Question

Given AX^2+BX+c=0
Derive the general solution for x

Answer 1

Would you use the quadratic formula?

Answer 2

no, i don't think so

Answer 3

have you learnt differentiation?

Answer 4

To a degree, yes. This is actually a physics problem.

Answer 5

Oh ok. The first D would be 2x or something, right?

Answer 6

just wait...
I've done a question like this last year for math methods, just trying to remember what I did

Answer 7

your fine, thanks for trying to help.

Answer 8

I know what the answer is, just trying to think of how to get there

Answer 9

what you can try to do, instead of differentiation, is to complete the square.
so, y=ax^2+bx+c
y=a(x^2+bx/a+c/a)
y=a(x^2+bx/a+(b/2a)^2-(b/2a)^2+c/a)
y=a(x+b/2a)^2-a(b^2/4a^2+c/a)
y=a(x+b/2a)^2+(c-b^2/4a)

Answer 10

Confusing right?

Answer 11

I think differentiation would be easier...

Answer 12

Then using the form: y=a(x-b)+c
turning point: (b,c)
then: (-b/2a, c-b^2/4a)

Answer 13

now let me try to figure it out differentiation style

Answer 14

right!!! I sorta remember now!!!

Answer 15

remembering is a good thing!

Answer 16

1. you need to differentiate the equation
y=ax^2+bx+c
so, dy/dx=2ax+b

Answer 17

Understand that step.

Answer 18

then, you need to find the minimum of the graph or the turning point

Answer 19

to do that, the 'gradient' or dy/dx must be zero

Answer 20

so, dy/dx=2ax+b
0=2ax+b
x=-b/2a

Answer 21

since it is the general solution, it means the turning point...forgot to point that out

Answer 22

Is that the critical point? If so, I understand, just beeb awhile.

Answer 23

*been

Answer 24

yes, the peak

Answer 25

your answer would be x=-b/2a
if you need to find y, just substitute -b/2a into the original equation y=ax^2+bx+c
then you should get y=c-b^2/2a

Answer 26

Is that it? Praise God!
Thank you soooo much!

Answer 27

sorry for the confusion earlier on, had to open my yr 11 textbook from last year!!!

Answer 28

Oh my goodness. I'm a 12th grader in Tennessee...

Answer 29

whoops!!!

y=c-b^2/4a
not y=c-b^2/2a

Answer 30

no, I'm only an 11th grader :) I was doing advanced maths last year
I'm from Australia

Answer 31

Our American system needs help. Thanks!

Answer 32

no, thank YOU!!! I needed the refreshment in differentiation

Answer 33

Mind if I ask a few questions? I really don't mean to interrupt or cause confusion, but I wonder what you guys are trying to prove here, the general solutions for formula for a quadratic equation as given in the problem set?

Or the equation of the min/max point.

Answer 34

As it was said, the extrema is:
\[\Large x=-\frac{b}{2a} \] Plugging that back into the original equation will give you the y-value of the extrema. Not the roots.

Answer 35

Right, that is what I needed.

Answer 36

That is, what chelsea04 provided.

Answer 37

You need the roots? Or the extrema? Please, pardon me.

But
\[\Large ax^2+bx+c=0 \] is a quadratic equation, if it asks you to find the x, I read this question as: find the x-value that makes the given statement true.

And not: Find the Maximum/Minimum / Zero-Slope.

Answer 38

I am perfectly fine with @Chelsea04 proof, I just wonder if everyone was talking about the same thing here (-: I clearly misunderstood this problem then. Pardon me.

Answer 39

Its cool! Have a great day :)

Answer 40

@Spacelimbus can you tell me what the determinant is? I have a problem and need to find the determinant of a matrix is by hand.

Answer 41

\[\Large \det(A)=ad-cb \]
Considering the entries.

\[\left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\]

Answer 42

ohhh yea. Thanks! I should have made another question so I could have given you a best response!

Answer 43

no worries at all, I am still wondering about the problem above *laughs* it just seems wrong to me to confuse the roots with the max.