And what I'm going to show is how you derive

the solution equation to the general quadratic polynomial.

So you've got an equation like this: X is the unknown.

They want to solve x in terms of a, b,

and c. How do you do that?

Well, the first thing you do is you divide out A if that is permitted,

so that's X-squared plus B over A X plus C over A is zero.

Or the other possibility, A is zero.

The second step of the prove is a bit technical and it's called completing the square.

You write X-squared plus two times B over

two A times X and then

you add a term which would make the first three terms a perfect square,

which is in this case,

B-squared over four A-squared.

Having added that term,

you subtract it again and you add it C over A which is zero.

Now, the first three term are

perfect three terms are perfect square which we express by writing

it as X plus B over two A-squared.

And if we multiply that out,

we get exactly X-squared plus two times B over two A times X plus B over two A-squared.

And there's another term,

which we will bring to the other side,

which is B-squared over four A-squared

minus four A times C over four A-squared,

so that we can actually bring it under

the same denominator later.

Now we can take left and right square roots and we get X plus B over two A

is either the square root of B-squared minus 4 A_C divided by 4 A-squared;

or X plus B over two A is minus that same square root.

And then we get the final form,

X is minus B over two A plus,

we can take four over A-squared out of the square root and

get two A the square root of B-squared minus 4 A_C;

or X is minus B over two A

minus one over two A the square root of B-squared minus 4 A_C.

Of course, we can do all this only if A is not equal to zero.

But we can deal with that case.

Then, A-squared X plus B_X plus C is zero is

actually a linear equation.

So that's not really the equation we wanted to solve.

Now, this is the famous solution formula of

the quadratic equation and we obtained it just by a number of steps.

Nothing more, nothing less.

And there was only one step where we had to take care was

the second step where we divided out A and if we divide out a number,

the number shouldn't be zero.

That is a proof for transformation.

The second proof technique is a proof by induction.

So as an example,

let's have the statements, P(n),

sum of k is 1 to n of K which is basically sum one plus two plus three plus etc.

plus K is equal to N plus one over two and well,

we can first ask our-self whether P zero is true.

The sum of k is 1-2-0 of K is by definition a zero.

This definition always holds if

the index starting value of the index is higher than the ending value.

So on the other hand,

n plus one divided by two,

if we take n as zero, is also equal to zero.

So we see that P zero is true.