Question

aflati doua numere pozitive stiind ca media lor aritmetica este de 12,5 iar media geometrica este 10

Answer 1

[tex]\mathrm{m_a= \frac{a+b}{2}} \\ \\ \mathrm{12,5= \frac{a+b}{2} \longrightarrow a+b=25} \\ \\ \mathrm{m_g= \sqrt{a*b}} \\ \\ \mathrm{10= \sqrt{a*b}~|^2 \longrightarrow a*b=100} \\ \\ \mathrm{20*5=100} \\ \\ \boxed{\mathrm{a=20}} \\ \\ \boxed{\mathrm{b=5}} \\ \\ \underline{\mathrm{Verificare:20+5=25}}[/tex]

Answer 2

(a+b)/2 = 12.5 => a+b = 25 
√a*b =10 => a*b = 10²=> a*b = 100 => a=100/b

a+b = 25=> 100/b+b=25 (inmultim toata fractia cu b)=> 100+ b²-25b=0
a=100/b

si acum avem ec de gradul 2 : b²-25b+100=0
Δ= 625 - (4*1*100) = 225 => √Δ= 15
a= (25+15)/2=20
b=(25-15)/2=5