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.

    Like

  2. tomcircle says:

    Zhao Shuang 趙爽

    Like

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