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$(y^2-2y)/3=(y(y-2)-1)/2+(8-4y)/3$ |
di Francesco Speciale
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$(y^2-2y)/3=(y(y-2)-1)/2+(8-4y)/3$
$(y^2-2y)/3=(y(y-2)-1)/2+(8-4y)/3$;
Il m.c.m. è $6$
$(2(y^2-2y))/6=(3y(y-2)-3+2(8-4y))/6$;
Moltiplicando ambo i membri per $6$ si ha:
$2(y^2-2y)=3y(y-2)-3+2(8-4y)$;
$2y^2-4y=3y^2-6y-3+16-8y$;
$(2-3)y^2+(-4+6+8)y+3-16=0$;
$-y^2+10y-13=0$; cioè
$y^2-10y+13=0$
$\Delta=b^2-4ac=(-10)^2-(4*13*1)=100-52=48$
$x_(1,2)=(-b+-sqrt(\Delta))/(2a)=(10+-sqrt(48))/2=(10+-(4sqrt3))/2 => x_1=5+2sqrt3 ^^ x_2=5-2sqrt3$.
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