## Sum of roots and Product of roots of Quadratic Equation

Given a quadratic equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$, we have: $\displaystyle\boxed{\alpha+\beta=\frac{-b}{a}}$ $\displaystyle\boxed{\alpha\beta=\frac{c}{a}}$

How do we prove this? It is actually due to the quadratic formula!

Recall that the quadratic formula gives the roots of the quadratic equation as: $\displaystyle\boxed{x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

Now, we can let $\displaystyle \alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}$ $\displaystyle \beta=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Hence, $\displaystyle \alpha+\beta=\frac{-2b}{2a}=\frac{-b}{a}$ $\begin{array}{rcl} \displaystyle \alpha\beta&=&\frac{-b+\sqrt{b^2-4ac}}{2a}\times\frac{-b-\sqrt{b^2-4ac}}{2a}\\ &=&\frac{b^2-(b^2-4ac)}{4a^2}\\ &=&\frac{4ac}{4a^2}\\ &=&\frac{c}{a}\end{array}$

In the above proof, we made use of the identity $(A+B)(A-B)=A^2-B^2$

The above formulas are also known as Vieta’s formulas (for quadratic). There we have it, this is how we prove the formula for the sum and product of roots!