Al-Khwarizmi's 6th quadratic case

[KevinMWHM]

The problem: I need to come up with a formula based on Al-Khwarizmi's 6th algebraic equation; bx+c=x^2.

I'm just having a definition problem that's holding me up from moving forward.

The first line of his solution is to "halve the number of roots". What is meant by "number of roots"? Number of roots for a square is just one, isn't it? Or am I defining "roots" wrong in this case?Thanks
-Kevin

 

[SteamKing]

The problem: I need to come up with a formula based on Al-Khwarizmi's 6th algebraic equation; bx+c=x^2.

I'm just having a definition problem that's holding me up from moving forward.

The first line of his solution is to "halve the number of roots". What is meant by "number of roots"? Number of roots for a square is just one, isn't it? Or am I defining "roots" wrong in this case?Thanks
-Kevin

The number of roots is always equal to the highest degree of the polynomial. The roots may be equal, or they may be different, as the case may be.

 

[Mark44]

Moved from Homework section. @KevinMWHM, if you post in the homework sections, you need to use the homework template.

 

[epenguin]

Antique text translated, they express themselves not in our language, in the ordinary and the mathematical sense.

I'll make a reasoned guess what may be meant could be "find the average of the roots". Which you can certainly do.
Then the distance from there to the roots is the same for both, so you are only having to find one thing.
Put it this way, if you can express the quadratic as the difference between the square of the mean and the square of a root, (m² - x²) then you can solve it. I don't know if you have the whole text or jusr this fragment, but in the former case there should be other indications of whether this is the idea.

 

[pbuk]

Google is your friend and quickly leads to this page: http://www-groups.dcs.st-and.ac.uk/history/Extras/Al-Khwarizmi_quadratics.html

From this example it is clear that the translation "halve the number of roots" should be interpreted as "halve the coefficient of x", which gives you the term $\frac b2$

 

[epenguin]

Looks like I was about right?