equation? Who said anything about x having a power of 1 and 2? I get it if that statement(s) is true, but where did you get the first part from?
Click to expand...

Are you blind? Twice.
a(X)2 + b(X) + c = 0

If you square, cube, etc. You are raising it to a certain power. 2 = power of two , 3 = power of three.

The first X is being squared, which is being raised to a power of two.
The second X is being raised to a power of one. (Not affected)

2 > 1 , which means the highest power ( 2 )
 
Yeah, it should be: x = -b +/- square root of (b^2 -4ac) over 2a

In a equation like:

x2 + 3x – 4 = 0

This quadratic happens to factor:

x2 + 3x – 4 = (x + 4)(x – 1) = 0

Using a = 1, b = 3, and c = –4

Just fill it in. IF you have a t83 or 84 calculator then there is a way to do this without writing it out. I can help you if you need more help. Computer Science major :wink:
 
I have the ti84 plus silver edition, it's just at home, 2 states away! It would help I know. I might get you to explain that again when I get home Brock.
 
I loved those math problems.
 
I guess when you simplify x = (-b +/- square root of (c^2 - 4ac)) / 2a it becomes ax^2 + bx + c = 0 which is a way to find the roots by factoring, so thats how you get to ax^2 + bx + c = 0, the highest power of an unknown number is a square.

I think I got it took me awhile.
 
I think you are right @[gilaga]. I had to really think about what you said, as well as what @[Brock] said, but I think I get it now.
 
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