Now we will give proof of the solution of the quadratic formula. Why? Because we can!
y=ax^{2}+bx+cWe will rewite this equation in the form y = ( x^2 + p x ) - q with y = 0 in order to be able to easily factorize the whole thing. In order to find the values where the x-axis is crossed y has to be set to 0.
ax^{2}+bx+c=0We want a to be 1, (or essentially just disappear in front of x^{2}), so we divide the whole thing by a.
x^{^{2}}+\frac{b}{a}x+\frac{c}{a}=0Now we factorize the whole thing. It’s not as hard as it seems: As long as you realize (and have practiced) that ( a + b )^2 equals ( a^2 + 2 a b + b^2 ), this shouldn’t be too hard.
(x+\frac{b}{2a})^{2}-\frac{b^{2}}{4a^{2}}+\frac{c}{a}=0The appearance of -\frac{b^{2}}{4a^{2}} should not be too surprising. It is essentially deducting the b^2 in ( a^2 + 2 a b + b^2 ) . Now for the sake of clarity let’s do things really dumb and rewrite the function as:
(x+\frac{b}{2a})^{2}+\frac{b^{2}}{-4a^{2}}+\frac{c}{a}=0Now we can rewrite \frac{c}{a} so it has the same denominator as \frac{b^{2}}{-4a^{2}} by mulitplying both the numerator and the denominator by - 4 a .
(x+\frac{b}{2a})^{2}+\frac{b^{2}}{-4a^{2}}+\frac{-4ac}{-4a^{2}}=0Which simplifies to
(x+\frac{b}{2a})^{2}+\frac{b^{2}-4ac}{-4a^{2}}=0Which in turn can be rewritten as
(x+\frac{b}{2a})^{2}-\frac{b^{2}-4ac}{4a^{2}}=0Now we move -\frac{b^{2}-4ac}{4a^{2}} to the right
(x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}Now we take the square root of the whole thing
x+\frac{b}{2a}=\frac{\pm\sqrt{b^{2}-4ac}}{2a}Now let’s move \frac{b}{2a} to the right, so x is finally isolated
x=\frac{-b}{2a}+\frac{\pm\sqrt{b^{2}-4ac}}{2a}The denominator turns out to be 2a in both cases. We are lazy people, so having directly put the minus sign in front of b and the plus minus sign in front of \sqrt{b^{2}-4ac} has further clarified things. Now it’s tine for the final simplification
x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}Marvel and behold!
As they say in Dutch: “A kid can do the laundry”, which means:
This is ez!
