Question

1) Să se rezolve sistemele de ecuatii:
a)
x + y = 3
x² + x = y
b)
x + y = -6
x · y = 8
2) Rezolvati in multime R, ecuatiile:
a) 10x²-7x+1=0
b) 6m²x²-mx-1=0, m∈R°
c) x²-4x+1=0
d) 2x²+3x+10=0
e) 4x²-12x+9=0
f)) 15m²x²-7mx-2=0, m∈R°
g) $\frac{2x-1}{x+3}$ =$\frac{4x+7}{3x-2}$
h) (2x+3)² + (x-5) = (2x-1) (x+4).

Answer 1

$\displaystyle 1a). \left \{ {{x+y=3} \atop {x^2+x=y}} \right. \Rightarrow \left \{ {{y=x^2+x} \atop {x+x^2+x=3}} \right. \\ x+x^2+x=3 \\ x^2+2x-3=0 \\ a=1,~b=2,~c=-3 \\ \Delta=b^2-4ac=2^2-4 \cdot 1 \cdot (-3)=4+12=16\ \textgreater \ 0 \\ x_1= \frac{-b- \sqrt{\Delta} }{2a} = \frac{-2- \sqrt{16} }{2 \cdot 1} = \frac{-2-4}{2} = \frac{-6}{2} =-3 \\ x_2= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-2+ \sqrt{16} }{2 \cdot 1} = \frac{-2+4}{2} = \frac{2}{2} =1$
$\displaystyle \left \{ {{x=x_1} \atop {y=x_1^2+x}} \right. sau \left \{ {{x=x_2} \atop {y=x_2^2+x}} \right. \\ \left \{ {{x=-3} \atop {y=(-3)^2-3=9-3=6}} \right. sau \left \{ {{x=1} \atop {y=1^2+1=1+1=2}} \right.$
$\displaystyle b). \left \{ {{x+y=-6 \Rightarrow y=-6-x} \atop {x \cdot y=8}} \right. \\ x \cdot y=8 \Rightarrow x \cdot (-6-x)=8 \Rightarrow -6x-x^2=8 \Rightarrow -x^2-6x-8=0 | \cdot (-1) \\ \Rightarrow x^2+6x+8=0 \\ a=1,~b=6,~c=8 \\ \Delta=b^2-4ac=6^2-4 \cdot 1 \cdot 8=36-32=4\ \textgreater \ 0 \\ x_1= \frac{-b- \sqrt{\Delta} }{2a} = \frac{-6- \sqrt{4} }{2 \cdot 1} = \frac{-6-2}{2} = \frac{-8}{2} =-4 \\ x_2= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-6+ \sqrt{4} }{2 \cdot 1} = \frac{-6+2}{2} = \frac{-4}{2} =-2$
$\displaystyle \left \{ {{x=x_1} \atop {y=-6-x_1}} \right. sau \left \{ {{x=x_2} \atop {y=-6-x_2}} \right. \\ \left \{ {{x=-4} \atop {y=-6-(-4)=-2}} \right. sau \left \{ {{x=-2} \atop {y=-6-(-2)=-4}} \right.$
$\displaystyle 2a).10x^2-7x+1=0 \\ a=10,~b=-7,~c=1 \\ \Delta=b^2-4ac=(-7)^2-4 \cdot 10 \cdot 1=49-40=9\ \textgreater \ 0 \\ x_1= \frac{-b- \sqrt{\Delta} }{2a} = \frac{-(-7)- \sqrt{9} }{2 \cdot10} = \frac{7-3}{20} = \frac{4}{20} ^{(4} = \frac{1}{5} \\ x_2= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-(-7)+ \sqrt{9} }{2 \cdot 10} = \frac{7+3}{20} = \frac{10}{20} ^{(10} = \frac{1}{2}$
$\displaystyle b).6m^2x^2-mx-1=0 \\ a=6m^2,~b=-m,~c=-1 \\ \Delta=b^2-4ac=(-m)^2-4 \cdot 6m^2 \cdot (-1)=m^2+24m^2 \\ x_1= \frac{-b- \sqrt{\Delta} }{2a} = \frac{-(-m)- \sqrt{m^2+24m^2} }{2 \cdot 6m^2} = \frac{m- \sqrt{25m^2} }{12m^2} = \\ = \frac{m-5m}{12m^2} = \frac{-4m}{12m^2} =- \frac{1}{3m}~~~~~~~~~~~~m \not=0 \\ x_2= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-(-m)+ \sqrt{m^2+24m^2} }{2 \cdot 6m^2} = \frac{m+ \sqrt{25m^2} }{12m^2} = \\ = \frac{m+5m}{12m^2} = \frac{6m}{12m^2} = \frac{1}{2m} ~~~~~~~~~~~~~~~~~~~m \not = 0$
