Xjyuk

I have tested the quadratic formula and I have found that it works, yet I am curious as to how it was created. Can anybody please tell me one of the ways that it was created?

We begin with the equation $ax^2 + bx + c = 0$, for which we want to find $x$. We divide through by $a$ first, and then bring the constant term to the other side:

$$x^2 + frac b a x = - frac c a $$

Next, we complete the square on the left-hand side. Remember, to complete the square, you take half of the coefficient of the linear term, square it, and add it both sides. This means we add $(b/2a)^2 = b^2/4a^2$ to both sides:

$$x^2 + frac b a x + fracb^24a^2 = fracb^24a^2 - frac c a$$

The left-hand side factors as a result, and we combine the terms on the right-hand side by getting a common denominator:

$$left(x + frac b 2a right)^2 = fracb^2 - 4ac4a^2$$

We now take the square root of both sides:

$$x + frac b 2a = pm sqrt fracb^2 - 4ac4a^2$$

Solve for $x$:

$$x =- frac b 2a pm sqrt fracb^2 - 4ac4a^2$$

Recall that $sqrta/b = sqrt a / sqrt b$. Using this property, the denominator of our root becomes $2a$, giving a common denominator with $-b/2a$. Thus,

$$x = frac -b pm sqrtb^2 - 4ac2a$$

yielding the quadratic formula we all know and love.

We have $(p+q)^2equiv p^2+2pq+q^2$.

If we want to solve $ax^2+bx+c=0$, we can rewrite the equation so that the square of some expression involving $x$ is equal to a constant. As $a$ is not necessarily a perfect square, we can multiply the whole equation by $a$ and make it $a^2x^2+abx+ac=0$. But we also want the middle term to be $2$ times something, we further multiply the equation by $4$, which is an even perfect square. So, we have

beginalign*
4a^2x^2+4abx+4ac&=0\
(2ax)^2+2(2ax)(b)+b^2+4ac&=b^2\
(2ax+b)^2&=b^2-4ac\
2ax+b&=pmsqrtb^2-4ac\
x&=frac-bpmsqrtb^2-4ac2a
endalign*

The way to get it is actually a beautiful one. The original proof is different since algebra was still in an early stage of development, but here's how I would do it today:

Let's consider the equation $ax^2+bx+c=0$ If a=0, we have a first degree equation. Otherwise, our equation can be written as $x^2+fracbax +fracca=0$

On the other hand, $(x+fracb2a)^2=x^2+fracbax+fracb^2a^2$. This looks like an idea completely out of the air, but you can get it by trying different ones and checking which one works. As you can see, the first two terms look like those form the equation, so we can say that $x^2+fracbax = (x+fracb2a)^2 - fracb^2a^2$

Now go back to the equation and replace $x^2+fracbax$ We get that $(x+fracb2a)^2 - fracb^2a^2 + fracca = 0$ From this equation, it's just basic algebra.

$(x+fracb2a)^2 = fracb^2a^2 - fracca$ Take the square root on both sides and enjoy your formula (unless I made a mistake, which I probably have)