How Al-Khwarizmi find the quadraic equation formula?

2. ## 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 $\displaystyle x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$.

3. ## Re: Quadratic equation history

Originally Posted by yyakob
How Al-Khwarizmi find the quadraic equation formula?
He used a divining rod.

4. ## Re: Quadratic equation history

Originally Posted by yyakob
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.

5. ## Re: Quadratic equation history

Gin soaked doubt follows.
Originally Posted by HallsofIvy
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.