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Thread: Quadratic equation history

  1. #1
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    Quadratic equation history

    How Al-Khwarizmi find the quadraic equation formula?
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  2. #2
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    Re: Quadratic equation history

    I can't speak for the actual history but it was probably by using "completing the square" since that is the easiest way to derive it:

    With the equation $\displaystyle ax^2+ bx+ c= 0$ we can write it as $\displaystyle a\left(x^2+ \frac{b}{a}x\right)+ c= 0$. We can complete the square by adding (and subtracting)$\displaystyle \left(\frac{b}{2a}\right)^2$ inside the parentheses. $\displaystyle a\left(x^2+ \frac{b}{a}x+ \frac{b^2}{4a^2}- \frac{b^2}{4a}\right)+ c$$\displaystyle = a\left(x^2+ \frac{b}{a}x+ \frac{b^2}{4a^2}\right)- \frac{b^2}{4a}+ c$$\displaystyle = \left(x+ \frac{b}{2a}\right)^2+ c- \frac{b^2}{4a^2}= 0$.

    $\displaystyle \left(x+ \frac{b}{2a}\right)^2= \frac{b^2}{4a^2}- c= \frac{b^2- 4ac}{4a^2}$.

    Taking the square root of both sides, $\displaystyle x+ \frac{b}{2a}= \pm\frac{\sqrt{b^2- 4ac}}{2a}$.

    So $\displaystyle x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$.
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    Re: Quadratic equation history

    Quote Originally Posted by yyakob View Post
    How Al-Khwarizmi find the quadraic equation formula?
    He used a divining rod.
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  4. #4
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    Re: Quadratic equation history

    Quote Originally Posted by yyakob View Post
    How Al-Khwarizmi find the quadraic equation formula?
    Your translator device has misspelled the name, it is al-Khwārizmī
    Go to that link. It has a great deal of information. If you have a good mathematics library look into An Introduction to The History Of Mathematics by Howard Eves.
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    Re: Quadratic equation history

    Gin soaked doubt follows.
    Quote Originally Posted by HallsofIvy View Post
    I can't speak for the actual history but it was probably by using "completing the square" since that is the easiest way to derive it:

    With the equation $\displaystyle ax^2+ bx+ c= 0$ we can write it as $\displaystyle a\left(x^2+ \frac{b}{a}x\right)+ c= 0$. We can complete the square by adding (and subtracting)$\displaystyle \left(\frac{b}{2a}\right)^2$ inside the parentheses. $\displaystyle a\left(x^2+ \frac{b}{a}x+ \frac{b^2}{4a^2}- \frac{b^2}{4a}\right)+ c$$\displaystyle = a\left(x^2+ \frac{b}{a}x+ \frac{b^2}{4a^2}\right)- \frac{b^2}{4a}+ c$$\displaystyle = \left(x+ \frac{b}{2a}\right)^2+ c- \frac{b^2}{4a^2}= 0$.

    $\displaystyle \left(x+ \frac{b}{2a}\right)^2= \frac{b^2}{4a^2}- c= \frac{b^2- 4ac}{4a^2}$.

    Taking the square root of both sides, $\displaystyle x+ \frac{b}{2a}= \pm\frac{\sqrt{b^2- 4ac}}{2a}$.

    So $\displaystyle x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$.
    Considering that their time period didn't have the convenience of our modern notation, it's rather hard for me to appreciate this probable conjecture.
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