Checkout this formula,

F'(x)=±sqrt(f(x)^2-4f(x))

The formula above derives the well-known quadratic formula with ease.

It is also,called "makasis formula " given the name of the person who developed it.

Try me if you want me to show you
How it works.


Please,i want to meet with people who wish( or are doing masters) to do masters in mathematical economics

how does it work my niqq*?

paulolee:
how does it work my niqq*?

It works by you differentiating the function and plugging the results into the formula for evaluation

makasimatics:


It works by you differentiating the function and plugging the results into the formula for evaluation

do it here and let's see

Apart from the known methods which could be used to derive the quadratic formula, a new formula has been developed by me, and the formula can proof the Almighty Formula. The formula appears magical by a simple glance on it, and even during its evaluation. Below is the magic formula in the mathematical capsule 1.0.

Capsule 1.0

F'(x)=±sqrt(f'(x)^2 -4f(x))


Given a standard form of the polynomial of the second degree as,
ax^2+bx+c=0

Now we let f(x)=X^2+(b/a)X+c/a....1

The above equation is a monic or reduced quadratic equation.

F'(x)=2X+b/a.......2

Now plug f(x) and f'(x) into the magic formula for evaluation.

Thus we have,

2X+b/a=±sqrt((2x+b/a)^2-4(x^2+b/ax+c/a) )

2x+b/a=±sqrt(4x^2+4b/ax+b^2/a^2-4x^2-4b/ax-4c/a)

Taking the like terms and finding the common denominator inside the root sign reduces to:

























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

Take the square root to have

2x+b/a=±sqrt(b^2-4ac) /a

Now,multiply by 1/2 to have

X+b/2a=±sqrt( b^2-4ac)/2a

Solve for x to have,and take the common divisor at the RHS to have

X=-b±sqrt(b^2-4ac) /2a


Have you seen this before? I have a proof for it
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Chiefupper:
do it here and let's see

nice one u get d concept

Kbdon:
nice one u get d concept

Thank you so much.