Question

Rezolvati in multimea numerelor naturale ecuatiile:A)x^2-x=56 B)x^2+x=42 C)x^2+5x=84

Answer 1

x² -x - 56 =0           Δ =1 + 224  = 225     ; √Δ = √225 =15
x₁ = ( 1 + 15) /2  = 8                      x₂= ( 1 - 15 ) /2 = - 7 

x² + x - 42 =0           Δ=1 + 168 =169             √Δ=√169 =13 
x₁ = ( -1 -13 ) /2 = -7                    x₂ = ( -1 + 13) /2 = 6 

x² + 5x - 84 =0               Δ=25 + 336=361             √Δ=√361 =19 
x₁ = ( -5 -19 ) /2 = - 12                    x₂ = ( - 5 + 19 ) / 2 = 7

Answer 2

$\displaystyle a).x^2-x=56 \\ x^2-x-56=0 \\ a=1,b=-1,c=-56 \\ \Delta=b^2-4ac=(-1)^2-4 \cdot 1 \cdot (-56)=1+224=225\ \textgreater \ 0 \\ x_1= \frac{1+ \sqrt{225} }{2 \cdot 1} = \frac{1+15}{2} = \frac{16}{2} =8 \\ \\ x_2= \frac{1- \sqrt{225} }{2 \cdot 1} = \frac{1-15}{2} = \frac{-14}{2} =-7 \\ \\ S=\{-7;8\}$

$\displaystyle b).x^2+x=42 \\ x^2+x-42=0 \\ a=1 ,b=1,c=-42 \\ \Delta=b^2-4ac=1^2-4 \cdot 1 \cdot (-42)=1+168=169\ \textgreater \ 0 \\ x_1= \frac{-1+ \sqrt{169} }{2 \cdot 1} = \frac{-1+13}{2} = \frac{12}{2} =6 \\ \\ x_2= \frac{-1- \sqrt{169} }{2 \cdot 1}= \frac{-1-13}{2} = \frac{-14}{2} =-7 \\ \\ S=\{-7;6\}$

$\displaystyle c).x^2+5x=84 \\ x^2+5x-84=0 \\ a=1,b=5,c=-84 \\ \Delta=b^2-4ac=5^2-4 \cdot 1 \cdot (-84)=25+336=361\ \textgreater \ 0 \\ x_1= \frac{-5+ \sqrt{361} }{2 \cdot 1} = \frac{-5+19}{2} = \frac{14}{2} =7 \\ \\ x_2= \frac{-5- \sqrt{361} }{2 \cdot 1} = \frac{-5-19}{2} = \frac{-24}{2} =-12 \\ \\ S=\{-12;7\}$