Indo-arabská matematika

Chvarizmí: „Číslo patriace do jednej z troch tried sa môže rovnať počtu iných tried. Môžete napríklad povedať:" 

„Štvorec sa rovná piatim koreňom“ [v súčasnej terminológii $x^2=5x$], čo je to isté ako
„Štvorec je dvadsaťpäť“ [... $x^2=25$] alebo
„Koreň je päť“ [... $x=5$] . 

Iný príklad 

„Päť štvorcov je rovné desiatim koreňom“ [... $5x^2=10x$]
„Jeden štvorec je rovný dvom koreňom“ [... $x^2=2x$]
„Štvorec je štyri“ [... $x^2=4$]
„Koreň je dva“ [... $x=2$].

1. Zadanie: $10 \rightarrow {\binom {x} {10-x} } \rightarrow {x(10-x)} \wedge { x^2}$.
2. Riešte rovnicu: $x^2=4[x(10-x)]$.
3. Pravidlo (pre riešenie rovnice prvého typu $x^2=8x$ ):
1. $x\rightarrow 10-x$ 
   
2. $x(10-x) \rightarrow 10x-x^2$ 
3. $4(10x-x^2 ) \rightarrow 40x-4x^2$
4. $x^2=40x-4x^2$
5. $5x^2=40x$
6. $x^2=8x(=64 )\rightarrow x=8$
7. $10-x=2$
   

1. Riešte rovnicu: $x^2+10x=39$.
2. Pravidlo (pre riešenie rovnice štvrtého typu $x^2+10x=39$ ):
1. $\frac{1}{2} z 10=5 \rightarrow 5 .5=25$ v dnešnej terminológii $(x+5) ^2-25=39$  
   
2. $39+25=64$ ... $(x+5) ^2=64$  
3. $\sqrt{64} =8$ ... $x+5=\sqrt{64}$
4. $8-5 = 3 \Rightarrow x=3$ ... $x = 8-5 \Rightarrow x=3$

Al-Chvarizmí dokazuje správnosť postupu. Odvodenie je čisto geometrické. Uvádza tu dva rôzne spôsoby dôkazu. V oboch prípadoch ide o doplnenie na štvorec.
V prvom prípade máme:

1. $x^2+10x=39$
2. $39+25=64$
3. $\sqrt{64} =8$ 
4. $8-5=3=x$    

V druhom prípade máme:

1. $x^2 + 4( \frac{10}{4} x)= x^2+10x=39$ 
2. $39+4(\frac{25}{4})= 64$ 
3. $\sqrt{64} =2(\frac{10}{4})+x$
4. $x=3$                       

Táto rovnica sa neskôr stala súčasťou snáď všetkých učebníc algebry až do novoveku.

I have divided ten into two parts; I have then multiplied each of them by itself, and when I had added the products together, the sum was fifty-eight dirhams. Computation: Suppose one of the two parts to be thing, and the other ten minus thing. Multiply ten minus thing by itself; it is a hundred and a square minus twenty things. Then multiply thing by thing; it is a square. Add both together. The sum is a hundred, plus two squares minus twenty things, which are equal to fifty-eight dirhams. Take now the twenty negative things from the hundred and the two squares, and add them to fifty-eight; then a hundred, plus two squares, are equal to fifty-eight dirhams and twenty things. Reduce this to one square, by taking the moiety of all you have. It is then: fifty dirhams and a square, which are equal to twenty-nine dirhams and ten things. Then reduce this, by taking twenty-nine from fifty; there remains twenty-one and a square, equal to ten things. Halve the number of the roots, it is five; multiply this by itself, it is twenty-five; take from this the twenty one which are connected with the square, the remainder is four. Extract the root, it is two. Subtract this from the moiety of the root$, namely, from five, there remain three. This is one of the portions; the other is seven. This question refers you to one of the six cases, namely "squares and numbers equal to roots.

1. Zadanie: $10 \rightarrow {\binom {x} {10-x} } \rightarrow {(10-x)^2} \wedge { x^2}$.
2. Riešte rovnicu: ${(10-x)^2} + { x^2}=58$.
3. Pravidlo – riešenie rovnice:
1. $x\wedge 10-x$ 
   
2. $(10-x)(10-x)=100+x^2-20x$ 
3. $x^2$
   
4. $100+x^2-20x+x^2=58\rightarrow$$100+2x^2-20x=58$
5. $100+2x^2=58+20x$
6. $50+x^2=29+10x$
7. $21+x^2=10x$
   
8. $\frac{1}{2} z 10 \rightarrow 5$
9. $\sqrt{25-21}=2 \rightarrow p=2$
   
10. $x=5-2=3$