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va rog ajutor matematica clasa9 xla puterea2=7x+12=0; 2xla puterea 2-5x+2=0 ;3x la puterea2-10x+3=0 ;4x la puterea 2-17x+4=0 va rog mult

Răspuns :

[tex]\displaystyle a).x^2+7x+12=0 \\ a=1,b=7,c=12 \\ \Delta=b^2-4ac=7^2-4 \cdot 1 \cdot 12=49-48=1\ \textgreater \ 0 \\ x_1= \frac{-7+1}{2} = \frac{-6}{2} =-3 \\ \\ x_2= \frac{-7-1}{2} = \frac{-8}{2} =-4 \\ \\ S=\{-3;-4\}[/tex]

[tex]\displaystyle b).2x^2-5x+2=0 \\ a=2,b=-5,c=2 \\ \Delta=b^2-4ac=5^2-4 \cdot 2 \cdot 2=25-16=9\ \textgreater \ 0 \\ x_1= \frac{5+ \sqrt{9} }{2 \cdot 2} = \frac{5+3}{4} = \frac{8}{4} =2 \\ \\ x_2= \frac{5- \sqrt{9} }{2 \cdot 2} = \frac{5-3}{4} = \frac{2}{4} = \frac{1}{2} \\ \\ S=\{2; \frac{1}{2} \}[/tex]

[tex]\displaystyle c).3x^2-10x+3=0 \\ a=3,b=-10,c=3 \\ \Delta=b^2-4ac=10^2-4 \cdot 3 \cdot 3=100-36=64\ \textgreater \ 0 \\ x_1= \frac{10+ \sqrt{64} }{2 \cdot 3} = \frac{10+8}{6} = \frac{18}{6} =3 \\ \\ x_2= \frac{10- \sqrt{64} }{2 \cdot 3} = \frac{10-8}{6} = \frac{2}{6} = \frac{1}{3} \\ \\ S=\{3; \frac{1}{3} \}[/tex]

[tex]\displaystyle d).4x^2-17x+4=0 \\ a=4,b=-17,c=4 \\ \Delta= b^2-4ac=17^2-4 \cdot 4 \cdot 4=289-64=225\ \textgreater \ 0 \\ x_1= \frac{17+ \sqrt{225} }{2 \cdot 4} = \frac{17+15}{8} = \frac{32}{8} =4 \\ \\ x_2= \frac{17- \sqrt{225} }{2 \cdot 4} = \frac{17-15}{8} = \frac{2}{8} = \frac{1}{4} \\ \\ S=\{4; \frac{1}{4} \}[/tex]