$\displaystyle c).x^2-4x+1=0 \\ a=1,~b=-4,~c=1 \\ \Delta=b^2-4ac=(-4)^2-4 \cdot 1 \cdot 1=16-4=12\ \textgreater \ 0 \\ x_1= \frac{-b- \sqrt{\Delta} }{2a} = \frac{-(-4)- \sqrt{12} }{2 \cdot 1} = \frac{4-2 \sqrt{3} }{2} = \frac{2\left(2- \sqrt{3} \right)}{2} =2- \sqrt{3} \\ x_2= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-(-4)+ \sqrt{12} }{2 \cdot 1} = \frac{4+2 \sqrt{3} }{2} = \frac{2\left(2+\sqrt{3} \right)}{2} =2+ \sqrt{3}$
$\displaystyle d).2x^2+3x+10=0 \\ a=2,~b=3,~c=10 \\ \Delta=b^2-4ac= 3^2-4 \cdot 2 \cdot 10=9-80=-71 \\ \Delta\ \textless \ 0 \Rightarrow Nu ~sunt~solutii~reale.~~~~~~~~~~S= \varnothing$
$\displaystyle e).4x^2-12x+9=0 \\ a=4,~b=-12,~c=9 \\ \Delta=b^2-4ac=(-12)^2-4 \cdot 4 \cdot 9=144-144=0 \\ \Delta=0 \Rightarrow x_1=x_2=- \frac{b}{2a} =- \frac{-12}{2 \cdot 4} = \frac{12}{8} = \frac{3}{2}$
$\displaystyle f).15m^2x^2-7mx-2=0 \\ a=15m^2,~b=-7m,~c=-2\\ \Delta=b^2-4ac=(-7m)^2-4\cdot 15m^2\cdot (-2)=49m^2+120m^2\\x_1= \frac{-b-\sqrt{\Delta} }{2a}=\frac{-(-7m)-\sqrt{49m^2+120m^2}}{2\cdot 15m^2}=\frac{7m-\sqrt{169m^2} }{30m^2}=\\=\frac{7m-13m}{30m^2} = \frac{-6m}{30m^2} =- \frac{1}{5m}~~~~~~~~~~~~m \not = 0\\x_2= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-(-7m)+ \sqrt{49m^2+120m^2} }{2 \cdot 15m^2} =\frac{7m+\sqrt{169m^2}}{30m^2}=\\=\frac{7m+13m}{30m^2} = \frac{20m}{30m^2} = \frac{2}{3m} ~~~~~~~~~~~~m\not=0$
$\displaystyle g). \frac{2x-1}{x+3} = \frac{4x+7}{3x-2} \\ (2x-1)(3x-2)=(x+3)(4x+7) \\ 6x^2-4x-3x+2=4x^2+7x+12x+21 \\ 6x^2-4x^2-4x-3x-7x-12x=21-2 \\ 2x^2-26x=19 \\ 2x^2-26x-19=0 \\ a=2,~b=-26,~c=-19\\ \Delta=b^2-4ac=(-26)^2-4 \cdot 2 \cdot (-19)=676+152=828\ \textgreater \ 0 \\ x_1= \frac{-b- \sqrt{\Delta} }{2a} = \frac{-(-26)- \sqrt{828} }{2 \cdot 2} = \frac{26-6 \sqrt{23} }{4} = \frac{2\left(13-3 \sqrt{23\right)} }{4} = \\ = \frac{13-3 \sqrt{23} }{2}$
$\displaystyle x_2= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-(-26)+ \sqrt{828} }{2 \cdot 2} = \frac{26+6 \sqrt{23} }{4} = \frac{2\left(13+3 \sqrt{23}\right) }{4} = \\ = \frac{13+3 \sqrt{23} }{2}$
$\displaystyle h).(2x+3)^2+(x-5)=(2x-1)(x+4) \\ (2x)^2+2 \cdot 2x \cdot 3+3^2+x-5=2x^2+8x-x-4 \\ 4x^2+12x+9+x-5=2x^2+8x-x-4 \\ 4x^2-2x^2+12x+x-8x+x=-4-9+5 \\ 2x^2+6x=-8 \\ 2x^2+6x+8=0 \\ a=2,~b=6,~c=8 \\ \Delta=b^2-4ac=6^2-4 \cdot 2 \cdot8=36-64=-28 \\ \Delta\ \textless \ 0 \Rightarrow Nu~ sunt ~solutii~ reale.~~~~~~~~~~~~~S=\varnothing$