about solving ax^2+bx+c=0?

Question

I notice that a solution can be given by x=x(min)+sqrt(-f(xmin))
this is where x(min) is the value of x where the function reaches its minimun value and f(xmin) is the function evaluated in xmin
and sqrt mean the square root of
can anyone turn this into the feneral formula for solving second order equation? (10points in it)
and by the way can anyone quote a demontration for the above?

Answer 1

No, this is not a general solution. it works only if a=1 (the coefficient of x^2 = 1)

that's the proof:

let f(x) = ax^2 + bx + c

f'(x) = 2ax + b ===> to find x(min) or x(max), let f'(x) =0

2ax + b=0 ==> 2ax = -b ==> x = -b/2a

x(min) or x(max) = -b/(2a)

f(xmin or xmax) = a(-b/(2a))^2 + b(-b/(2a)) + c = (-b^2)/(4a) + c

ur solution for this function was:

x=x(min)+sqrt(-f(xmin)) = -b/(2a) +or- sqrt( -(-b^2)/(4a) + c ) =

= -b/(2a) +or- sqrt( (b^2 - 4ac)/(4a) ) =

= -b/(2a) +or- sqrt( (b^2 - 4ac)/(***4a***) )

but the real solution is -b/(2a) +or- sqrt( (b^2 - 4ac)/(***4a^2***) )

the difference between the solutions is shown between the stars "***"

urs if 4a, but the another is 4a^2

if we assume a=1, then your solutions works because 4(1) = 4(1^2)

good job, at least u found a general solution for second order equation with a=1


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u can create a general solution this way:

x=x(min)+sqrt(-f(xmin)/a) #- i add an a.

Answer 2

The "general formula" you are referring to is the quadratic formula. Your solution has a major flaw. It only works when a = 1. I spent some time to work out the derivations. I hope you will see the problem.

First, note that only for a parabola that's concave up (where a is a positive number) do you have an x mininum. In the other case you have an x maximum, but either max or min, it does not matter here.

There is one formula you have to know, that is
x min (or max) = -- b / (2a)

so to solve for f(xmin), simply plug in that function

ax ^ 2 + bx + c = 0
so
a (( -b/(2a))^2) + b (-b/(2a)) + c = f(xmin)
after simplifying and finding the common denominator:
f (xmin) = (4ac - b^2 ) / (4a)
take the sqrt (-f(xmin)) you get
sqrt( (b^2 - 4ac) / (4a))
TAKE NOTE that the denominator is 4a instead of 4a^2
IF IT WERE 4a ^2, then you can add that entire square root to the first part of "your solution"--- x(min), which is -b/(2a), and that will give you the familiar QUADRATIC FORMULA.
BUT your solution gives 4a instead of 4a^2, so your solution is not exactly the same as the quadratic formula.

So why does it work when a = 1? obviously
when a = 1, 4a = 4a^2

I hope this helps

Answer 3

First of all, you must be talking about a case when a>0 because if a<0 then there is no min of the function. Anyway, if a>0 and
f(x) = ax^2 + bx + c, then
f'(x) = 2ax + b
f'(x) = 0 when x = -b/2a.
x = -b/2a is x(min).

sqrt(-f(xmin)) = sqrt(-f(-b/2a))
= sqrt(-(ab^2/4a^2 - b^2/2a + c)
= sqrt(-(b^2/4a - b^2/2a + c)
I don't think this is going to go anywhere useful because there is no guarantee that the argument of sqrt() is going to be positive.

It is possible that there are some equations for which your formula works, but it isn't a general formula. Case in point, the answer above mine provides a counter-example.

There is already a formula for solving a second order equation. It is called the quadratic formula and it is derived by taking the equation ax^2 + bx + c = 0 through the algorithm of completing the square. The result is:

x = [-b (+or-) sqrt(b^2 - 4ac)]/2a

Answer 4

I don't think that your observation is correct. It only works for a = 1. The correct version would be

[1] ... x = xmin +/- sqrt(-f(xmin) / a)

Write p = xmin and q = f(xmin). The equation ax^2 + bx + c = 0 is equivalent to

[2] ... a(x - p)^2 + q = 0

with

[3] ... p = -b/2a
[4] ... q = c - ap^2 = c - b^2/4a.

Equation [2] is easy to solve:

[2] ... a(x - p)^2 + q = 0
[5] ... a(x - p)^2 = -q
[6] ... (x - p)^2 = -q/a
[7] ... x - p = +/- sqrt(-q/a)
[8] ... x = p +/- sqrt(-q/a)

and this is the corrected version of the formula you found.

Formulas [3], [4] and [8] together form a general solution. We can combine them:

[8] ... x = p +/- sqrt(-q/a)
[9] ... = (-b/2a) +/- sqrt(-[c - b^2/4a]/a)
[10] ... = -b/2a +/- sqrt([b^2 - 4ac]/4a^2)
[11] ... = -b/2a +/- sqrt(b^2 - 4ac) / 2a
[12] ... = [-b +/- sqrt(b^2 - 4ac)] / 2a

which is the famous "quadratic formula".

Answer 5

x^2+2x+1=0
(x+1)^2
x=-1
The minimum in this parabola is 0
so according to you solutions it would be x(0)+Sqr(-f(0))
this gives x=Sqrt(-1) which is wrong since the correct answer is -1

Answer 6

uh.........no