Question

To completing square form.
Steps are Must.
-0.035 x^2+1.4 x+6.

Answer 1

divide all by -0.035 first...

Answer 2

x^2 -40x -171.43

Answer 3

wait, doesn't this have to be equal to zero to do this?

Answer 4

it is equal to zero.

Answer 5

-0.035 x^2+1.4 x+6=0 ?????????

Answer 6

Yes^^

Answer 7

Hello, Sat.

Answer 8

no it just has to be the same as what you start with. does not have to be equal to zero

Answer 9

hello

Answer 10

but you have to put it in that form at some point, or at least start with knowing what it is equal to.

Answer 11

if it is equal to zero then to what turing said, divide by the leading coefficient. if it is just an expression you have to factor it out

Answer 12

it is equal to zero, this is bcz i myself made that equation.

So, after divinf i get:
x^2 -40x -171.43

Answer 13

dividing*

Answer 14

in other words if you start with
\[y=-0.035 x^2+1.4 x+6\] you have to end with
\[y=\text{ expression} \] that is the same as the initial one

Answer 15

that is an equation, coz i myself made it.

Answer 16

BRB

Answer 17

x^2-40x=171.43
b=-40
completing the square means you add (1/2b)^2 to both sides now
so add 0.000156 to both sides
at this point I frankly would throw out the idea of completing the square since in this case the decimal thing is getting silly, but theoretically
x^2-40x+(1/6400)=(6/0.035)+(1/6400)
(x-1/80)^2=(6/0.035)+(1/6400)
\[x=1/80+\sqrt{(6/0.035)+(1/6400)}\]
I decided to keep fractions here as they are more reasonable than a decimal addition like 0.000156 which would hardly change 171.43 in any noticable way

Answer 18

\[y=-.035x^2+.14x+6\]
\[y=-.035(x^2-4)+6\]
\[y=-.035(x^2-4x+4)+6+.035\times 4\]
\[y=-.035(x^2-4x+4)+6.14\]
\[y=-.035(x-2)^2+6.14\]

Answer 19

sorry it should be\[1/80 \pm \sqrt{(6/0.035)+(1/6400)}\] if y=0

Answer 20

the method i used above is just rewriting it as you would any quadratic you wanted to change from the form
\[y=ax^2+bx+c\] into
\[y=a(x+h)^2+k\] i certainly would not use it to find the zeros, although you could

Answer 21

in this form you can see that the vertex is
\[(2,6.14)\] but you could find the vertex by computing
\[-\frac{b}{2a}\]as well

Answer 22

in fact that is a shorter way of completing the square. find
\[-\frac{b}{2a}\] in your case it is 2.
the write as
\[y=-.035(x-2)^2+\text{ something }\] and you find the "something" by replacing x by 2 in the original expression and see what you get

Answer 23

Ah, I see satellite's method is more general and efficient than mine, but for the problem posted with y=0 am I not correct in my method too?

Answer 24

sure if it is set equal to zero you do not have to keep it the same and are free to divide by the leading coefficient

Answer 25

if
\[ax^2+bx+c=0\iff x^2+\frac{b}{a}x+\frac{c}{a}=0\]

Answer 26

but if you need to end with what you start with, then you have to "factor it out" and cannot divide by it. because
\[ax^2+bx+c=0\neq x^2+\frac{b}{a}x+\frac{c}{a}=0\]

Answer 27

that is incorrect form which sat gave me! :/