Question

va rog mult ajutatima...
1.$\left \{ {{x+y=8} \atop {x-y=1}} \right.$
2.$\left \{ {{3x+5y=21} \atop {2x-y=1}} \right.$
3.$x^{2} -6x+8=0$
4.$x^{2} +10x+9=0$
5.$x^{2} -x-2=0$

Answer 1

$\displaystyle 1). \left \{ {{x+y=8} \atop {x-y=1}} \right. \\ 2x/=9 \\ x= \frac{9}{2} \\ \frac{9}{2} +y=8 \\ 9+2y=2 \cdot 8 \\ 9+2y=16 \\ 2y=16-9 \\ 2y=7 \\ y= \frac{7}{2}$

$\displaystyle 2). \left \{ {3x+5y=21} \atop {2x-y=1|\cdot5}} \right. \\ \left \{ {{3x+5y=21} \atop {10x-5y=5}} \right. \\ 13x/=26 \\ x= \frac{26}{13} \\ x=2 \\ 3 \cdot 2+5y=21 \\ 6+5y=21 \\ 5y=21-6 \\ 5y=15 \\ y= \frac{15}{5} \\ y=3$

$\displaystyle 3).x^2-6x+8=0 \\ a=1,b=-6,c=8 \\ \Delta=b^2-4ac=6^2-4 \cdot 1 \cdot 8=36-32=4\ \textgreater \ 0 \\ x_1= \frac{6+ \sqrt{4} }{2 \cdot 1} = \frac{6+2}{2} = \frac{8}{2} =4 \\ \\ x_2= \frac{6- \sqrt{4} }{2 \cdot 1} = \frac{6-2}{2} = \frac{4}{2} =2$

$\displaystyle 4).x^2+10x+9=0 \\ a=1,b=10,c=9 \\ \Delta=b^2-4ac=10^2-4 \cdot 1 \cdot 9=100-36=64\ \textgreater \ 0 \\ x_1= \frac{-10+ \sqrt{64} }{2 \cdot 1} = \frac{-10+8}{2} = \frac{-2}{2} =-1 \\ \\ x_2= \frac{-10- \sqrt{64} }{2 \cdot 1} = \frac{-10-8}{2} = \frac{-18}{2} =-9$

$\displaystyle 5).x^2-x-2=0 \\ a=1,b=-1,c=-2 \\ \Delta=b^2-4ac=1^2-4 \cdot1 \cdot (-2)=1+8=9\ \textgreater \ 0 \\ x_1= \frac{1+ \sqrt{9} }{2 \cdot1} = \frac{1+3}{2} = \frac{4}{2} =2 \\ \\ x_2= \frac{1- \sqrt{9} }{2 \cdot1} = \frac{1-3}{2} = \frac{-2}{2} =-1$