Question

Another question... rearrange into quadratic form ax^2+bx+c=0

0.20= x^2/(25-x)

where a=1, and identify values of b and c

Answer 1

multiply both side by (25-x), and move all the terms to the desired side. coefficient before x^2 is a, coefficient before x is b, and constant term is c.

Answer 2

then you need to divide the whole eqn by a number to yield a= 1. so our final b and final c is divided from the original b and c's by that amount.

Answer 3

Got to here then I got confused:
0.20=x^2/ (25-x)
0.20 (25-x)=x^2
5- 0.20x=x^2

Answer 4

lets have all the babies on our right side of the equal sign. moving everything to the right side, you get x^2+0.20x-5=0

Answer 5

Shouldn't your 0.20x be negative?

Answer 6

Because you distribute the 0.20 to (25-x)
So wouldn't it read x^2-0.20+5?

Answer 7

\[\large \color{red}ax^2 +\color{red}bx+\color{red}c=0\]\[\large 0.20 = \frac{x^2}{25-x}\]\[\large 0.20(25-x) = x^2 \]\[\large 0.20(25) - 0.20x = x^2 \]\[\large \color{red}{1}x^2 + \color{red}{0.20}x -\color{red}{0.20(25)}=0\]

Answer 8

It's not negative because you're moving it over to the other side of the equal sign.

Answer 9

but you have:
\[0.20(25)-0.20x\]
on one line then:
\[+0.20x-0.20(25)\]

Answer 10

\[\large 0.20(25) - 0.20x = x^2\] get everything to the right side. \[\large 0.2(25) -(0.20(25)) -0.20x =x^2\]\[\large -0.20x = x^2 -0.20(25)\]\[\large -0.20x + 0.20x = x^2 -0.20(25) +0.20x \]\[\large 0 = x^2 +0.20x -0.20(25)\]

Answer 11

Understand it now?

Answer 12

I'm still very confused. Thank you for trying to help me, though. I'm a little fuzzy because I had surgery today. I guess I'll just skip over it and try coming back another time. Thank you again

Answer 13

What part are you confused with? What we have to do is try making the left side 0 by subtracting each term by its opposite.

Answer 14

how you got from \[0.20(25)-0.20x=x^2\]
to \[0.20(25)-(0.20(25))-0.20x=x2\]

Answer 15

I want to eliminate the "0.20(25)" from the left side by SUBTRACTING it from the left. Whatever i do to the LEFT side i have to do to the RIGHT side.

Answer 16

It's already on the left.. Why is it being eliminated?

Answer 17

So now were moving everything to the right?

Answer 18

it's being eliminated from the left to the right because we are trying to fit in the format of \(\large ax^2 +bx +c = 0\) in which case our 0 will be on the left and all the other numbers will be on the right

Answer 19

Thats why we've written our solutions as \(\large 0 = x^2 +0.20x -0.20(25)\)

Answer 20

Okay let me write this down

Answer 21

Okay. NOW I understand

Answer 22

\[0=x^2-0.20(25)-0.20x\]
is now
\[0=1^2-0.20(25)-0.20(c)\]
Because it says a=1?

Answer 23

No the x doesn't just disappear. 1 is a coefficient of x^2 meaning its the integer that is multiplied to the variable x^2. So it would be \(\large 1x^2\) or simply \(\large x^2\)