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!

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2 Responses to Sum of roots and Product of roots of Quadratic Equation

1. tomcircle says:

http://zh.wikipedia.org/zh-sg/%E8%B5%B5%E7%88%BD

The Chinese mathematician Zhou
Shuang 趙爽 (222 AD) of the 吳国 (Wu state) in 3 Kingdoms era (三国) had discovered 1,300 years earlier than the French mathematician Vièta.

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2. tomcircle says:

Zhao Shuang 趙爽

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