₦airaland Forum

Do you know that,

F'(x)=±2squrt(f'(x)^2 -4f(x) / 4)

Can be used to derive Almighty Formula.

This is my new finding.

Wow....who knew?

Enlighten me pls on how you would derive it from what you wrote up there.

sholay2011:
Wow....who knew?

Enlighten me pls on how you would derive it from what you wrote up there.

Very fine. For simplicity, hope you knew that -b±squrt(b^2-4ac) / 2a are the solutions or roots of a quadratic equation? If so,then let me use normal numbers to denote the coefficients if x and constant of the quadratic equation given below:

X^2-x-6=0.............A
The above equation A is a reduced or monic quadratic equation since a=1.
Note: the sign "^" denotes raised to power as shown in equation A.

We write out the formula.

F'(x)=±2squrt(f'(x)^2 -4f(x)/4)

Here there is a nominator,and denominator inside the squrt.

Now, we shall plug f'(x) and f(x) into the numerator of the formula.

Let f(x)= x^2 -x -6
then, f'(x)= 2x-1

Therefore, we substitute the above into the formula and evaluate.

Thus we have,

2x-1=±2squrt( (2x-1)^2 -4(x^2-x-6)/4 )
Holding the LHS constant, we deal with the RHS. That is, we expand the binomial inside the square root sign,and open the other expression by -4. All happen at the numerator.
Thus we have :

(2x -1)(2x -1) = 4x^2-4x+1

-4(x^2-x-6)= -4x^2 -4x -24

Then we plug our results after expansion into the formula to have.

2x-1=±2squrt(4x^2-4x+1-4x^2+4x+24 )/4 )

The terms having the same variable as well as power will cancel out inside the square root,while the constant terms will add up to yield 25 inside the same square root sign so that we have 25/4 inside the squrt.



2x-1=±2squrt(25/4)
2x-1=±2(5/2)

(since the squrt of 25 is 5 ,and that of 4 is 2)

Thus, 2x-1=2(±5/2)

2x-1:=±5 (since 2/2=0)

We multiply by 1/2 to have,

X -1/2 = ±5/2.

Solve for x to get.

X=±5/2+1/2

The common denominator yields

X=±5+1/2.

Therefore, for plus or minus we have

X=6/2=3

Or x=-5+1 / 2 =-4/2=-2.

X=-2 or 3