question

Using the quadratic formula, we obtain two solutions :

alpha =   [ - b + √[ b^2 - 4ac] ] / {2a]       and   beta  =  [ - b - √[ b^2 - 4ac] ] / {2a]

So

alpha + beta   =      [ - b + √[ b^2 - 4ac] ] / {2a]   +   [ - b -√[ b^2 - 4ac] ] / {2a]   =  [ - 2b] / [2a]  =   -b/a

And  alpha * beta    =

[ - b + √[ b^2 - 4ac] ] / {2a]   *   [ - b + √[ b^2 - 4ac] ] / {2a]  =  [   b^2  - [b^2 - 4ac] ]  / [ 4a^2]  =

[ 4ac ] / [ 4a^2]  =   c / a

Here, the solutions are the x values that make this equation true.

ax² + bx + c = 0

We can use the quadratic formula to solve for x.

x = \({-b \pm \sqrt{b^2-4ac} \over 2a}\)

So..the two solutions are

\(\alpha=\frac{-b+\sqrt{b^2-4ac}}{2a} \quad \text{and} \quad \beta=\frac{-b-\sqrt{b^2-4ac}}{2a} \)

And...

\(\alpha+\beta=\frac{-b+\sqrt{b^2-4ac}+-b-\sqrt{b^2-4ac}}{2a} = \frac{-2b}{2a}=\frac{-b}{a}\)

and

\(\alpha*\beta=\frac{(-b+\sqrt{b^2-4ac})(-b-\sqrt{b^2-4ac}) }{(2a)(2a)} =\frac{b^2-b\sqrt{b^2-4ac}+b\sqrt{b^2-4ac}-(b^2-4ac)}{4a^2} = \frac{4ac}{4a^2}=\frac{c}{a}\)